prove using contraposition that if a and b are integers:
$a^2-4b-2\neq0$
So the contrapositive:
$a^2-4b-2=0$ implies that at least one of either a or b are not integers
Working from the above, I got:
$a^2=4b+2$
$a^2=2(2b+1)$
If both a and b are integers, $2(2b+1)$ is a perfect square.
In other words:
$\sqrt{2(2b+1)}=k$
Where k is an integer
I also realised that 2b+1 is an odd number so:
$k=\sqrt{2}\sqrt{2b+1}$
I ended my proof by saying that since 2b+1 is odd, it does not have a multiple of 2 so k will always have multiple or $\sqrt{2}$ which means that it cannot be an integer.
$\endgroup$ 22 Answers
$\begingroup$You're almost done. You can conclude your proof as follows:
$a^2=2(2b+1)$ immediately implies, you have $a=2k, \thinspace k\in\mathbb Z$. Therefore, we have
$$4k^2=2(2b+1)\iff 2k^2=2b+1$$
which gives a contradiction.
$\endgroup$ 2 $\begingroup$If $a$ was an integer, then it would be odd or even. If it was odd, then $a^2$ would be odd too, and then it cannot be of the form $2(2b+1)$ for some integer $b$. And if $a$ is even, then $a^2$ is multiple of $4$, and therefore it also cannot be of the form $2(2b+1)$ for some integer $b$.
I don't know what is it that you mean when you write that “$k$ will always have an irrational multiple, namely $\sqrt2$”.
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