racodelitusmn

2022-07-12

Proof verification for proof by contraposition
prove using contraposition that if a and b are integers:
${a}^{2}-4b-2\ne 0$
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: ${a}^{2}=4b+2$
${a}^{2}=2\left(2b+1\right)$
If both a and b are integers, $2\left(2b+1\right)$ is a perfect square.
In other words: $\sqrt{2\left(2b+1\right)}=k$
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.

pompatzz8

Expert

Explanation:
You're almost done. You can conclude your proof as follows:
${a}^{2}=2\left(2b+1\right)$ immediately implies, you have $a=2k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{Z}$. Therefore, we have $4{k}^{2}=2\left(2b+1\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}2{k}^{2}=2b+1$ which gives a contradiction.

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