Direct proof of n^2 is even implies n is even

Reply Fri 12 Apr, 2013 07:15 pm
SO far, I have been able to come up with these steps:
1. if n^2 is even, n^2=2k (definition of an even number)
2. n*n=2k
3. n=2k/n
4. k is a multiple of n, so n= 2*n*x/n
5. n=2x

Therefore, n is even...

The problem is, how do I prove step 4 (that k is a multiple of n)?
Reply Fri 12 Apr, 2013 09:42 pm
lampiaodecapoeira wrote:

The problem is, how do I prove step 4 (that k is a multiple of n)?

How indeed...

You have k = (n/2)n, but to conclude from that that k is a multiple of n is tantamount to saying that (n/2) is an integer, which is begging the question, since if you know n/2 is an integer, then you already know n is even.

Instead of your route, you could trying branching off in another direction after step 2.

You have: 2 | n*n.

Are you allowed to use the little theorem that says: if you have prime p, and integers a and b such that p | ab, then p | a or p | b ?
Reply Sat 13 Apr, 2013 03:38 pm
If n^2 is even, then n^2 - 1 is odd, therefore the product (n+1)(n-1) is odd.
Now n+1 and n-1 are consecutive integers on either side of n, which are either both even or both odd. So since their product is odd, they are both odd, which proves n is even,
0 Replies
Reply Sun 14 Apr, 2013 02:13 am
Yes, I can use it. So what would be the next step?

Thank you!
Reply Sun 14 Apr, 2013 09:41 am
If you can use it, then you're basically done.

According to that theorem, if 2 | n*n, then (2 | n or 2 | n), since 2 is prime.
(You're letting 2 be p, and you're letting n be both a and b.)

2 | n => there exists integer m such that n = 2m.
Reply Mon 15 Apr, 2013 06:40 am
right. ok, thank you.

But supposing I can't use this theorem? (for the fun of mathematics)
Reply Tue 16 Apr, 2013 02:53 pm
Well, I guess I'll defer to fresco in that case.

(His answer doesn't use that theorem.)
0 Replies

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