Proving that (Even Numbers)^n = Even Numbers

[number0]

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.

 

[micromass]

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...

 

[number0]

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?

 

[micromass]

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

 

[number0]

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

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

 

[micromass]

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]

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(2^n-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.

 

[number0]

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]

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.

Start with the base case, n = 2.
If a² is even, prove that a is even.

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

... and prove true for n = k + 1:
a^{k + 1} is even -> ?