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What is a Proof by Contradiction, and how to prove by contradiction that $6 - \sqrt{2}$ is an irrational number?

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    $\begingroup$ Suppose that $6-\sqrt 2$ is rational. Then... $\endgroup$ – abiessu Mar 23 '15 at 17:56
  • $\begingroup$ then what should I do... $\endgroup$ – Hussain Halai Mar 23 '15 at 18:00
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    $\begingroup$ $HussainHalai What have you tried? This is not that difficult of a problem. Edit in how far you've gotten and we'll see if we can help get you past wherever you're stuck. $\endgroup$ – user137731 Mar 23 '15 at 18:01
  • $\begingroup$ Hint: if $r$ is rational, then $6-r$ is rational. $\endgroup$ – shalop Mar 23 '15 at 18:01
  • $\begingroup$ What if I had 6 - 2 ^(1/7) would I use the same method... $\endgroup$ – Hussain Halai Mar 23 '15 at 19:02
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A proof by contradiction is assuming something then building on it and finding that it leads to contradiction, concluding that the assumed statement is false.

Assume $x=6-\sqrt{2}=\frac{p}{q}$

$x^2=8-12\sqrt{2}=\frac{p^2}{q^2}$

$8q^2=p^2+12\sqrt{2}q^2$

$8q^2-p^2=12\sqrt{2}q^2$

$8q^2-p^2$ is rational, and $q^2$ is rational, thus $12\sqrt{2}$ is rational.

However, we know that that is not true, and thus, $6-\sqrt{2}$ is irrational.

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    $\begingroup$ Needlessly complicated! If $6-\sqrt 2 = \frac{p}{q}$, then $\sqrt 2 = 6 - \frac{p}{q}$ would be rational. $\endgroup$ – TonyK Mar 23 '15 at 18:17
  • $\begingroup$ @TonyK Now I feel retarded... I have never done such an exercise, so yeah, I made useless steps. Thanks though. :) $\endgroup$ – Hasan Saad Mar 24 '15 at 14:36
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If $6-\sqrt{2}$ is rational, then $\sqrt{2}$ is rational. Then just prove it for $\sqrt{2}$ which is easy.

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The method of proof by contradiction is done by assuming the negation of a given proposition is true, then finding a contradiction with that assumption, hence proving the original statement. For instance, suppose $6-\sqrt{2}$ is rational, then we can say that for $a,b\in \mathbb{Z}$ we have $$6-\sqrt{2}=\frac{a}{b}$$ where $a$ and $b$ are coprime. Now, we can do some algebra and find a contradiction.

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