Can someone please tell me what the contrapositive to this statement is: the square of an even number is an even number.

I thought it was if a number is not even, then its square is not even.

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- Jul 1st 2008, 10:38 AMsjenkinsContrapositive
Can someone please tell me what the contrapositive to this statement is: the square of an even number is an even number.

I thought it was if a number is not even, then its square is not even. - Jul 1st 2008, 10:50 AMMoo
mmmm

Let $\displaystyle P ~:~ \text{n is an even number}$

$\displaystyle Q ~:~ \text{its square is even}$

The first sentence is $\displaystyle P \implies Q$ with some different syntax, but though the same meaning. I'll try to explain it... Grammatically, the "square of" comes__after__the "even number", in the sense that you have in a first time "an even number" and then, you get its "square". This is why you consider the square part as a consequence...

The contraposive is $\displaystyle \bar{Q} \implies \bar{P}$, that is to say :

$\displaystyle \text{the square of a number is not even} \implies \text{the number is not even}$.

In other words : "If the square of a number is not even, then this number is not even".

Is it clear ? Does it look correct to you ? (Worried) - Jul 1st 2008, 11:00 AMsjenkins
Yes, thanks so much Moo! I appreciate your constant willingness to help me!!!

- Jul 1st 2008, 12:16 PMSoroban
Hello, sjenkins!

Quote:

Can someone please tell me what the contrapositive to this statement is:

The square of an even number is an even number.

"The square of an even number is an even number."

$\displaystyle \text{This says: }\;\text{If }\underbrace{x\text{ is even,}}_p\text{ then }\underbrace{x^2\text{ is even.}}_q$

We want: .$\displaystyle \sim\: q \to \sim p$

$\displaystyle \text{This is: \;If }\underbrace{x^2\text{ is not even,}}_{\sim q}\text{ then } \underbrace{x\text{ is not even.}}_{\sim p}$

In other words: If the square of a number is odd, then the number is odd.

Too fast for me, Moo!

. - Jul 1st 2008, 12:21 PMsjenkinsMore contrapositive help
I just don't think I was thinking it through. Now, can anyone tell me if the following is a proof of this statement???

Assume that x is not an even number

Then x is not a multiple of 2

In other words, x=2k+1 for some integer k

So x^2=(2k+1)^2=4k^2+1=2(2k^2)+1=2n where n=2k^2+1

We check that n is also an integer which it is since k is an integer and the integers are closed under multiplication

So, x is not a multiple of 2

Therefore, x^2 is not even - Jul 1st 2008, 12:25 PMMoo
- Jul 1st 2008, 12:33 PMsjenkins
Why do I have such a difficult time understanding all of this. I am really trying here and nothing I do seems to work out.

- Jul 1st 2008, 02:42 PMalgebraic topology
- Jul 1st 2008, 02:50 PMPlato
- Jul 1st 2008, 03:17 PMalgebraic topology
I beg Moo’s pardon, the bar didn’t show up clearly on my browser (and it’s also the first time I’ve ever seen this notation being used.)

- Jul 1st 2008, 04:11 PMsjenkinsTrying this proof again...
Here's try number 2, does this work to prove if the square of a number is not even, then the number is not even.

Assume that x^2 is not an even number

Then x^2 is not a multiple of 2

In other words, x^2=(2k+1)^2 for some integer k

It follows that x=2k+1

Therefore x is not even - Jul 1st 2008, 04:34 PMReckoner
Your proof is okay, except for one little thing:

This does not work. Suppose $\displaystyle x = -1$. We can write $\displaystyle x^2 = (-1)^2 = (1)^2 = (2\cdot0 + 1)^2 = (2k + 1)^2,\;k=0\in\mathbb{Z}$, but $\displaystyle x = -1\neq 2k + 1 = 1$. In other words, when taking the square root of a square, you have to be aware that $\displaystyle \sqrt{x^2} = \lvert x\rvert$. Of course, this is easy to rectify.

However, I think a direct proof of the original statement would be a lot simpler: Let $\displaystyle n$ be an even integer, i.e., $\displaystyle \exists p\in\mathbb{Z},\;n = 2p$, and then show, through a chain of implications, that $\displaystyle n^2$ must also be even. - Jul 1st 2008, 04:37 PMsjenkins
I agree that a direct proof of the original statement may have been easier, unfortunately the book said I had to create an indirect proof (Doh)

- Jul 1st 2008, 04:39 PMsjenkins
Can I ask what you would write to fix this problem...I'm confused!

- Jul 1st 2008, 04:44 PMReckoner
Okay, then. Your proof is almost there; just fix the flaw I pointed out. This may help you, however: if $\displaystyle n^2$ is not even, then 2 does not divide it. But if $\displaystyle n^2 = n\cdot n$ is not divisible by 2, neither is any of its factors, so 2 does not divide $\displaystyle n$.