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
The first sentence is 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 , that is to say :
.
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 ?
Hello, sjenkins!
We must carefully write the statement as an implication.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."
We want: .
In other words: If the square of a number is odd, then the number is odd.
Too fast for me, Moo!
.
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
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
Your proof is okay, except for one little thing:
This does not work. Suppose . We can write , but . In other words, when taking the square root of a square, you have to be aware that . Of course, this is easy to rectify.
However, I think a direct proof of the original statement would be a lot simpler: Let be an even integer, i.e., , and then show, through a chain of implications, that must also be even.