racodelitusmn

Answered

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: ${a}^{2}=4b+2$

${a}^{2}=2(2b+1)$

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.

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: ${a}^{2}=4b+2$

${a}^{2}=2(2b+1)$

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.

Answer & Explanation

pompatzz8

Expert

2022-07-13Added 11 answers

Explanation:

You're almost done. You can conclude your proof as follows:

${a}^{2}=2(2b+1)$ immediately implies, you have $a=2k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{Z}$. Therefore, we have $4{k}^{2}=2(2b+1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2{k}^{2}=2b+1$ which gives a contradiction.

You're almost done. You can conclude your proof as follows:

${a}^{2}=2(2b+1)$ immediately implies, you have $a=2k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{Z}$. Therefore, we have $4{k}^{2}=2(2b+1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}2{k}^{2}=2b+1$ which gives a contradiction.

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