# Proving that (Even Numbers)^n = Even Numbers

## Homework Statement

Let "a" be an element of any integer. Let "n" be an element of any natural number that is greater than or equal to 2. The integer "a^n" is an even number if and only if "a" is an even number. Prove this bi-conditional statement.

## Homework Equations

Definition of an even number: Let x be an even number. Then there exists an integer y such that x = 2y.

Definition of an odd number: Let x be an odd number. Then there exists an integer y such that x = 2y + 1.

## The Attempt at a Solution

Note that it is a bi-conditional statement, this means that we have to prove:
1. If a^n is an even number, then a is an even number.
2. If a is an even number, then a^n is an even number.

The most straightforward approach I had was proving it directly (I only know direct proof, contrapositive, and contradiction for now):

1. Since a^n is even, then there is an integer y such that a^n = 2y. This brings forth the following relationships:

a^n = 2y
a^n = 2 * (2^(n-1) * k^n) - where k is an integer.
a^n = 2^n * k^n
a = 2k

Thus, a is even.

2. Since a is even, then there is an integer y such that a = 2y. This brings forth the following relationships:

a^n = (2y)^n
(2^n)*(y^n) = 2 * (2^(n-1)) * y^n
a^n = 2 * (2^(n-1) * y^n)
a^n = 2m - where m is an integer

Thus, a^n is even.

The reason I substitute y = (2^(n-1) * k^n) or m = (2^(n-1)) * y^n) was under the closure property of integers.

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I am not even sure if I proved this statement correctly. Furthermore, I am also having doubts about how I applied the closure property of integers. Can anyone give me feedback?

On a side note, I already know that I can easily prove (2.) by using contradiction.

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a^n = 2y
a^n = 2 * (2^(n-1) * k^n) - where k is an integer.
I don't see where the 2^(n-1) comes from...

Maybe you should try the contrapositive...

I don't see where the 2^(n-1) comes from...

Maybe you should try the contrapositive...
I tried doing the contrapositive. Here is what I got so far:

If a is odd, then a^n is odd:

a = 2x + 1 - in this case, x is odd.
a^n = (2x + 1)^n

Then I got stuck... any tips?

Try working out the (2x+1)^n...

Try working out the (2x+1)^n...
Sorry to sound stupid, but I do not really know how to expand (2x + 1)^n...

You do this by applying the binomial theorem: http://en.wikipedia.org/wiki/Binomial_theorem

Otherwise, you could also try a proof by induction, but you mentioned that you can't use it. However, I see no other way of proving this thing...

gb7nash
Homework Helper
Let "a" be an element of any integer. Let "n" be an element of any natural number that is greater than or equal to 2. The integer "a^n" is an even number if and only if "a" is an even number. Prove this bi-conditional statement.
The backward direction should be straightforward. Let a be an even integer. So a = 2k for some integer k. Looking at an, we have:

an = (2k)n = 2nkn = 2(2n-1)(kn). So an is even.

For the forward direction, I think your best bet is to prove the contrapositive, like the previous poster suggested.

You do this by applying the binomial theorem: http://en.wikipedia.org/wiki/Binomial_theorem

Otherwise, you could also try a proof by induction, but you mentioned that you can't use it. However, I see no other way of proving this thing...
Well, after having contacted my professor, he said it would be acceptable if I use induction to prove this. And after studying induction for a good three hours or so, I still do not know how to use induction on this problem. Can anyone give me a leeway? Thanks.

eumyang
Homework Helper
Well, after having contacted my professor, he said it would be acceptable if I use induction to prove this. And after studying induction for a good three hours or so, I still do not know how to use induction on this problem. Can anyone give me a leeway? Thanks.
If a2 is even, prove that a is even.

Then, assume true for n = k:
If ak is even, then a is even.

... and prove true for n = k + 1:
ak + 1 is even -> ?