Demonstrating if a number is even

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

I'm looking at a problem, and want to check if my attempt is right:
Problem
For n an integer number, consider the following definition:
Def. 1:

is even $\Longleftrightarrow (\exists k \in \mathbb{Z})(n=2k)$

Prove that:
nÂ²

is even $\Longleftrightarrow n$ is even.

My attempt

is even
$\Longleftrightarrow (\exists k \in \mathbb{Z})(n=2k)$ using Def. 1.
$\Longleftrightarrow (\exists k \in \mathbb{Z})(nÂ²=2(2kÂ²))$
$\Longleftrightarrow nÂ²$ is even since $2kÂ²\in \mathbb{Z}$

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

In your attempt, you have only proven the forward direction, which is "if n is even then n^2 is also even."

But now you have to do the backward direction, "if n^2 is even, then n is also even."
If n^2 = 2k for some integer k, then can you explain why n = 2p for some other integer p?

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

Thanks for the help.

Ok, so here is my new attempt.
What I'm not sure about is if writing $p = \sqrt{kÂ²}$ is enough to justify.

" $\Rightarrow$ ":
If

is even
$\Rightarrow (\exists k \in \mathbb{Z})(n=2k)$ using Def. 1
$\Rightarrow (\exists k \in \mathbb{Z})(nÂ²=2(2kÂ²)$
$\Rightarrow nÂ²$ since $2kÂ²\in \mathbb{Z}$

" $\Leftarrow$ ":
If nÂ²

is even
$\Rightarrow (\exists k\in \mathbb{Z})(nÂ²=2k)$ using Def. 1
$\Rightarrow (\exists p\in \mathbb{Z})(n=2p)$ , because $p=\sqrt{kÂ²}$
$\Rightarrow n$ is even

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

I found another way of approaching this.

is even $\Rightarrow n^2$ is even:
If

is even
$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = 2(2k^2))$
$\Rightarrow n^2$ is even

(Using the contrapositive)

is not even $\Rightarrow n^2$ is not even:
If

is not even
$\Rightarrow (\nexists k \in \mathbb{Z})(n \neq 2k)$
$\Rightarrow (\forall k \in \mathbb{Z})(n \neq 2k)$
$\Rightarrow (\forall k \in \mathbb{Z})(n^2 \neq 2(2k^2))$
$\Rightarrow n^2$ is not even

$\therefore n^2$ is even $\Longleftrightarrow n$ is even.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

I am not really convinced by the reverse direction.
How does $(\forall k \in \mathbb{Z})(n \neq 2k)$ imply $(\forall k \in \mathbb{Z})(n^2 \neq 2(2k^2))$ ?

Instead, you might find it easier to prove: "if n is odd, then n^2 is odd."

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

Ok, your point is convincing. Will do it that way. Thanks.

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

is even $\Rightarrow n^2$ is even:
If

is even
$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = 2(2k^2))$
$\Rightarrow n^2$ is even

(Using the contrapositive)

is not even $\Rightarrow n^2$ is not even $\equiv n$ is odd $\Rightarrow n^2$ is odd.
If

is odd
$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd

$\therefore n^2$ is even $\Longleftrightarrow n$ is even.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

mathtinkerer wrote:

$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd.

How do you know that n^2 = (2k + 1)(2k + 1)

is an odd number?

Shadow · **Posted:** Tue, 5 Apr 2011 19:00:54 UTC

Matt wrote:

mathtinkerer wrote:

$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd.

How do you know that n^2 = (2k + 1)(2k + 1)

is an odd number?

He doesn't, he hasn't yet decided to let k'=2k^2+k

daveyinaz · **Joined:** Sat, 16 Aug 2008 04:47:19 UTC **Posts:** 208

Shadow wrote:

Matt wrote:

mathtinkerer wrote:

$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd.

How do you know that n^2 = (2k + 1)(2k + 1)

is an odd number?

He doesn't, he hasn't yet decided to let k'=2k^2+k

perhaps you mean k' = 2k^2 + 2k

?

Shadow · **Posted:** Tue, 5 Apr 2011 22:07:16 UTC

daveyinaz wrote:

Shadow wrote:

Matt wrote:

mathtinkerer wrote:

$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd.

How do you know that n^2 = (2k + 1)(2k + 1)

is an odd number?

He doesn't, he hasn't yet decided to let k'=2k^2+k

perhaps you mean k' = 2k^2 + 2k

?

Yes of course, didn't you read my mind to see what I was thinking?

mathtinkerer · **Joined:** Sun, 20 Mar 2011 05:26:38 UTC **Posts:** 25

daveyinaz wrote:

Shadow wrote:

Matt wrote:

mathtinkerer wrote:

$\Rightarrow (\exists k \in \mathbb{Z})(n = 2k+1)$
$\Rightarrow (\exists k \in \mathbb{Z})(n^2 = (2k+1)(2k +1))$
$\Rightarrow n^2$ is odd.

How do you know that n^2 = (2k + 1)(2k + 1)

is an odd number?

He doesn't, he hasn't yet decided to let k'=2k^2+k

perhaps you mean k' = 2k^2 + 2k

?

Yes, I still need to do that. Thanks everybody.

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