# Can someone explain what this paragraph is saying more clearly?

In the last bullet, it says l must be even and provides an explanation. I don't understand the explanation, however. Why does it have to be even?

• Mhh it is essentially saying that $\sqrt{l}$ is integer, which does not need to be...so I think there is a flaw in the argument. Dec 16 '12 at 20:47
• Strange explanation! Suppose to the contrary that $a$ is odd, say $a=2t+1$. Then $a^2=2(2t^2+2t)+1$, making $a^2$ odd, false. So $a$ is even. Dec 16 '12 at 20:48
• which textbook/problems book is this from? Dec 16 '12 at 21:01
• Badly written, too: "is $\in\mathbb Z$". Dec 17 '12 at 7:09

I don't really follow the argument given. Here's a correct argument...

If $a^2 = 2l$, then $a^2$ is divisible by 2. But 2 is prime, so $a$ must be divisible by 2 too - say $a = 2b$. Then $(2b)^2 = 2l$, i.e. $2b^2 = l$, and so $l$ is divisible by 2.

• Could you phrase that last sentence a little differently? What does the primality of 2 have to do with it? Dec 16 '12 at 20:54
• @DougSmith It follows from the fundamental theorem of arithmetic: en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic Dec 16 '12 at 21:05
• I don't see how that relates. I'm really close to understanding what the guy is saying, I just don't get his example. Or why. Dec 16 '12 at 21:13
• Suppose 2 divides $xy$, where $x$ and $y$ are integers. Then the prime factorisation of $xy$ contains a 2. Then either $x$ is divisible by 2, or $y$ is (or both). Because, if neither of them is divisible by 2, then neither of them has $2$ in their prime factorisation, so when you multiply them together, the prime factorisation still doesn't have a 2 in. This contradicts the fact that the prime factorisation of $xy$ does have a 2 in (because prime factorisation is unique). Now let $x = y = a$. Dec 16 '12 at 22:06

They mean that in order to have $a=\sqrt{2}\sqrt{l}$ an integer, we must have that $l$ is an even number, say $2r$. In this case $a=\sqrt{2}\sqrt{2r}=2\sqrt{r}$. If $l$ is odd, we can not spilt of the $\sqrt{2}$ from $\sqrt{l}$, thus the $\sqrt{2}$ in $a$ remains, and $a$ will be irrational.

• A.Schulz: I think you mean to write $a = \sqrt{2}\sqrt{2r} = \sqrt{4}\sqrt{r} = 2\sqrt{r}$ Dec 16 '12 at 22:06
• @amWhy: Thanks for pointing out the typo. Fixed. Dec 17 '12 at 7:04