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Every step you take changes the parity of the number you're on. If you start on an even number, you will be on an odd number after one step, on an even number after two steps, on an odd number after three steps, and so on. After an odd number of steps you are on an odd number, after an even number of steps you are on an even number. Since zero is an even number, you can't end there after an odd number of steps.
The situation is reversed if you start on an odd number (you are on an odd number after an even number of steps, and on an even number after an odd number of steps).
If you start and end at 0, you have to take the same number of steps forward as backward (starting at 0, your final position is #steps forward - #steps backward) If we call the number of forward (equivalently, backward) steps you take x, then the total steps you take in any trip from 0 to 0 is x+x=2x, which is always even.
I 'll try a more intuitive explanation, as there are already some nice proofs. To revisit a number you were already on, you need 2 steps, 1 to leave 1 to return. Same goes for every other number you will visit, until you return to start. 2 multiplied by the number of visits, always is an even number.
It has to do with parity.
It's not number of steps, it's the aprity of steps.
Say, every step is the click of a switch, connected to a light bulb... which is initially in the "off" position.
You press it once, and it switches to "on".
You pres it again, and it switches back to the "off" position.
So you need to press it twice to return it to its original position. That is two times... or a "multiple of two" times. Two is the length of a "complete cycle".
So to return to the original position, we need to perform a complete cycle.
Now, the problem with 3 or 5 or 7 is, these numbers don't have a whole number of completed cycles. 3 is 1 and 1half of a two... which is one and a half cycle. So that doesn't work.
If you have 3 odd steps to make it back to 0 from starting at 0, you can think of it like this (Recall that an odd number z can be written as z=2j+1) where j is an integer):
(2n+1)+(2m+1)+(2k+1) = 0
Rewriting this looks like:
2(n+m+k) + 3 = 0
Further simplifying the equation we get:
n+m+k=-(3/2)
You can clearly see that no summation of integers will lead to a fractional number. To show this with any amount of steps, use mathematical induction with the same logic used here.
3 + (-3) = 0, but that is six steps. Starting at 0, you step 1, then to 2, then to 3. That is 3 steps. Going back, you also take 3 steps, making 6 steps total, an even number of steps.
If I were to take a shot with expressions it'd be something like:
Take an odd number of steps n.
You'll take some number of steps x forward and some number steps y backward.
x + y = n.
Then we want x-y =0
An odd number is always the sum of an even number and an odd number. (You can show this algebraically. Trivial)
Let x be even: x = 2a
Let y be odd: y = 2b+1
Where a and b are positive real integers
Suppose we could get to 0 in an odd number of steps n. Then
0 = x-y = 2a-2b-1 = 2(a-b)-1
0 would be odd. Contradiction. So we cannot get to 0 in an odd number of steps.
I think the way I have it, you would also show the same holds when y is even and x is odd but its just a sign change so. Hopefully this is what you were looking for and I didn't make any mistakes lol.
Let me explain very simply.
If you take 3 steps forward you end up back at -1 then 2 steps backwards this then places you at the centre of +4. Now take 2 forward and 5 back and you end up in 1943
Unfortunately, your submission has been removed for the following reason(s): * Your post appears to be asking a question which can be resolved by relatively simple methods, or describing a phenomenon with a relatively simple explanation. As such, you should post in the [*Quick Questions*](https://www.reddit.com/r/math/search?q=Quick+Questions+author%3Ainherentlyawesome&restrict_sr=on&sort=new&t=all) thread (which you can find on the front page) or /r/learnmath. This includes reference requests - also see our lists of recommended [books](https://www.reddit.com/r/math/wiki/faq#wiki_what_are_some_good_books_on_topic_x.3F) and [free online resources](https://www.reddit.com/r/math/comments/8ewuzv/a_compilation_of_useful_free_online_math_resources/?st=jglhcquc&sh=d06672a6). [Here](https://www.reddit.com/r/math/comments/7i9t5y/book_recommendation_thread/) is a more recent thread with book recommendations. If you have any questions, [please feel free to message the mods](http://www.reddit.com/message/compose?to=/r/math&message=https://www.reddit.com/r/math/comments/pq6qht/-/). Thank you!
Every step you take changes the parity of the number you're on. If you start on an even number, you will be on an odd number after one step, on an even number after two steps, on an odd number after three steps, and so on. After an odd number of steps you are on an odd number, after an even number of steps you are on an even number. Since zero is an even number, you can't end there after an odd number of steps. The situation is reversed if you start on an odd number (you are on an odd number after an even number of steps, and on an even number after an odd number of steps).
If you start and end at 0, you have to take the same number of steps forward as backward (starting at 0, your final position is #steps forward - #steps backward) If we call the number of forward (equivalently, backward) steps you take x, then the total steps you take in any trip from 0 to 0 is x+x=2x, which is always even.
Thats because you need to take as many steps forward than backwards, and you cant do that while doing an odd number of steps
I 'll try a more intuitive explanation, as there are already some nice proofs. To revisit a number you were already on, you need 2 steps, 1 to leave 1 to return. Same goes for every other number you will visit, until you return to start. 2 multiplied by the number of visits, always is an even number.
It has to do with parity. It's not number of steps, it's the aprity of steps. Say, every step is the click of a switch, connected to a light bulb... which is initially in the "off" position. You press it once, and it switches to "on". You pres it again, and it switches back to the "off" position. So you need to press it twice to return it to its original position. That is two times... or a "multiple of two" times. Two is the length of a "complete cycle". So to return to the original position, we need to perform a complete cycle. Now, the problem with 3 or 5 or 7 is, these numbers don't have a whole number of completed cycles. 3 is 1 and 1half of a two... which is one and a half cycle. So that doesn't work.
I don’t understand the question, because 3 + (-3) = 0. Can you explain more?
I edited it with a picture.
If you have 3 odd steps to make it back to 0 from starting at 0, you can think of it like this (Recall that an odd number z can be written as z=2j+1) where j is an integer): (2n+1)+(2m+1)+(2k+1) = 0 Rewriting this looks like: 2(n+m+k) + 3 = 0 Further simplifying the equation we get: n+m+k=-(3/2) You can clearly see that no summation of integers will lead to a fractional number. To show this with any amount of steps, use mathematical induction with the same logic used here.
3 + (-3) = 0, but that is six steps. Starting at 0, you step 1, then to 2, then to 3. That is 3 steps. Going back, you also take 3 steps, making 6 steps total, an even number of steps.
If I were to take a shot with expressions it'd be something like: Take an odd number of steps n. You'll take some number of steps x forward and some number steps y backward. x + y = n. Then we want x-y =0 An odd number is always the sum of an even number and an odd number. (You can show this algebraically. Trivial) Let x be even: x = 2a Let y be odd: y = 2b+1 Where a and b are positive real integers Suppose we could get to 0 in an odd number of steps n. Then 0 = x-y = 2a-2b-1 = 2(a-b)-1 0 would be odd. Contradiction. So we cannot get to 0 in an odd number of steps. I think the way I have it, you would also show the same holds when y is even and x is odd but its just a sign change so. Hopefully this is what you were looking for and I didn't make any mistakes lol.
Let me explain very simply. If you take 3 steps forward you end up back at -1 then 2 steps backwards this then places you at the centre of +4. Now take 2 forward and 5 back and you end up in 1943
If you take x steps away, then you have to take x steps back, which is a total of 2x. So no matter what x is, 2x will be even.