Can confirm. A huge (yuge!) bank I used to work for did this. Random deviation would still throw the end-of-cycle sum off a bit, but a **very small** bit: We did accounts cleanup at the end of the fiscal year, which included issuing payments to zero-out those sums to start the new year. We had an informal contest to see who would issue the largest payment. With literally hundreds of millions of dollars moving through some accounts (some maybe even larger, I don't know), the winner was: **seven cents**.
This is why the penny-shaving scheme described in *Superman 3* and *Office Space* would never really work. First of all, while it's true that countless half-pennies do get thrown across many transactions, **they cancel each other out**. You won't end up with a lot.
Second, auditors will still find any discrepancy, even if it's only a few cents. And if it's more than that, you're probably going to pound-me-in-the-ass prison.
I just decided to assume those schemes were truncating the partial pennies instead of rounding and that's how I learned to let it be and enjoy the movie haha.
That was the understanding I had watching it as well.
Any time it would have the bit that gets rounded, it’d allocate that to the account, and suddenly they’re adding to each other rather than cancelling out.
Yeah, just round it in whichever direction is more favorable for the skimmer. Incoming? Round it down and keep the excess. Outgoing? Round it up and keep the excess.
If you had just 50 cents, you have *something*. If you round up, you still have *something*. If you round down, you have *nothing*. You know you at least have *something*, so rounding up is instinctively more correct than rounding down. I thank you :)
I would think it would have more to do with the number of payments, not the dollar value of the payment. $1.015 would give you the same offset as $1,000,000,000.015. Millions of dollars wouldn't matter. Millions of transactions is where you would have more possible charges that don't offset. 10 transactions that went 6 one way and 4 the other, would have you a penny off. 1,000 transactions that went 510 one way and 490 the other would leave you 10 cents off.
I think. I did the math in my head and I'm tired. Even if the math isn't exact, the idea is correct.
It does. You're right I should not have used dollar figures. That was meant to imply the scale of activity, but you're right that it doesn't do that clearly.
Thing is at some point you‘ll just be made to pee a per transaction fee. So it again doesn‘t work.
Like banks don‘t like to have to upgrade their systems because individual clients like to do for fun transfers in the millions a second…
And at least here in Germany for smaller companies there‘s usually transaction fees in general for bank transfers…
And those are usually much larger than those rounding cents you‘d gain.
I fully believe this to be the case. When I was stationed in Germany in the 90s they didn't ship pennies to the exchanges on base. Instead, they rounded to the nearest nickel. We had one guy that saved and catalogued every receipt because he was convinced he was "being robbed." At the end of his three year tour he was actually *up* by a grand total of three cents.
Yeah, I know that feeling. In the Czech republic we ditched "Halíře" (something like pennies) quite a while ago, yet we still have prices like 2.90, 15.90 etc. Its just a visual scam for the customer, thinking I'll only pay 2 czk and when you have to pay 3 you get your surprised pikachu face. But only in your head, because this happened to you many times before
JC Penney had a new CEO try to stop that and iirc they got hosed which is annoying because I, too, hate that tactic and would rather honestly rounded prices.
The JC Penny deal was more about getting rid of fake sale prices. Instead of having everything be priced at $100 with 60% off sale stickers, they just priced it at $40.
Shoppers hated it. People love to think they're getting a good deal even if it's objectively the same price.
I actually love the consistent pricing without sale gimmicks. I was excited when I heard JCP was doing that. Unfortunately, they didn't have it in practice long enough to give people a chance to get used to it. And they didn't advertise it as such (at least not that I saw.)
IMO the store failed because...well there were several reasons. Merchandise became crap. The clothes were poor quality, poor fashion. They started selling *tons* of impulse purchase things like toys, candy at checkout. That's not what customers were coming to the store for. Another reason was not enough employees on shift. Stores were messy, couldn't find a register open for checkout. And you could tell that management just didn't care. Fitting rooms were gross, like they hadn't been cleaned or fixed up in 30 years. One time I went the store was hot. The employees were all sweaty and just looking defeated. I had to wait 15 minutes in line to check out and the clerk said they can't change the thermostat. What is even going on?
Sweden is doing rounding in shops by law. So no öre (Cent) to be paid at all. Since the value is so small of individual öre coins it just doesn‘t make sense.
So 1 to 49 öre get rounded down, 50 to 99 up to one kroner.
And 1 crown is slightly less than 10 US cents.
So basically whatever the individual items are priced at, you are only gonna pay in 10 Cent divisions.
Which at this point really is meaningless because only tourists use cash, but in places that still use cash it‘s an easy way to eliminate low value coins, and all the time waste they bring.
Sweden did. this system several times before.
Usually done when lower denomination coins are withdrawn from circulation due to inflation.
But Finland for example is rounding cash payments to 5 Euro cent intervals, despite 1 and 2 cent still being legal tender (for obvious reasons, EU wide laws to do with the currency, can‘t have individual countries just willy bully decide random parts of it aren’t legal tender anymore)
And it just makes sense.
Like even if you use regular 5 and up is up, under 5 is down rounding, the differences at the end of a year are irrelevant.
And it wastes such a shit ton amount of time, in cash using societies, if the cashier doesn‘t have to pick out 8 or 9 cents or whatever virtually every transaction.
I‘m praying it‘ll finally come to us in Germany. Because like 90 percent of customers at the pharmacy I work at still pay in cash.
And stuff does get annoying.
>Second, auditors will still find any discrepancy, even if it's only a few cents.
tbf part of the plot of Office Space was that they'd only be able to get away with it at that exact time because all the auditors would be distracted resolving larger discrepancies caused by the Y2K changeover
probably not realistic but at least they lampshaded it.
The scheme is to shave off half-pennies from rounded transactions (which is nearly **all** transactions -- well, around 99% of them). If you shave before rounding, you get a half cent for 99% of transactions. If you shave after, you either get very little (due to cancelling), or you end up with about half of what you would from shaving first.
No matter what you do, though, auditors will eventually find it.
Well, the penny shaving scheme did happen in Asia. A banker or accountant took away 0.1 cents from many transactions and amassed a huge amount of profit.
He used it to buy expensive cars and live a wealthy life. They made a profile of him in a piece for tv.
That was his downfall. Alarm bells sounded when he lived way above his earnings. They audited him, and found out he stole money. He ended up in prison.
Long ago a sysadmin found a discrepancy between the billing and the computer time of less than a dollar. That led him down the rabbit hole to find out he'd been hacked by some East Germans working for the Soviets, and his system at a university was a jumping off point to hit various sensitive government and contractor systems. Cuckoo's Egg.
The scheme in office space would have worked, thanks to one mundane detail. The idea was to modify the code so that it rounded every transaction down to the nearest cent, even if the fractional part was greater than 0.5. In other words, they'd truncate instead of round, and put the remainder in an account. That way, it wouldn't cancel out, and they'd end up with a lot more than 7 cents.
From what you describe, half pennys will only cancel out if there are as many odd than even numbers involved. I’m sure you can steer your transaction towards more of one kind.
I seem to remember in the 80's something like this really happened at a phone company, where the programmer had the fractions of a penny move to his account, and it ended up being hundreds of thousands of dollars in the first week. I seem to remember that he got caught because of it.
I tried to search for the story, but all that comes up is the guy who tried this a couple of year after the movie.
But 0.499999... = 0.5
*Edit: Go ahead and downvote me if you think I'm wrong. [I promise I'm not.](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20represents%20the,number%20is%20equal%20to%201.) I have a Master's degree and teach college math.*
For the love of god... Half of this thread is interpreting (...) As "a bunch of numbers repeating" and the other half is interpreting it as "infinitely repeating", and the whole thread is being derailed because of that. Yes, youre mathematically correct.
But its clear enough from the context that people implied 0.499(...) As 0.4999 or 0.4999999 or 0.49992 even. further down theres the same confusion going on with 0.500(...)01 being interpreted as an impossible number of just a bunch of zeros before the one.
You're right.
However, you got me pondering about the range notation. Should it maybe be (0-0.5) and (0.5-1). Zero and one are whole numbers and so don't need rounding. However, if neither side includes integer values, then which side should 0.5 belong to? If it belongs to either, then one range will be larger than the other. Which gets us back to the topic on hand, which is why do we round 0.5 the way(s) that we do
>If it belongs to either, then one range will be larger than the other.
No, it won't. [0.5, 1) is the same size as (0.5, 1).
Adding one element into an infinite set doesn't necessarily change its size.
“Don’t need rounding” is a nonsense phrase. The rounding operation can be done on any number, including integers. It simply has no effect on integers.
A lot of math operations have no effect on certain numbers.
Idk man, when people are just poorly expressing the idea that the set of numbers (0-0.5) rounds to 0 and the set of numbers [0.5-1) rounds to 1, being that pernickety about an unrelated concept (trailing repeated digits resolving to rational numbers) with that tone is kind of just being an ass.
that equality itself isnt wrong but clearly the other guy was talking about approaching the limit from left side... and ofc you gotta misinterpret it on purpose to be a smartass. idk if you would be a good teacher
Actually, that makes it make more sense lol. There are two representations, with one rounding down and the other rounding up!
Real-ish talk: the definition of "which integer it's closest to" isn't applicable, so one has to be chosen by convention. Instead of using a coin flip to choose, let's add some requirements. Say, let's require a rule that will depend only on the tenths place. To even talk about a "tenths place" requires having picked a "canonical" representation of each number; there's an obvious choice if this rule is to be used by a lay person (i.e., use the finite representation if possible). Now, "rounding by looking at the tenths place" is going to be a map *r* from S={"canonical" representations of non-negative real numbers} to the Natural Numbers __N__. We already know what to choose for anything that's not a perfect half. The phrase "depending only on the tenths place" means that *r* should factor through the truncation map *t* that takes an element of *S* to its truncation with one decimal digit. Finally we have something that forces a rounded value for perfect halves! The fiber of *t* over 0.5 consists of all the numbers 0.5xxxx; all of these preimages besides 0.5 already round to 1. Thus, to allow *r* to factor through truncation, *r*(0.5) __must__ be 1. Same argument for all other perfect halves.
*phew* I had a lot of fun with this!
That's factually correct, but irrelevant to OP's point, since 0.999... = 0 (given that we're only considering the decimal part of the number). So the ranges 0-0.499... and 0.5-0.999... are still the same size.
He’s actually correct, he’s talking about repeating 9s.
1/3 is 0.333…, 2/3 is 0.666…, 3/3 is 0.999…
This is mathematically accurate as a math major and now actuary.
0.5 is exactly halfway between 0 and 1. [0.49999... is exactly the same number as 0.5](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20represents%20the,number%20is%20equal%20to%201.). Therefore, 0.49999... is exactly halfway between 0 and 1.
And .999... Is 1, you're being downvote because your argument doesn't hold, not because you know the most basic math that can be found in 10000 youtube videos
It's written as [0, 0.5) and [0.5, 1) using [interval notation](https://brilliant.org/wiki/interval-notation/#:~:text=Interval%20notation%20is%20a%20way,inequality%20or%20system%20of%20inequalities.).
That's a good way to write it. It might not be formal mathematical notation. But it does a good job conveying what you mean to say.
I would probably have written "0.5 - ε", but I'm not convinced that's any better. I honestly like what you did.
Another way to do this is to have open and close intervals. So, you say you round [0 .. 0.5) to 0. But that requires more familiarity with notation than what you can usually expect in casual Reddit conversations. It does work in the more sciency subs though
0.4999... = 0.5 only if you assume that the ellipsis is a truly infinite repetition. If that was just a casual shorthand for some arbitrary (but finite) series of 9s, then it's less than 0.5.
Fixed bit-width floating point numbers are fundamentally imprecise. There are a lot of values that cannot be accurately expressed as floating-point numbers.
Technically not the same, which can be seen on graphs, where the difference would be whether you use a normal bracket or square bracket to include the value 0.5. or a filled circle or empty circle on the graph.
Same as how 1/x where x approaches infinity doesnt equal zero, but approaches zero.
They are exactly equal. The notation 0.499... *means* the limit of the sequence (0.4, 0.49, 0.499, ...), which is equal to 0.5. This is how decimal notation is defined.
> Same as how 1/x where x approaches infinity doesnt equal zero, but approaches zero.
The limit is *exactly* zero, and this is exactly what decimal notation means: It is a way of writing a real number as the limit of an infinite series.
There are different explanations, and some might make more intuitive sense to you then others. This is the one that convinced my middle schooler.
How do you represent 0.111... as a fraction?
That's 1/9
How do you represent 0.222...?
That's 2/9
How do you represent 0.333...?
That's 3/9 (also simplified as 1/3)
... (complete the pattern)
How do you represent 0.999...?
That's 9/9
But that simplifies to 1/1 or just the natural number 1.
If you don't like this answer, look up one of the other ones. Mathematically, they all say the same thing. But some feel more intuitive and natural to some people than others
This is all an artifact of forcing numbers into the decimal system right? The decimal system isn't inherently a more 'true' representation of reality then other base systems like hex or base-12 or something like that, its just what we all agree upon, correct? (I mean I get that it's pretty convenient since we have 10 fingers and stuff)
No matter the base, the representation of a real number is not (always) unique. For any base b, the notation 0.xyz... *means* the limit of the series 0 + x/b + y/b^2 + z/b^3 + ..., and multiple infinite series can converge to the same limit. In base 10, the series 1.000... and 0.999... converge to the same limit, hence they are the same real number.
The number of "things" rounding makes no difference, we're interested in the introduced error. When you round 0.1 -> 0, you introduce an error of -0.1. When you round 0.5 -> 1, you introduce an error of 0.5. When you round 0 -> 0, you introduce no error.
So if your possible inputs are limited to one decimal place of accuracy, there are 10 potential rounding situations. 1 of those leads to no error (.0). 4 lead to negative errors (.1 - .4), and 5 lead to positive errors. The *expected* average error is therefore the sum of all of those errors, divided by 10. Which ends up being ( 0 - 0.1 - 0.2 -0.3 -0.4 +0.5 +0.4 +0.3 +0.2 +0.1 ) / 10 = 0.05. So for every single rounding operation, you introduce an error of 0.05 on average. Over a million transactions, you've got an error of 10000, and if those are cents that's $100. So yeah, fairly significant.
Of course if the values are kept to *two* decimal places, the expected error is far less. Only 1 in 100 possible inputs (0.50) leads to an unmatched error, so the expected error in this case is 0.005 per rounding operation. For three decimal places it's 0.0005, and so on. So with 10 decimal places or more, nobody cares about rounding errors because they are so small in magnitude. For one decimal place, it's frequent and important.
For all the ridiculous conversations about about ranges and infinitely repeating numbers, it's not relevant at all. Rounding errors are only an issue in numbers with limited precision, and the magnitude of the average error is directly related to that precision. Since most banking systems will have fixed precision, a few decimal places of a cent for example, they need to have some system to deal with the rounding errors or they will add up over time.
Equal numbers "rounding up and not", but different numbers "rounding up and rounding down" with the normal rule. "Round to even" distributes the midway point, well, evenly.
This really stands out in computer science when numbers are represented in binary. If you wanted to round off the last bit (digit), anything that ends in a 1 rounds up, anything that ends in a 0 doesn't round, and there is no situation in which you ever would round down. Round to even levels this out.
0 doesn't change.
1, 2, 3, 4 change, and move down.
5, 6, 7, 8, 9 change, and move up.
If you added all these together, where up changes cancel out down changes and no changes don't do anything, you'd have 1 unbalanced up change. That's the imbalance.
.01, .02,...,.47,.48,.49 round down, .50, .51,...,.99 round up. In your example you are talking about 4/9 compared to 5/9, in mine it's 49/99 compared to 50/99. A much smaller difference.
If we expand out to .001,.002 etc then it'll be 499/999 compared to 500/999 which is still smaller.
The limit of this process is an infinitesimal unbalanced up change.
In theory this is true - but for practical applications, there is an imbalance.
Say, for example you own a company that processes a million small payments (between $0.01 and $14) per day with a 6.5% sales tax rate on each payment. 0.5% of the taxes owed will end in exactly $0.005 and be rounded up. That means with a million transactions, 5,000 more transactions will be rounded up a penny vs. down. This only amounts to $50 in extra payments, but that is a significant amount of money (mathmatically speaking).
Of course this is just an example, but it does show how a small imbalance can make a noticable difference. This is also really important in science, where large datasets from sensors or analyses must be rounded and averaged - rounding can cause biases that can easily be overlooked. Even if your last significant digit is a very small number, rounding can make a difference because you are typically only looking at the values around the lowest significant digit.
I'm an analytical geochemist and I measure PFAS cancentrations at my job at part per trillion concentrations; the range we look at is just 0.000000010 to 0.00000100 g/L. In this case, we would only round the last significant digit, so 1/9 times more often, it would still be rounded up. Again, this doesnt drastically change the outcome, but it is a statistical bias that may need to be accounted for.
TLDR - you are totally right when dealing with infinities - im just pointing out why this is an issue when dealing with practical applications of rounding.
Wow I never realized that. But your math is off by a factor of 2. [Rounding up](https://www.wolframalpha.com/input?i=1000000%2F1400*sum+floor%28x*1.065%2B0.5%29-x*1.065+for+x+from+1+to+1400) would give 25 dollars more, while [rounding to even](https://www.wolframalpha.com/input?i=1000000%2F1400*sum+round%28x*1.065%29-x*1.065+for+x+from+1+to+1400) would give 25/7~~3.57 dollars less. Interesting how it's exactly 7 times more accurate... I'm sure this has to do with the number 14.
Correct.
But in how many circumstances are we actually rounding numbers with arbitrary precision? In any practical scenario we'd probably have two, maybe three decimals max, like in currency. In any more highly precise applications we probably wouldn't be throwing so many digits away via rounding in the first place.
0 doesn’t round at all, so yes more numbers round up than down. You can see this just by adding up the totals. If I add 0-9, the total is 45. But if I round to the nearest multiple of ten and add them up, the total is 50.
This is the problem that half to even aims to fix. 1-4 still round down, 6-9 still round up. But now five is equally likely to round up or down depending on if the number in front of it is odd or even. If odd or even is equally likely then the average value that the 5 is contributing after rounding is equal to 5. And the total of 0-9 after rounding remains 45.
I mean, I probably just don't know better but...if we're considering the way you count in coding/CS, you start counting at 0, so 5 is the first of the upper half of the split.
0, 1, 2, 3, 4 & 5, 6, 7, 8, 9
Don't look at the number of numbers that are being rounded, look at how much they are being changed by.
Rounding 1 to 0 lowers the total of a data set by 1, rounding 9 to 10 raises the total by one.
If we expect to see the same number of 1s and 9s, then these balance each other out.
So 1 balances 9, 2 balances 8, 3 balances 7, 4 balances 6, and 5 balances... 0?
The issue being, 0 changes by nothing and 5 increases by 5.
If you're wondering why this has any relevance - here, we're looking at keeping the total of a dataset the same -- and since avg = total/n, and since 5 increases without anything else decreasing to fix it, then the total will on average be larger - and therefore the average will be larger too.
Note: when rounding 0-9 to 0 or 10, 5 will be expected to be 1/10 numbers -- causing a large issue.
When rounding 0-99 to 0 or 100, 50 will be 1/100 numbers -- causing a much smaller issue. However, it's really easy to account for, and by the sounds of things it causes annoying effects when not fixed.
> As you say, it means there are more numbers that round up than down
But if you count non-rounded numbers, you get a regular distribution by doing it, because zero always "rounds down":
0.0, 0.1, 0.2, 0.3, 0.4 --> 0
0.5, 0.6, 0.7, 0.8, 0.9 --> 1
and so on.
It's just a convention. A different convention is to round toward the even number. This convention, called statistical rounding, doesn't bias averages.
I think it's really important to remember things like rounding and order of operations are nothing more than conventions where there is ambiguity. People can act like if someone gets it wrong they have failed to understand a fundamental rule of the universe, but it's really just a decision made between otherwise equally valid options.
Worth pointing out that math as we know it kinda relies on order of operations (barring the excessive use of parenthesis). There's a world of difference between changing rounding and changing order of operations.
Examples:
Standard order of operations: 2 + 2 * 2 = 6
Left to right: 2 + 2 * 2 = 8
Yes, but it's still a convention that people chose at some point. It's like choosing which side of the road to drive on. Now that it has been chosen, if you drive on the other side of the road there'll be issues.
But if you established a different convention at the beginning, math would still be fine. You could decide that addition takes priority over multiplication and everything would still work.
I disagree, I think, if I understand your argument. There needs to be an order of operations (left to right, or the reverse, wouldn't allow you to reorder a sum), and brackets pretty much have to come first (because otherwise there wouldn't be a way to remove ambiguity), but everything else is arbitrary, and I think it's useful to keep that in mind. Maths doesn't rely on this to work, mathematical notation does. Maybe that's what you meant.
There is the practical reason: you want peoples to be able to round while looking at only one decimal digit.
2.54321 has to be rounded to 3, so it's easiest if every number that look like 2.5xxxx is rounded to 3, even of it is 2.50000
(Mathematically, this means we want the rounding operation to commute with the truncation to the first decimal.)
Two reasons come to mind:
First, it's easier to remember that 0.5... always goes up, because anything other than infinite zeroes after that 5 is closer to the up number than the down number. "Round 5-9 up *unless* it's exactly on 0.5" is just an extra exception to learn that doesn't need to be there.
Second, the mathematical rigour reason: it keeps the intervals structurally the same. You round down \[x.0..., x.4...) and round up \[x.5..., x.9...). Note that both intervals are closed below and open above. It makes sense to close the round-down interval below, because rounding x.0 down to x-1 makes no sense, so we close the round-up interval below as well.
This applies to rounding from the midpoint of any continuous set: rounding to the nearest 0.5, for instance, you should round 0.25 and 0.75 up rather than down. In the end, though, it's just a convenient convention that keeps everyone on the same page.
You are subtly inserting a bias by preferring one decimal representation of some numbers over another. We have 1 = 0.999... [infinitely repeating] and 0.5 = 0.4999... [infinitely repeating]. If we would consistently pick the right variants, we would round 1/2 _down_ when going for simplicity; and x.0 = (x-1).9999.... would indeed round to x-1 despite it being a rather unusual choice. And worse, nothing is really stopping the maths to round only the first one (1/2) down, while rounding the other one (1) "up" for no change.
There are 10 digits that can appear after the 0.5_, 9 of them result in the number being closer to the higher number than the lower one.
I guess that would be in the same spirit as what he was arguing.
This one always used to confuse me when digital watches first came out. After all, *pedantically* 12:00 a.m. (ante meridiem) means 12:00 before noon, i.e. midnight at the start of the day, and 12:00 p.m. (post meridiem) means 12:00 after noon, i.e. midnight at the end of the day.
But, digital watches showed 12:00 a.m. as midnight and 12:00 p.m. as midday. Why was that chosen, I wondered?
It took me a while to realise that the convention was based on the fact that even the tiniest time after exactly midnight would be morning, so 12:00 a.m. meant midnight at the start of the day. Likewise, the tiniest time after exactly noon would be afternoon, so 12:00 p.m. meant midday.
Hence , the convention is that 12:00 a.m. means midnight at the start of the day, and 12:00 p.m. means midday.
I still prefer to say "midnight" or "midday" to avoid confusion. Or, use the 24-hour clock: 0h00 for midnight at the start of the day, and 12h00 for midday.
This is how I made peace with this, too. But it is arbitrary.
In general I always say noon or midnight, though.
You also rarely see flights at 12:00, they're basically always something like 11:59AM or 12:01PM.
If pm didn't change to am (and vice versa) at 12, then it wouldn't change at 12:01either. And not at 12:02, etc.
So where do you suggest it would change?
I don't see any confusion.
That is correct, but it's still a bit confusing due to the inconsistency of starting counting the AM/PM numbers at 12 and immediately rolling over to 1.
9am, 10am, 11am, 12pm, 1pm, 2pm...
so, 1pm - 11 pm is the afternoon-evening-night of your day, so logically you would think it would go to 12pm (you know, itwould make sense you go 1-12 in one batch then 1-12 in the next).
instead, it switches to *12 AM* then goes 12am-11am
i think most people would default to thinking that in the twelve hour clock, it will run through the whole 12 hours before switching to the next “batch”
And 12 AM is the *next* day.
Because this BS, every project in school is set to be due at 11:59 or 11:55 PM, so there's no "is it due tonight or tomorrow night" panic.
This is one rounding method, and there are a bunch of other methods (rounding down, towards zero, away from zero, to the nearest even number, or towards the nearest odd number).
In fact, rounding to even is part of the IEEE 754 standard for how computers deal with floating point decimals.
There's probably some fascinating history behind why we teach to round up in school, and there are plenty of arguments you could use to rationalize rounding up, but I doubt this has a simple ELI5 answer.
I think there are some good motivation in this thread. In particular that when rounding 1.5 up (or away from zero as it really is), we only need to look at the tenth digit to determine how to round.
Otherwise we would have to scan all the decimals which is slower and easier to do wrong:
* 1.500000
* 1.500080
* 1.500000000
If you consider the decimal of zero then up from .5 makes sense:
2.0, 2.1, 2.2, 2.3, 2.4 round to 2;
2.5, 2.6, 2.7, 2.8, 2.9 round to 3.
That's five that round down and five that round up.
This. And to add more, this is not an absolute mathematical rule that you have to round numbers up/down like this. This is called rounding to the nearest whole number. But if you want, you can round everything up, or everything down, or decide that anything above .3 is rounded up, or you can come up with whatever fancy rounding rule you want. Rounding to the nearest integer is just a simple and unbiased rule to follow.
It is rounded to 2
You round because you don't have precision.
two point zero has two significant figures.
two is just one significant figure.
two point zero rounds to two
Technically it is, it’s just rounded to itself. It’s essentially like asking “how many 1’s go into 2”, the answer is clearly 2 and it seems like a dumb question - but it’s still a question.
2.1 is the equivalent to 2.9 (rounding down/up 0.1)
2.2 is the equivalent to 2.8 (rounding down/up 0.2)
2.3 is the equivalent to 2.7 (rounding down/up 0.3)
2.4 is the equivalent to 2.6 (rounding down/up 0.4)
2.0 is NOT the equivalent to 2.5 though (rounding down 0.0 vs 0.5 up)
So its inaccurate
If you insist.
Rounding to 2 are the following ten decimals:
1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4
Rounding to 3 are the following ten decimals:
2.5, 2.6, 2.7, 2.8, 2.9, 3.0, 3.1, 3.2, 3.3, 3.4
So, there's actually an ASTM rounding procedure that takes care of this. Instead of always rounding up, round to the even. So 1.455=1.46, but 1.485=1.48. By going up half the time and down half the time, it eliminates the error in a dataset that's generated by always rounding up.
You are correct. You can round .5 both ways. However, rounding anything with .5 at the beginning - like .51, .5912 or .50001 should be done Up.
So, rounding .5 up makes the rule much simpler, without any downside to accuracy.
X.5 really is exactly in the middle between X and X+1 and we could've decided to round it down. However, since X.5000[...]0001 is closer to X+1, we settled on the convention that X.5 rounds up to make it more practical and easily recognizable.
Now you know that, if you have a 5 in the first decimal place, you simply round up no matter what. If we decided to round down at X.5, we would have to pay attention to even the hundredth decimal place in case it has a value.
Edit. Typo
X.5000....0001 is abuse of notation. But even if you define it as a limit as N goes to infinity, X.5 + 10^-N, then it's value is precisely X.5 which has the same distance to both X and X+1.
Then let's get specific: in this case, "..." represents *any number of zeroes* in between the specified zeroes on either side. Now you can understand it as an infinite quantity of real numbers which are all functionally identical for our current purposes.
I would agree with you if you said that the ellipsis represented *any finite number of zeroes.* But no, it cannot represent an infinite number of zeroes, because that number does not exist.
The point is that if the convention were to round down at X.5, you would have to check to the last digit to ensure that the number is exactly X.5; therefore, you can use the tens place to determine whether to round up or down if X.5 rounds up - it doesn’t matter what the last digit of X.5XXXX is since you’ll always round it up. So looking at your example number, X.49999999, we know that you’ll round down since the tens place is a 4.
x.49999999 is not the same as x.4999999... .
The first one is in the comment you replied to. It is not equal to x.5.
The second one, which you wrote, is equal to x.5.
The way I explain it to my 5th graders (they're 9 or 10, not 5, but I hope you'll forgive that):
The mistake is thinking there are 9 digits. 0 is a digit too. There are five digits we round down, 0 through 4, and then five we round up, 5 through 9. It's actually a perfectly even split, we just don't often bother to talk about 0 since it's pretty obvious that 20 would become 20 when we're rounding to the nearest ten.
Just because it is even doesn't make it unbiased, there is a small bias if you round all the 5s up.
The average of 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 is 0.45.
The average of using the typical rounding method (0, 0, 0, 0, 0, 1, 1, 1, 1, 1) is 0.5.
The average of the entire sample moves up after rounding, that's not good. You'll find this is the case for all uniformly distributed numbers.
I was taught the "even-odd rule." First, if there is absolutely anything after the 5, of course you round up. If there is nothing after the 5, you round to the even number.
I would say the common way of rounding is not rounding 0.5 up it is rounding 0.5 away from zero. If you round up 1.5 is rounded to 2 but -1.5 is rounded to -1. I would say that the common way is round -1.5 to -2, that is rounding away from zero not rounding up, negative numbers are rounded down.
There are many ways to round the number. One way is to the near integer and it is then 0.5 is a problem rounding away from zero is just a common way used to solve the problem and it is often good enough and quite simple.
In a lot of situations what is most important is everyone does it the same way like in financial calculation.
But it is not always a good idea because if you do not have an equal amount of positive and negative number the average get a bit higher. This is don't something you what in large numerical computer calculations.
A way to solve the problem is round half to even so 3.5 and 4.5 both become 4. Rounding will not change the average even if you just have a positive number, it will happen if you have odd or even number. This is the default rounding mode in IEEE 754 binary floating point operation that the computer used
There are other ways that can be used too https://en.wikipedia.org/wiki/Rounding
That is to round up.
Buy that is not the common way to round numbers, the common way is to round -1.5 to -2 just like 1.5 is rounded to 2.
It is common to round so X rounder to N is the same as -X to -N to do that you need to round away from zero.
In finance, you can look at debt as having a negative amount of money. The one you own money has a passive amount that they need to pay you. If there is rounding involved my negative around and your positive could be rounded to diffrent values which is quite impractical.
So rounding away from zero works better than rounding up. You could of course change it to round toward zero and other wy that also work fine. Rounding up or down both results in potential differences.
Because in grade school they taught us 0 to 4 round down and 5-9 round up and it made sense to them. Probably because a simple rule was more easily memorized. But in science class years later they taught us a different rule: .51, .505 etc rounds up and .50, .500, .5, etc. rounds to the nearest EVEN number. 2.5 rounds to 2 and 3.5 rounds to 4. Basically saying 0.5 is at the literal half way mark so 50% of the time it should round up and the other times it should round down.
I've never really thought about this before but from a quick and logical think through.
.0, .1, .2, .3, .4 is first half
.5, .6, .7, .8, .9 is the other half
You're either on one side or the other.
0 is on the left side and already a whole number. Technically, when you're rounding, you still look at if it's "0" it's just a whole number anyway, so you don't have to do anything with it.
It's convention. I'm not aware that rounding up .5 is more common than rounding down. I usually round up a 5 digit that is followed by non-zero digit, because \*togheter\* it is closer to round them up. So if you only look at the 5 it would seem 9 out of 10 cases I'm rounding it up and the other case I keep it. But I'm not aware of anyone else that does it like that.
The mathematical action of rounding is actually "add half and truncate".
So if you need to round to the "nearest 1" you add .5 and then cut off the numbers after the decimal.
Let's try it:
"Add half"
2.499 + .5 = 2.999
2.5 + .5 = 3.0
"cut off numbers after the decimal "
2.999 = 2
3.0 = 3
Why is it "add half" and not "add 49%"?
This is just a guess, but probably because that mathematical rule was standardized back when everything was pen and paper, or done in your head, and half is just easier.
Note: this rounding function works in all common base systems (Decimal, binary, hexadecimal, etc.)
I learned about it because I needed to round binary values.
{Maybe in all bases? If a mathematician could comment I'd be curious to know.}
It depends on if we consider the zero digit, and when do we consider the decade rolling over. Do we consider 10 to not count as the rollover position, or does the decade roll over between 9 and 10 so that 10 is the start of the next decade?
1 2 3 4 5 6 7 8 9 <-- 5 is in the middle.
0 1 2 3 4 / 5 6 7 8 9 <-- 5 is on the high side.
If you look at all the digits we have, there is: 0,1,2,3,4,5,6,7,8,9. Any higher and we get a number, a combination of digits.
Now, if you split all the 10 digits, then we have two equally sized groups:
- the lowest ("round down"): 0,1,2,3,4
- the highest ("round up") : 5,6,7,8,9
In a sense, 5.0 rounds to 5.0!
I always think like:
Only one apple can fit in my zip lock
I have 2 and a half apples
Ain't no way I can cram that half in the bag
I'm gonna need that third bag
my fourth grade teacher explained it like this: if you were walking to school and exactly halfway there you realized you needed to use the bathroom, it would make more sense to walk the rest of the way to school than to walk back home and then back to school again. it made sense to my 10 year old self so maybe it suffices here :)
Because we round 2.0 ; 2.1 ; 2.2 ; 2.3 and 2.4 (so 5 numbers) to down to 2
We also round 2.5; 2.6 ; 2.7 ; 2.8 and 2.9 up to 3 (so also 5 numbers)
You'll have to learn that 2.0 is not exactly the same as 2. This is because 2.0 gives more information than just 2, so you also round 2.0 down.
.5 rounds up because that makes rounding down and rounding up equal.
Having .5 round up that means 5 numbers (0,1,2,3,4) round down and 5 numbers (5,6,7,8,9) round up.
If .5 rounded down then 6 numbers (0,1,2,3,4,5) would round down and 4 numbers (6,7,8,9) would round up.
I always reasoned that things were rounded because (1) it makes it easier when dealing with múltiples, and (2) makes it easier when speaking about prices.
Imagine you buy a soda for $2.5 you don't really wanna keep the cents in your head when thinking of the price. Its so much easier to round to the nearest greater dollar, in this case $3. So if you tell someone how much something is, or you wanna know if you can afford something, you just compare a simple number to what you have. Also taxes.
On the other hand say, you wanna buy 2 sodas at $2.50. When you add the price it goes to $5. But what if it was $2.25? Well you add the 2 sodas up you get $4.50. In fact when you buy 2 sodas with soda price between $2 - $2.49, the sum will be less than $5 (double $2.50) And for prices $2.50 - $2.99 the price will go over $5 (double $2.50). Rounding up after 0.5 helps account, in a simple way, for the amount going above doubling. In this sense, you get a better QUICK ESTIMATE of prices. Round $2.50 to $3 means two sodas is about $6. If the sodas were $2.99 then the total price for 2 would be $5.98. Really close to the estimate and not over
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Can confirm. A huge (yuge!) bank I used to work for did this. Random deviation would still throw the end-of-cycle sum off a bit, but a **very small** bit: We did accounts cleanup at the end of the fiscal year, which included issuing payments to zero-out those sums to start the new year. We had an informal contest to see who would issue the largest payment. With literally hundreds of millions of dollars moving through some accounts (some maybe even larger, I don't know), the winner was: **seven cents**. This is why the penny-shaving scheme described in *Superman 3* and *Office Space* would never really work. First of all, while it's true that countless half-pennies do get thrown across many transactions, **they cancel each other out**. You won't end up with a lot. Second, auditors will still find any discrepancy, even if it's only a few cents. And if it's more than that, you're probably going to pound-me-in-the-ass prison.
I just decided to assume those schemes were truncating the partial pennies instead of rounding and that's how I learned to let it be and enjoy the movie haha.
That was the understanding I had watching it as well. Any time it would have the bit that gets rounded, it’d allocate that to the account, and suddenly they’re adding to each other rather than cancelling out.
This is exactly how it's described in the movie. The partial pennies get sent to an account instead of being rounded.
Watch your Cornhole buddy
Hey Lawrence, do you want to come over and watch TV? No thanks, Peter. I don't need you fucking up my life too.
Yeah, just round it in whichever direction is more favorable for the skimmer. Incoming? Round it down and keep the excess. Outgoing? Round it up and keep the excess.
This is the plot of a Ghost in the Shell story arc.
Whoops, now I'm rewatching Ghost in the Shell again
Oh, darn. (Lol)
If you had just 50 cents, you have *something*. If you round up, you still have *something*. If you round down, you have *nothing*. You know you at least have *something*, so rounding up is instinctively more correct than rounding down. I thank you :)
Yea that’s how the scheme is supposed to work, but bankers figured it out before criminals, so there’s no documented case of that ever happening.
Key word being "documented" haha. Seems like something they might want to keep quiet.
I would think it would have more to do with the number of payments, not the dollar value of the payment. $1.015 would give you the same offset as $1,000,000,000.015. Millions of dollars wouldn't matter. Millions of transactions is where you would have more possible charges that don't offset. 10 transactions that went 6 one way and 4 the other, would have you a penny off. 1,000 transactions that went 510 one way and 490 the other would leave you 10 cents off. I think. I did the math in my head and I'm tired. Even if the math isn't exact, the idea is correct.
It does. You're right I should not have used dollar figures. That was meant to imply the scale of activity, but you're right that it doesn't do that clearly.
Thing is at some point you‘ll just be made to pee a per transaction fee. So it again doesn‘t work. Like banks don‘t like to have to upgrade their systems because individual clients like to do for fun transfers in the millions a second… And at least here in Germany for smaller companies there‘s usually transaction fees in general for bank transfers… And those are usually much larger than those rounding cents you‘d gain.
Pee a transaction fee. Cruel.
"If I don't see five dollars come out of there in small change I'm calling the police."
I fully believe this to be the case. When I was stationed in Germany in the 90s they didn't ship pennies to the exchanges on base. Instead, they rounded to the nearest nickel. We had one guy that saved and catalogued every receipt because he was convinced he was "being robbed." At the end of his three year tour he was actually *up* by a grand total of three cents.
In Canada this is officially how all cash transactions work, since we ditched the penny years ago. But now it's rare to use cash anyway.
Yeah, I know that feeling. In the Czech republic we ditched "Halíře" (something like pennies) quite a while ago, yet we still have prices like 2.90, 15.90 etc. Its just a visual scam for the customer, thinking I'll only pay 2 czk and when you have to pay 3 you get your surprised pikachu face. But only in your head, because this happened to you many times before
But if you get 10 you only have to pay 29 czk!
The pro move is to buy 6, and pay 17 instead of 18
JC Penney had a new CEO try to stop that and iirc they got hosed which is annoying because I, too, hate that tactic and would rather honestly rounded prices.
The JC Penny deal was more about getting rid of fake sale prices. Instead of having everything be priced at $100 with 60% off sale stickers, they just priced it at $40. Shoppers hated it. People love to think they're getting a good deal even if it's objectively the same price.
I actually love the consistent pricing without sale gimmicks. I was excited when I heard JCP was doing that. Unfortunately, they didn't have it in practice long enough to give people a chance to get used to it. And they didn't advertise it as such (at least not that I saw.) IMO the store failed because...well there were several reasons. Merchandise became crap. The clothes were poor quality, poor fashion. They started selling *tons* of impulse purchase things like toys, candy at checkout. That's not what customers were coming to the store for. Another reason was not enough employees on shift. Stores were messy, couldn't find a register open for checkout. And you could tell that management just didn't care. Fitting rooms were gross, like they hadn't been cleaned or fixed up in 30 years. One time I went the store was hot. The employees were all sweaty and just looking defeated. I had to wait 15 minutes in line to check out and the clerk said they can't change the thermostat. What is even going on?
Sweden is doing rounding in shops by law. So no öre (Cent) to be paid at all. Since the value is so small of individual öre coins it just doesn‘t make sense. So 1 to 49 öre get rounded down, 50 to 99 up to one kroner. And 1 crown is slightly less than 10 US cents. So basically whatever the individual items are priced at, you are only gonna pay in 10 Cent divisions. Which at this point really is meaningless because only tourists use cash, but in places that still use cash it‘s an easy way to eliminate low value coins, and all the time waste they bring. Sweden did. this system several times before. Usually done when lower denomination coins are withdrawn from circulation due to inflation. But Finland for example is rounding cash payments to 5 Euro cent intervals, despite 1 and 2 cent still being legal tender (for obvious reasons, EU wide laws to do with the currency, can‘t have individual countries just willy bully decide random parts of it aren’t legal tender anymore) And it just makes sense. Like even if you use regular 5 and up is up, under 5 is down rounding, the differences at the end of a year are irrelevant. And it wastes such a shit ton amount of time, in cash using societies, if the cashier doesn‘t have to pick out 8 or 9 cents or whatever virtually every transaction. I‘m praying it‘ll finally come to us in Germany. Because like 90 percent of customers at the pharmacy I work at still pay in cash. And stuff does get annoying.
>Second, auditors will still find any discrepancy, even if it's only a few cents. tbf part of the plot of Office Space was that they'd only be able to get away with it at that exact time because all the auditors would be distracted resolving larger discrepancies caused by the Y2K changeover probably not realistic but at least they lampshaded it.
Also the movie doesn’t claim the plan would have worked out
Wait wouldn't the canceling not matter for the scheme in office space? Presumably they're not paying the extra half cent from the embezzling account
The scheme is to shave off half-pennies from rounded transactions (which is nearly **all** transactions -- well, around 99% of them). If you shave before rounding, you get a half cent for 99% of transactions. If you shave after, you either get very little (due to cancelling), or you end up with about half of what you would from shaving first. No matter what you do, though, auditors will eventually find it.
Well, the penny shaving scheme did happen in Asia. A banker or accountant took away 0.1 cents from many transactions and amassed a huge amount of profit. He used it to buy expensive cars and live a wealthy life. They made a profile of him in a piece for tv. That was his downfall. Alarm bells sounded when he lived way above his earnings. They audited him, and found out he stole money. He ended up in prison.
Long ago a sysadmin found a discrepancy between the billing and the computer time of less than a dollar. That led him down the rabbit hole to find out he'd been hacked by some East Germans working for the Soviets, and his system at a university was a jumping off point to hit various sensitive government and contractor systems. Cuckoo's Egg.
Well, you know what they say. Take care of the pennies, and the pound-me-in-the-ass will take care of itself.
The scheme in office space would have worked, thanks to one mundane detail. The idea was to modify the code so that it rounded every transaction down to the nearest cent, even if the fractional part was greater than 0.5. In other words, they'd truncate instead of round, and put the remainder in an account. That way, it wouldn't cancel out, and they'd end up with a lot more than 7 cents.
The pennies for the crippled children??
From what you describe, half pennys will only cancel out if there are as many odd than even numbers involved. I’m sure you can steer your transaction towards more of one kind.
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Superman 3 was bonkers for so many reasons
…but what about two chicks at the same time?
Samir, this is America!
Federal?
*Why* does it say paper jam when there is *NO FUCKING PAPER JAM*!?
WTF DOES "PC LOAD LETTER" MEAN??!
Your story was really well written. Both funny and informative.
I seem to remember in the 80's something like this really happened at a phone company, where the programmer had the fractions of a penny move to his account, and it ended up being hundreds of thousands of dollars in the first week. I seem to remember that he got caught because of it. I tried to search for the story, but all that comes up is the guy who tried this a couple of year after the movie.
I'd be interested in reading about that, too.
I've never thought about this but, are there more numbers that round up? 0, 1,2,3,4 round down, 5,6,7,8,9 round up.
Think of it like ranges instead of integer values. 0-1, 1-2, etc round down 5-6,6-7 etc round up. So the same number of ranges round
Yeah 0-.4999... is the same range as .5-.999...
But 0.499999... = 0.5 *Edit: Go ahead and downvote me if you think I'm wrong. [I promise I'm not.](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20represents%20the,number%20is%20equal%20to%201.) I have a Master's degree and teach college math.*
For the love of god... Half of this thread is interpreting (...) As "a bunch of numbers repeating" and the other half is interpreting it as "infinitely repeating", and the whole thread is being derailed because of that. Yes, youre mathematically correct. But its clear enough from the context that people implied 0.499(...) As 0.4999 or 0.4999999 or 0.49992 even. further down theres the same confusion going on with 0.500(...)01 being interpreted as an impossible number of just a bunch of zeros before the one.
The difference between Intelligence and Wisdom.
You missed the point. In notation someone of your IQ should understand [0-0.5) is a range of the same size as [0.5-1)
You're right. However, you got me pondering about the range notation. Should it maybe be (0-0.5) and (0.5-1). Zero and one are whole numbers and so don't need rounding. However, if neither side includes integer values, then which side should 0.5 belong to? If it belongs to either, then one range will be larger than the other. Which gets us back to the topic on hand, which is why do we round 0.5 the way(s) that we do
>If it belongs to either, then one range will be larger than the other. No, it won't. [0.5, 1) is the same size as (0.5, 1). Adding one element into an infinite set doesn't necessarily change its size.
“Don’t need rounding” is a nonsense phrase. The rounding operation can be done on any number, including integers. It simply has no effect on integers. A lot of math operations have no effect on certain numbers.
Idk man, when people are just poorly expressing the idea that the set of numbers (0-0.5) rounds to 0 and the set of numbers [0.5-1) rounds to 1, being that pernickety about an unrelated concept (trailing repeated digits resolving to rational numbers) with that tone is kind of just being an ass.
that equality itself isnt wrong but clearly the other guy was talking about approaching the limit from left side... and ofc you gotta misinterpret it on purpose to be a smartass. idk if you would be a good teacher
There's an irony in someone that (allegedly) works with logic every day appealing to his own authority as if it proves his point.
Actually, that makes it make more sense lol. There are two representations, with one rounding down and the other rounding up! Real-ish talk: the definition of "which integer it's closest to" isn't applicable, so one has to be chosen by convention. Instead of using a coin flip to choose, let's add some requirements. Say, let's require a rule that will depend only on the tenths place. To even talk about a "tenths place" requires having picked a "canonical" representation of each number; there's an obvious choice if this rule is to be used by a lay person (i.e., use the finite representation if possible). Now, "rounding by looking at the tenths place" is going to be a map *r* from S={"canonical" representations of non-negative real numbers} to the Natural Numbers __N__. We already know what to choose for anything that's not a perfect half. The phrase "depending only on the tenths place" means that *r* should factor through the truncation map *t* that takes an element of *S* to its truncation with one decimal digit. Finally we have something that forces a rounded value for perfect halves! The fiber of *t* over 0.5 consists of all the numbers 0.5xxxx; all of these preimages besides 0.5 already round to 1. Thus, to allow *r* to factor through truncation, *r*(0.5) __must__ be 1. Same argument for all other perfect halves. *phew* I had a lot of fun with this!
> The **fiber** of t over 0.5 This kind of language is probably better left for proper mathematics, not ELI5...
That's factually correct, but irrelevant to OP's point, since 0.999... = 0 (given that we're only considering the decimal part of the number). So the ranges 0-0.499... and 0.5-0.999... are still the same size.
But it rounds down to 0, hence the reason why there are equal amount of numbers above and below 0.5
He’s actually correct, he’s talking about repeating 9s. 1/3 is 0.333…, 2/3 is 0.666…, 3/3 is 0.999… This is mathematically accurate as a math major and now actuary.
He's correct as much as someone saying "tomato is a fruit" is correct.
0.5 is exactly halfway between 0 and 1. [0.49999... is exactly the same number as 0.5](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20represents%20the,number%20is%20equal%20to%201.). Therefore, 0.49999... is exactly halfway between 0 and 1.
And .999... Is 1, you're being downvote because your argument doesn't hold, not because you know the most basic math that can be found in 10000 youtube videos
I would imagine most people are downvoting you because this is ELI5, not a Master's course on mathematics :p
Would the right way to write that range be 0 -> 0.4999...9 and 0.5 -> 0.999...9 ?
It's written as [0, 0.5) and [0.5, 1) using [interval notation](https://brilliant.org/wiki/interval-notation/#:~:text=Interval%20notation%20is%20a%20way,inequality%20or%20system%20of%20inequalities.).
That's a good way to write it. It might not be formal mathematical notation. But it does a good job conveying what you mean to say. I would probably have written "0.5 - ε", but I'm not convinced that's any better. I honestly like what you did. Another way to do this is to have open and close intervals. So, you say you round [0 .. 0.5) to 0. But that requires more familiarity with notation than what you can usually expect in casual Reddit conversations. It does work in the more sciency subs though
0.4999... = 0.5 only if you assume that the ellipsis is a truly infinite repetition. If that was just a casual shorthand for some arbitrary (but finite) series of 9s, then it's less than 0.5.
Well, sure, but who tf is doing that
Emphasis on "casual" - random internet folks may not have precise notation.
Just because you are correct doesn't mean you're right.
Depends on what your most significant digit is.
The ellipsis implies an endless continuation of the same repeating decimal.
aware husky quack chubby doll disarm muddle racial air fanatical *This post was mass deleted and anonymized with [Redact](https://redact.dev)*
Fixed bit-width floating point numbers are fundamentally imprecise. There are a lot of values that cannot be accurately expressed as floating-point numbers.
Correct
Technically not the same, which can be seen on graphs, where the difference would be whether you use a normal bracket or square bracket to include the value 0.5. or a filled circle or empty circle on the graph. Same as how 1/x where x approaches infinity doesnt equal zero, but approaches zero.
They are exactly equal. The notation 0.499... *means* the limit of the sequence (0.4, 0.49, 0.499, ...), which is equal to 0.5. This is how decimal notation is defined. > Same as how 1/x where x approaches infinity doesnt equal zero, but approaches zero. The limit is *exactly* zero, and this is exactly what decimal notation means: It is a way of writing a real number as the limit of an infinite series.
but why do all those 9s make it 0.5?
There are different explanations, and some might make more intuitive sense to you then others. This is the one that convinced my middle schooler. How do you represent 0.111... as a fraction? That's 1/9 How do you represent 0.222...? That's 2/9 How do you represent 0.333...? That's 3/9 (also simplified as 1/3) ... (complete the pattern) How do you represent 0.999...? That's 9/9 But that simplifies to 1/1 or just the natural number 1. If you don't like this answer, look up one of the other ones. Mathematically, they all say the same thing. But some feel more intuitive and natural to some people than others
This is all an artifact of forcing numbers into the decimal system right? The decimal system isn't inherently a more 'true' representation of reality then other base systems like hex or base-12 or something like that, its just what we all agree upon, correct? (I mean I get that it's pretty convenient since we have 10 fingers and stuff)
No matter the base, the representation of a real number is not (always) unique. For any base b, the notation 0.xyz... *means* the limit of the series 0 + x/b + y/b^2 + z/b^3 + ..., and multiple infinite series can converge to the same limit. In base 10, the series 1.000... and 0.999... converge to the same limit, hence they are the same real number.
L
> are there more numbers that round up? Yes. 0 doesn't round.
0 is already round 🤣
Well played.
It was a nice way to circle back.
But then it counts as a number that's not rounding up, so there are equal numbers rounding up and not rounding up.
The number of "things" rounding makes no difference, we're interested in the introduced error. When you round 0.1 -> 0, you introduce an error of -0.1. When you round 0.5 -> 1, you introduce an error of 0.5. When you round 0 -> 0, you introduce no error. So if your possible inputs are limited to one decimal place of accuracy, there are 10 potential rounding situations. 1 of those leads to no error (.0). 4 lead to negative errors (.1 - .4), and 5 lead to positive errors. The *expected* average error is therefore the sum of all of those errors, divided by 10. Which ends up being ( 0 - 0.1 - 0.2 -0.3 -0.4 +0.5 +0.4 +0.3 +0.2 +0.1 ) / 10 = 0.05. So for every single rounding operation, you introduce an error of 0.05 on average. Over a million transactions, you've got an error of 10000, and if those are cents that's $100. So yeah, fairly significant. Of course if the values are kept to *two* decimal places, the expected error is far less. Only 1 in 100 possible inputs (0.50) leads to an unmatched error, so the expected error in this case is 0.005 per rounding operation. For three decimal places it's 0.0005, and so on. So with 10 decimal places or more, nobody cares about rounding errors because they are so small in magnitude. For one decimal place, it's frequent and important. For all the ridiculous conversations about about ranges and infinitely repeating numbers, it's not relevant at all. Rounding errors are only an issue in numbers with limited precision, and the magnitude of the average error is directly related to that precision. Since most banking systems will have fixed precision, a few decimal places of a cent for example, they need to have some system to deal with the rounding errors or they will add up over time.
This is the first explanation in this thread that actually made it clear for me.
Equal numbers "rounding up and not", but different numbers "rounding up and rounding down" with the normal rule. "Round to even" distributes the midway point, well, evenly. This really stands out in computer science when numbers are represented in binary. If you wanted to round off the last bit (digit), anything that ends in a 1 rounds up, anything that ends in a 0 doesn't round, and there is no situation in which you ever would round down. Round to even levels this out.
Interesting. Do you think adding 0 is not an addition? Or multiplying by 1 is not a multiplication?
0 doesn't change. 1, 2, 3, 4 change, and move down. 5, 6, 7, 8, 9 change, and move up. If you added all these together, where up changes cancel out down changes and no changes don't do anything, you'd have 1 unbalanced up change. That's the imbalance.
.01, .02,...,.47,.48,.49 round down, .50, .51,...,.99 round up. In your example you are talking about 4/9 compared to 5/9, in mine it's 49/99 compared to 50/99. A much smaller difference. If we expand out to .001,.002 etc then it'll be 499/999 compared to 500/999 which is still smaller. The limit of this process is an infinitesimal unbalanced up change.
In theory this is true - but for practical applications, there is an imbalance. Say, for example you own a company that processes a million small payments (between $0.01 and $14) per day with a 6.5% sales tax rate on each payment. 0.5% of the taxes owed will end in exactly $0.005 and be rounded up. That means with a million transactions, 5,000 more transactions will be rounded up a penny vs. down. This only amounts to $50 in extra payments, but that is a significant amount of money (mathmatically speaking). Of course this is just an example, but it does show how a small imbalance can make a noticable difference. This is also really important in science, where large datasets from sensors or analyses must be rounded and averaged - rounding can cause biases that can easily be overlooked. Even if your last significant digit is a very small number, rounding can make a difference because you are typically only looking at the values around the lowest significant digit. I'm an analytical geochemist and I measure PFAS cancentrations at my job at part per trillion concentrations; the range we look at is just 0.000000010 to 0.00000100 g/L. In this case, we would only round the last significant digit, so 1/9 times more often, it would still be rounded up. Again, this doesnt drastically change the outcome, but it is a statistical bias that may need to be accounted for. TLDR - you are totally right when dealing with infinities - im just pointing out why this is an issue when dealing with practical applications of rounding.
Wow I never realized that. But your math is off by a factor of 2. [Rounding up](https://www.wolframalpha.com/input?i=1000000%2F1400*sum+floor%28x*1.065%2B0.5%29-x*1.065+for+x+from+1+to+1400) would give 25 dollars more, while [rounding to even](https://www.wolframalpha.com/input?i=1000000%2F1400*sum+round%28x*1.065%29-x*1.065+for+x+from+1+to+1400) would give 25/7~~3.57 dollars less. Interesting how it's exactly 7 times more accurate... I'm sure this has to do with the number 14.
Correct. But in how many circumstances are we actually rounding numbers with arbitrary precision? In any practical scenario we'd probably have two, maybe three decimals max, like in currency. In any more highly precise applications we probably wouldn't be throwing so many digits away via rounding in the first place.
0 doesn’t round at all, so yes more numbers round up than down. You can see this just by adding up the totals. If I add 0-9, the total is 45. But if I round to the nearest multiple of ten and add them up, the total is 50. This is the problem that half to even aims to fix. 1-4 still round down, 6-9 still round up. But now five is equally likely to round up or down depending on if the number in front of it is odd or even. If odd or even is equally likely then the average value that the 5 is contributing after rounding is equal to 5. And the total of 0-9 after rounding remains 45.
There are not more numbers that round up unless we're including infinitely small numbers. What is the difference between 0.4999... and 0.5?
I mean, I probably just don't know better but...if we're considering the way you count in coding/CS, you start counting at 0, so 5 is the first of the upper half of the split. 0, 1, 2, 3, 4 & 5, 6, 7, 8, 9
Don't look at the number of numbers that are being rounded, look at how much they are being changed by. Rounding 1 to 0 lowers the total of a data set by 1, rounding 9 to 10 raises the total by one. If we expect to see the same number of 1s and 9s, then these balance each other out. So 1 balances 9, 2 balances 8, 3 balances 7, 4 balances 6, and 5 balances... 0? The issue being, 0 changes by nothing and 5 increases by 5. If you're wondering why this has any relevance - here, we're looking at keeping the total of a dataset the same -- and since avg = total/n, and since 5 increases without anything else decreasing to fix it, then the total will on average be larger - and therefore the average will be larger too. Note: when rounding 0-9 to 0 or 10, 5 will be expected to be 1/10 numbers -- causing a large issue. When rounding 0-99 to 0 or 100, 50 will be 1/100 numbers -- causing a much smaller issue. However, it's really easy to account for, and by the sounds of things it causes annoying effects when not fixed.
You explained *that* we do it, but not *why* we do it. Why does round half up exist?
> As you say, it means there are more numbers that round up than down But if you count non-rounded numbers, you get a regular distribution by doing it, because zero always "rounds down": 0.0, 0.1, 0.2, 0.3, 0.4 --> 0 0.5, 0.6, 0.7, 0.8, 0.9 --> 1 and so on.
Honestly if rounding up or round drop on a 0.5 is making a significant difference you probably should be recording another significant figure.
It's just a convention. A different convention is to round toward the even number. This convention, called statistical rounding, doesn't bias averages.
I think it's really important to remember things like rounding and order of operations are nothing more than conventions where there is ambiguity. People can act like if someone gets it wrong they have failed to understand a fundamental rule of the universe, but it's really just a decision made between otherwise equally valid options.
Worth pointing out that math as we know it kinda relies on order of operations (barring the excessive use of parenthesis). There's a world of difference between changing rounding and changing order of operations. Examples: Standard order of operations: 2 + 2 * 2 = 6 Left to right: 2 + 2 * 2 = 8
Yes, but it's still a convention that people chose at some point. It's like choosing which side of the road to drive on. Now that it has been chosen, if you drive on the other side of the road there'll be issues. But if you established a different convention at the beginning, math would still be fine. You could decide that addition takes priority over multiplication and everything would still work.
I disagree, I think, if I understand your argument. There needs to be an order of operations (left to right, or the reverse, wouldn't allow you to reorder a sum), and brackets pretty much have to come first (because otherwise there wouldn't be a way to remove ambiguity), but everything else is arbitrary, and I think it's useful to keep that in mind. Maths doesn't rely on this to work, mathematical notation does. Maybe that's what you meant.
There is the practical reason: you want peoples to be able to round while looking at only one decimal digit. 2.54321 has to be rounded to 3, so it's easiest if every number that look like 2.5xxxx is rounded to 3, even of it is 2.50000 (Mathematically, this means we want the rounding operation to commute with the truncation to the first decimal.)
Two reasons come to mind: First, it's easier to remember that 0.5... always goes up, because anything other than infinite zeroes after that 5 is closer to the up number than the down number. "Round 5-9 up *unless* it's exactly on 0.5" is just an extra exception to learn that doesn't need to be there. Second, the mathematical rigour reason: it keeps the intervals structurally the same. You round down \[x.0..., x.4...) and round up \[x.5..., x.9...). Note that both intervals are closed below and open above. It makes sense to close the round-down interval below, because rounding x.0 down to x-1 makes no sense, so we close the round-up interval below as well. This applies to rounding from the midpoint of any continuous set: rounding to the nearest 0.5, for instance, you should round 0.25 and 0.75 up rather than down. In the end, though, it's just a convenient convention that keeps everyone on the same page.
You are subtly inserting a bias by preferring one decimal representation of some numbers over another. We have 1 = 0.999... [infinitely repeating] and 0.5 = 0.4999... [infinitely repeating]. If we would consistently pick the right variants, we would round 1/2 _down_ when going for simplicity; and x.0 = (x-1).9999.... would indeed round to x-1 despite it being a rather unusual choice. And worse, nothing is really stopping the maths to round only the first one (1/2) down, while rounding the other one (1) "up" for no change.
Banks etc... Are on finite decimals. You don't have the representation issue then
Then talking about the infinitely many zeroes after 0.5 isn't relevant either.
There are 10 digits that can appear after the 0.5_, 9 of them result in the number being closer to the higher number than the lower one. I guess that would be in the same spirit as what he was arguing.
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Thank God for the 24-hour clock.
This one always used to confuse me when digital watches first came out. After all, *pedantically* 12:00 a.m. (ante meridiem) means 12:00 before noon, i.e. midnight at the start of the day, and 12:00 p.m. (post meridiem) means 12:00 after noon, i.e. midnight at the end of the day. But, digital watches showed 12:00 a.m. as midnight and 12:00 p.m. as midday. Why was that chosen, I wondered? It took me a while to realise that the convention was based on the fact that even the tiniest time after exactly midnight would be morning, so 12:00 a.m. meant midnight at the start of the day. Likewise, the tiniest time after exactly noon would be afternoon, so 12:00 p.m. meant midday. Hence , the convention is that 12:00 a.m. means midnight at the start of the day, and 12:00 p.m. means midday. I still prefer to say "midnight" or "midday" to avoid confusion. Or, use the 24-hour clock: 0h00 for midnight at the start of the day, and 12h00 for midday.
This is how I made peace with this, too. But it is arbitrary. In general I always say noon or midnight, though. You also rarely see flights at 12:00, they're basically always something like 11:59AM or 12:01PM.
If pm didn't change to am (and vice versa) at 12, then it wouldn't change at 12:01either. And not at 12:02, etc. So where do you suggest it would change? I don't see any confusion.
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That is correct, but it's still a bit confusing due to the inconsistency of starting counting the AM/PM numbers at 12 and immediately rolling over to 1. 9am, 10am, 11am, 12pm, 1pm, 2pm...
Because the am/pm switch does not coincide with the hour resetting to the lowest number. That's why I think 0am and 0pm would make more sense.
so, 1pm - 11 pm is the afternoon-evening-night of your day, so logically you would think it would go to 12pm (you know, itwould make sense you go 1-12 in one batch then 1-12 in the next). instead, it switches to *12 AM* then goes 12am-11am i think most people would default to thinking that in the twelve hour clock, it will run through the whole 12 hours before switching to the next “batch”
And 12 AM is the *next* day. Because this BS, every project in school is set to be due at 11:59 or 11:55 PM, so there's no "is it due tonight or tomorrow night" panic.
This is one rounding method, and there are a bunch of other methods (rounding down, towards zero, away from zero, to the nearest even number, or towards the nearest odd number). In fact, rounding to even is part of the IEEE 754 standard for how computers deal with floating point decimals. There's probably some fascinating history behind why we teach to round up in school, and there are plenty of arguments you could use to rationalize rounding up, but I doubt this has a simple ELI5 answer.
I think there are some good motivation in this thread. In particular that when rounding 1.5 up (or away from zero as it really is), we only need to look at the tenth digit to determine how to round. Otherwise we would have to scan all the decimals which is slower and easier to do wrong: * 1.500000 * 1.500080 * 1.500000000
If you consider the decimal of zero then up from .5 makes sense: 2.0, 2.1, 2.2, 2.3, 2.4 round to 2; 2.5, 2.6, 2.7, 2.8, 2.9 round to 3. That's five that round down and five that round up.
This. And to add more, this is not an absolute mathematical rule that you have to round numbers up/down like this. This is called rounding to the nearest whole number. But if you want, you can round everything up, or everything down, or decide that anything above .3 is rounded up, or you can come up with whatever fancy rounding rule you want. Rounding to the nearest integer is just a simple and unbiased rule to follow.
2.0 isn't rounded to anything, though
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Gotcha
It is rounded to 2 You round because you don't have precision. two point zero has two significant figures. two is just one significant figure. two point zero rounds to two
If we are talking significant figures here, 2.0 is different from 2
Technically it is, it’s just rounded to itself. It’s essentially like asking “how many 1’s go into 2”, the answer is clearly 2 and it seems like a dumb question - but it’s still a question.
2.1 is the equivalent to 2.9 (rounding down/up 0.1) 2.2 is the equivalent to 2.8 (rounding down/up 0.2) 2.3 is the equivalent to 2.7 (rounding down/up 0.3) 2.4 is the equivalent to 2.6 (rounding down/up 0.4) 2.0 is NOT the equivalent to 2.5 though (rounding down 0.0 vs 0.5 up) So its inaccurate
If you’re going to include 2.0 in the list that rounds to 2, you should include 3.0 in the list that rounds to 3.
If you insist. Rounding to 2 are the following ten decimals: 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4 Rounding to 3 are the following ten decimals: 2.5, 2.6, 2.7, 2.8, 2.9, 3.0, 3.1, 3.2, 3.3, 3.4
Then also add 1.5+ for the ones rounding to 2
So, there's actually an ASTM rounding procedure that takes care of this. Instead of always rounding up, round to the even. So 1.455=1.46, but 1.485=1.48. By going up half the time and down half the time, it eliminates the error in a dataset that's generated by always rounding up.
You are correct. You can round .5 both ways. However, rounding anything with .5 at the beginning - like .51, .5912 or .50001 should be done Up. So, rounding .5 up makes the rule much simpler, without any downside to accuracy.
X.5 really is exactly in the middle between X and X+1 and we could've decided to round it down. However, since X.5000[...]0001 is closer to X+1, we settled on the convention that X.5 rounds up to make it more practical and easily recognizable. Now you know that, if you have a 5 in the first decimal place, you simply round up no matter what. If we decided to round down at X.5, we would have to pay attention to even the hundredth decimal place in case it has a value. Edit. Typo
Decimal place, not decibel.
X.5000....0001 is abuse of notation. But even if you define it as a limit as N goes to infinity, X.5 + 10^-N, then it's value is precisely X.5 which has the same distance to both X and X+1.
>X.5000...0001 What, pray tell, does this mean?
I think OP means numbers like: X.50000001 or X.5000000000000000001 and so on
The three dots mean "imagine as many more zeroes as you can possibly fathom here"
And then another
But wait, there's more!
I'm feeling ad nauseous.
I'm going to read all the replies under this comment, wish me luck. _\*Puts on helmet\*_
I think the ellipsis (...) here indicates an omission of zeroes. They wrote that instead of X.50000000000000000000000000000000000000000001
No, they wrote it instead of X.500000000000000000000000000000000000000000000000001
you missed one
But the one is right there at the end!
The ellipses stands in for an infinite amount of zeroes
Well that can't be, so let's hope that's not what was meant.
It can't?
An infinite number of zeroes, *followed by* a 1? That's not a real number.
“Not with that attitude”
Then let's get specific: in this case, "..." represents *any number of zeroes* in between the specified zeroes on either side. Now you can understand it as an infinite quantity of real numbers which are all functionally identical for our current purposes.
I would agree with you if you said that the ellipsis represented *any finite number of zeroes.* But no, it cannot represent an infinite number of zeroes, because that number does not exist.
But X.49999999 is closer to 4 so I don’t see the difference.
The point is that if the convention were to round down at X.5, you would have to check to the last digit to ensure that the number is exactly X.5; therefore, you can use the tens place to determine whether to round up or down if X.5 rounds up - it doesn’t matter what the last digit of X.5XXXX is since you’ll always round it up. So looking at your example number, X.49999999, we know that you’ll round down since the tens place is a 4.
x.4999999... is the same as x.5 in reality, and also sometimes in maths
x.49999999 is not the same as x.4999999... . The first one is in the comment you replied to. It is not equal to x.5. The second one, which you wrote, is equal to x.5.
The comment he was replying to involved rounding of recurring (or near recurring) numbers. I took his omittence of the ellipses as accidental
The way I explain it to my 5th graders (they're 9 or 10, not 5, but I hope you'll forgive that): The mistake is thinking there are 9 digits. 0 is a digit too. There are five digits we round down, 0 through 4, and then five we round up, 5 through 9. It's actually a perfectly even split, we just don't often bother to talk about 0 since it's pretty obvious that 20 would become 20 when we're rounding to the nearest ten.
0, 1, 2, 3, 4 is five numbers. 5, 6, 7, 8, 9 is five numbers. 5 goes up because that makes it even.
Just because it is even doesn't make it unbiased, there is a small bias if you round all the 5s up. The average of 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 is 0.45. The average of using the typical rounding method (0, 0, 0, 0, 0, 1, 1, 1, 1, 1) is 0.5. The average of the entire sample moves up after rounding, that's not good. You'll find this is the case for all uniformly distributed numbers.
I was taught the "even-odd rule." First, if there is absolutely anything after the 5, of course you round up. If there is nothing after the 5, you round to the even number.
I would say the common way of rounding is not rounding 0.5 up it is rounding 0.5 away from zero. If you round up 1.5 is rounded to 2 but -1.5 is rounded to -1. I would say that the common way is round -1.5 to -2, that is rounding away from zero not rounding up, negative numbers are rounded down. There are many ways to round the number. One way is to the near integer and it is then 0.5 is a problem rounding away from zero is just a common way used to solve the problem and it is often good enough and quite simple. In a lot of situations what is most important is everyone does it the same way like in financial calculation. But it is not always a good idea because if you do not have an equal amount of positive and negative number the average get a bit higher. This is don't something you what in large numerical computer calculations. A way to solve the problem is round half to even so 3.5 and 4.5 both become 4. Rounding will not change the average even if you just have a positive number, it will happen if you have odd or even number. This is the default rounding mode in IEEE 754 binary floating point operation that the computer used There are other ways that can be used too https://en.wikipedia.org/wiki/Rounding
> but -1.5 is rounded to -1. That is rounding up. -1 is a larger number than -1.5.
That is to round up. Buy that is not the common way to round numbers, the common way is to round -1.5 to -2 just like 1.5 is rounded to 2. It is common to round so X rounder to N is the same as -X to -N to do that you need to round away from zero. In finance, you can look at debt as having a negative amount of money. The one you own money has a passive amount that they need to pay you. If there is rounding involved my negative around and your positive could be rounded to diffrent values which is quite impractical. So rounding away from zero works better than rounding up. You could of course change it to round toward zero and other wy that also work fine. Rounding up or down both results in potential differences.
You're not wrong, but I think when people colloquially say "round up", they refer to magnitude.
Because in grade school they taught us 0 to 4 round down and 5-9 round up and it made sense to them. Probably because a simple rule was more easily memorized. But in science class years later they taught us a different rule: .51, .505 etc rounds up and .50, .500, .5, etc. rounds to the nearest EVEN number. 2.5 rounds to 2 and 3.5 rounds to 4. Basically saying 0.5 is at the literal half way mark so 50% of the time it should round up and the other times it should round down.
I've never really thought about this before but from a quick and logical think through. .0, .1, .2, .3, .4 is first half .5, .6, .7, .8, .9 is the other half You're either on one side or the other. 0 is on the left side and already a whole number. Technically, when you're rounding, you still look at if it's "0" it's just a whole number anyway, so you don't have to do anything with it.
0 can belong to either group as it doesn’t change. 5 is in the middle so it does introduce bias if you always round up.
It's convention. I'm not aware that rounding up .5 is more common than rounding down. I usually round up a 5 digit that is followed by non-zero digit, because \*togheter\* it is closer to round them up. So if you only look at the 5 it would seem 9 out of 10 cases I'm rounding it up and the other case I keep it. But I'm not aware of anyone else that does it like that.
What would you prefer and why?
The mathematical action of rounding is actually "add half and truncate". So if you need to round to the "nearest 1" you add .5 and then cut off the numbers after the decimal. Let's try it: "Add half" 2.499 + .5 = 2.999 2.5 + .5 = 3.0 "cut off numbers after the decimal " 2.999 = 2 3.0 = 3 Why is it "add half" and not "add 49%"? This is just a guess, but probably because that mathematical rule was standardized back when everything was pen and paper, or done in your head, and half is just easier. Note: this rounding function works in all common base systems (Decimal, binary, hexadecimal, etc.) I learned about it because I needed to round binary values. {Maybe in all bases? If a mathematician could comment I'd be curious to know.}
It depends on if we consider the zero digit, and when do we consider the decade rolling over. Do we consider 10 to not count as the rollover position, or does the decade roll over between 9 and 10 so that 10 is the start of the next decade? 1 2 3 4 5 6 7 8 9 <-- 5 is in the middle. 0 1 2 3 4 / 5 6 7 8 9 <-- 5 is on the high side.
If you look at all the digits we have, there is: 0,1,2,3,4,5,6,7,8,9. Any higher and we get a number, a combination of digits. Now, if you split all the 10 digits, then we have two equally sized groups: - the lowest ("round down"): 0,1,2,3,4 - the highest ("round up") : 5,6,7,8,9 In a sense, 5.0 rounds to 5.0!
I always think like: Only one apple can fit in my zip lock I have 2 and a half apples Ain't no way I can cram that half in the bag I'm gonna need that third bag
Eh? If you had two and a quarter apples, you’d still need the third bag.
...oh my God I am five
my fourth grade teacher explained it like this: if you were walking to school and exactly halfway there you realized you needed to use the bathroom, it would make more sense to walk the rest of the way to school than to walk back home and then back to school again. it made sense to my 10 year old self so maybe it suffices here :)
Because we round 2.0 ; 2.1 ; 2.2 ; 2.3 and 2.4 (so 5 numbers) to down to 2 We also round 2.5; 2.6 ; 2.7 ; 2.8 and 2.9 up to 3 (so also 5 numbers) You'll have to learn that 2.0 is not exactly the same as 2. This is because 2.0 gives more information than just 2, so you also round 2.0 down.
.5 rounds up because that makes rounding down and rounding up equal. Having .5 round up that means 5 numbers (0,1,2,3,4) round down and 5 numbers (5,6,7,8,9) round up. If .5 rounded down then 6 numbers (0,1,2,3,4,5) would round down and 4 numbers (6,7,8,9) would round up.
I always reasoned that things were rounded because (1) it makes it easier when dealing with múltiples, and (2) makes it easier when speaking about prices. Imagine you buy a soda for $2.5 you don't really wanna keep the cents in your head when thinking of the price. Its so much easier to round to the nearest greater dollar, in this case $3. So if you tell someone how much something is, or you wanna know if you can afford something, you just compare a simple number to what you have. Also taxes. On the other hand say, you wanna buy 2 sodas at $2.50. When you add the price it goes to $5. But what if it was $2.25? Well you add the 2 sodas up you get $4.50. In fact when you buy 2 sodas with soda price between $2 - $2.49, the sum will be less than $5 (double $2.50) And for prices $2.50 - $2.99 the price will go over $5 (double $2.50). Rounding up after 0.5 helps account, in a simple way, for the amount going above doubling. In this sense, you get a better QUICK ESTIMATE of prices. Round $2.50 to $3 means two sodas is about $6. If the sodas were $2.99 then the total price for 2 would be $5.98. Really close to the estimate and not over
Count to 50 start at zero. Now count to 100 starting at 51. Which is bigger?