I know that if $n^2$ is even then $n$ is even, but take $n^2 = 6$, which is even. However, $n = \sqrt 6$ is not even? So what is going on?
Is it because $P \implies Q$ is false when $P$ is true and $Q$ is false?
Asked
Active
Viewed 82 times
-2
-
10"even" and "odd" are properties of integers, not arbitrary real numbers. The statement you wuote is about integers. That is the particular inferno at issue here. – Arturo Magidin Oct 10 '22 at 02:02
-
We can in fact extend the notion of parity to $\Bbb Z[\sqrt 6],,$ where $\sqrt 6$ is even. – Bill Dubuque Oct 10 '22 at 02:18
-
5In other words: first choose an integer n (this is the part you're missing). Now look at whether $n^2$ is even. If it is, then $n$ is even. – Théophile Oct 10 '22 at 02:23
-
I agree Arturo. – Peter Oct 10 '22 at 10:55
1 Answers
2
The theorem is that if $n$ is an INTEGER such that $n^2$ is even, then $n$ itself is even. Of course, if you just give me any even number, it's square root doesn't have to be even, because it's square root doesn't have to be an integer in the first place, as your example shows. However, if the square root is an integer, then it will have to be even.
Btw, you can replace every instance of the word even with the word odd in what I just said, and it will still be true verbatim.
Carl Chaanin
- 498