Contrapositive

**sjenkins** · July 1st 2008, 10:38 AM

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.

**Moo** · July 1st 2008, 10:50 AM

Originally Posted by sjenkins

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.

mmmm

Let $P ~:~ \text{n is an even number}$
$Q ~:~ \text{its square is even}$

The first sentence is $P \implies Q$ 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 $\bar{Q} \implies \bar{P}$ , that is to say :
$\text{the square of a number is not even} \implies \text{the number is not even}$ .

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 ?

**sjenkins** · July 1st 2008, 11:00 AM

Yes, thanks so much Moo! I appreciate your constant willingness to help me!!!

**Soroban** · July 1st 2008, 12:16 PM

Hello, sjenkins!

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

We must carefully write the statement as an implication.

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

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

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

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

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

Too fast for me, Moo!
.

**sjenkins** · July 1st 2008, 12:21 PM

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

**Moo** · July 1st 2008, 12:25 PM

Originally Posted by sjenkins

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

$(a+b)^2=a^2+2ab+b^2 \neq a^2+b^2$ !!!

By the way, what you proved here is :

$\text{x is not even} \implies \text{the square of x is not even}$ , but this is not the contraposive, nor the statement.

**sjenkins** · July 1st 2008, 12:33 PM

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.

**algebraic topology** · July 1st 2008, 02:42 PM

Originally Posted by Moo

The contraposive is $\bar{Q} \implies \bar{P}$

That is not the contrapositive. The contrapositive of $P\implies Q$ is $\neg\,Q\implies\neg\,P$ .

$Q\implies P$ is the converse of $P\implies Q$ .

**Plato** · July 1st 2008, 02:50 PM

Originally Posted by algebraic topology

That is not the contrapositive. The contrapositive of $P\implies Q$ is $\neg\,Q\implies\neg\,P$ .

$Q\implies P$ is the converse of $P\implies Q$ .

Did you not see that Moo used $\overline Q$ for $\neg Q$ ?

**algebraic topology** · July 1st 2008, 03:17 PM

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.)

**sjenkins** · July 1st 2008, 04:11 PM

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

**Reckoner** · July 1st 2008, 04:34 PM

Your proof is okay, except for one little thing:

Originally Posted by sjenkins

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

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

However, I think a direct proof of the original statement would be a lot simpler: Let $n$ be an even integer, i.e., $\exists p\in\mathbb{Z},\;n = 2p$ , and then show, through a chain of implications, that $n^2$ must also be even.

**sjenkins** · July 1st 2008, 04:37 PM

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

**sjenkins** · July 1st 2008, 04:39 PM

Can I ask what you would write to fix this problem...I'm confused!

**Reckoner** · July 1st 2008, 04:44 PM

Originally Posted by sjenkins

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

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

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