The square root of 2 is irrational

The square root of 2 is irrational is the claim in number theory that there is no rational number that when multiplied by itself equals the number 2.

One proof by reductio ad absurdum consists of first assuming that the square root of 2 can be written as a rational number. Thus, there is a pair of coprime integers [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] such that their ratio [math]\displaystyle{ \frac{p}q = \sqrt 2 }[/math] (Every rational number has an irreducible representation). And so, [math]\displaystyle{ \frac{p^2}{q^2} = 2 }[/math], and [math]\displaystyle{ p^2=2q^2 }[/math]. As [math]\displaystyle{ p^2 }[/math] has a factor of [math]\displaystyle{ 2 }[/math], it is even, and thus [math]\displaystyle{ p }[/math] is also even. Thus [math]\displaystyle{ p }[/math] can be written as [math]\displaystyle{ 2k }[/math], where [math]\displaystyle{ k }[/math] is an integer. It then follows that [math]\displaystyle{ (2k)^2=2q^2 }[/math], [math]\displaystyle{ 4k^2=2q^2 }[/math], [math]\displaystyle{ 2k^2=q^2 }[/math], and thus [math]\displaystyle{ q^2 }[/math] also has a factor of [math]\displaystyle{ 2 }[/math], and thus [math]\displaystyle{ q }[/math] is also even. If [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are both even, then they share a common factor of [math]\displaystyle{ 2 }[/math], and thus are not coprime, leading to a contradiction as we have already established that [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are coprime. And so, no such [math]\displaystyle{ p,q }[/math] can exist, and the square root of 2 is irrational.

It's not part of the definition, but the proof is easy.

Suppose one has a rational number <nobr aria-hidden="true">mn</nobr>[math]\displaystyle{ \lt mfrac\gt \lt mi\gt m\lt /mi\gt \lt mi\gt n\lt /mi\gt \lt /mfrac\gt }[/math]<script type="math/tex" id="MathJax-Element-6">\frac{m}{n}</script> that is reducible. We can assume <nobr aria-hidden="true">n≥1</nobr>[math]\displaystyle{ \lt mi\gt n\lt /mi\gt \lt mo\gt ≥\lt /mo\gt \lt mn\gt 1\lt /mn\gt }[/math]<script type="math/tex" id="MathJax-Element-7">n \ge 1</script> (divide the numerator and denominator by <nobr aria-hidden="true">−1</nobr>[math]\displaystyle{ \lt mo\gt −\lt /mo\gt \lt mn\gt 1\lt /mn\gt }[/math]<script type="math/tex" id="MathJax-Element-8">-1</script> otherwise).

Take an integer <nobr aria-hidden="true">k≥2</nobr>[math]\displaystyle{ \lt mi\gt k\lt /mi\gt \lt mo\gt ≥\lt /mo\gt \lt mn\gt 2\lt /mn\gt }[/math]<script type="math/tex" id="MathJax-Element-9">k \ge 2</script> which divides both of <nobr aria-hidden="true">m</nobr>[math]\displaystyle{ \lt mi\gt m\lt /mi\gt }[/math]<script type="math/tex" id="MathJax-Element-10">m</script> and <nobr aria-hidden="true">n</nobr>[math]\displaystyle{ \lt mi\gt n\lt /mi\gt }[/math]<script type="math/tex" id="MathJax-Element-11">n</script>, and so <nobr aria-hidden="true">m′=m/k</nobr>[math]\displaystyle{ \lt msup\gt \lt mi\gt m\lt /mi\gt \lt mo\gt ′\lt /mo\gt \lt /msup\gt \lt mo\gt =\lt /mo\gt \lt mi\gt m\lt /mi\gt \lt mrow class="MJX-TeXAtom-ORD"\gt \lt mo\gt /\lt /mo\gt \lt /mrow\gt \lt mi\gt k\lt /mi\gt }[/math]<script type="math/tex" id="MathJax-Element-12">m' = m/k</script> and <nobr aria-hidden="true">n′=n/k</nobr>[math]\displaystyle{ \lt msup\gt \lt mi\gt n\lt /mi\gt \lt mo\gt ′\lt /mo\gt \lt /msup\gt \lt mo\gt =\lt /mo\gt \lt mi\gt n\lt /mi\gt \lt mrow class="MJX-TeXAtom-ORD"\gt \lt mo\gt /\lt /mo\gt \lt /mrow\gt \lt mi\gt k\lt /mi\gt }[/math]<script type="math/tex" id="MathJax-Element-13">n'=n/k</script> are smaller integers satisfying <nobr aria-hidden="true">m′n′=mn</nobr>[math]\displaystyle{ \lt mfrac\gt \lt msup\gt \lt mi\gt m\lt /mi\gt \lt mo\gt ′\lt /mo\gt \lt /msup\gt \lt msup\gt \lt mi\gt n\lt /mi\gt \lt mo\gt ′\lt /mo\gt \lt /msup\gt \lt /mfrac\gt \lt mo\gt =\lt /mo\gt \lt mfrac\gt \lt mi\gt m\lt /mi\gt \lt mi\gt n\lt /mi\gt \lt /mfrac\gt }[/math]<script type="math/tex" id="MathJax-Element-14">\frac{m'}{n'} = \frac{m}{n}</script>.

Repeat this process as long as the fraction remains reducible.

One cannot repeat forever, because the sequence of denominators is a sequence of positive integers that gets smaller and smaller, and no such sequence can continue infinitely (essentially one is using mathematical induction in this step).

When the process stops, one has an irreducible fraction representing the same rational number

Claim

Statement of the claim	The square root of 2 is irrational
Level of certainty	Proven
Nature	Theoretical
Counterclaim
Dependent on	Every rational number has an irreducible representation
Dependency of