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?
Can someone explain what this paragraph is saying more clearly?
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discrete-mathematics
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0Badly written, too: "is $\in\mathbb Z$". – 2012-12-17
2 Answers
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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.
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0Suppose$2$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 $x = y = a$. – 2012-12-16
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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.
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0@amWhy: Thanks for pointing out the typo. Fixed. – 2012-12-17