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Is this proof correct?

Suppose that $\sqrt{2}=\frac{a}{b}$, where $a,b \in \mathbb{N}$ and $a$ is as small as possible. Then $\sqrt{2}b=a$ which means $2b=\sqrt{2} a$. So we rewrite $\sqrt{2}=\frac{a}{b}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}a-a}{\sqrt{2}b-b}=\frac{2b-a}{a-b}.\,$ Note $\,2b-a=a(\sqrt{2}-1)<a$. So this fraction has a smaller numerator than the one we had. So this is a contradiction.

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    $\begingroup$ Yes, this is a well-known proof. $\endgroup$ – Lord Shark the Unknown Dec 8 '18 at 13:15
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    $\begingroup$ @user42493 This is essentially the same as this section with $a=m$, $b=n$, $k=2$, $q=1$. At least you were creative enough to 'discover' another proof. $\endgroup$ – Toby Mak Dec 8 '18 at 13:18
  • $\begingroup$ You have a typographical error. ab-a should read $2b-a$. $\endgroup$ – Ben W Dec 8 '18 at 13:44
  • $\begingroup$ Note that $2b-a<a$ is equivalent to $2b<2a$, which is true because $a>b$. $\endgroup$ – egreg Dec 8 '18 at 13:57
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It is a correct well-known proof. Essentially it uses denominator descent by the division algorithm, though that is obfuscated . Below I clarify this viewpoint for the generalization below. The proof in the question is exactly the special case $\, k = 2\,$ and $\,q = {\rm floor}(\sqrt 2) = 1\,$ of the proof below.

Irrationality of $\sqrt k\,$ if it is not an integer (excerpted from Wikipedia, slightly edited)

For an integer $k>0$, suppose $\sqrt k$ is not an integer, but is rational and can be expressed as $\frac{a}b$ for natural numbers $a$ and $b$, and let $q$ be the largest integer no greater than $\sqrt k.\,$ Then

\begin{aligned}{\sqrt {k}}&={\frac {a}{b}}\\[8pt]&={\frac {a({\sqrt {k}}-q)}{b({\sqrt {k}}-q)}}\\[8pt]&={\frac {a{\sqrt {k}}-aq}{b{\sqrt {k}}-bq}}\\[8pt]&={\frac {(b{\sqrt {k}}){\sqrt {k}}-aq}{b({\frac {a}{b}})-bq}}\\[8pt]&={\frac {bk-aq}{a-bq}}\end{aligned}

The numerator and denominator were each multiplied by $(\sqrt k − q)\,$ — which is positive but less than $1$ and then simplified independently. So the two resulting products, say $a'$ and $b'$, are themselves integers, which are less than $a$ and $b$ respectively. Therefore, no matter what natural numbers $a$ and $b$ are used to express $\sqrt k$, there exist smaller natural numbers $a' < a$ and $b' < b$ that have the same ratio. But infinite descent on the natural numbers is impossible, so this disproves the original assumption that $\sqrt k$ could be expressed as a ratio of natural numbers.

We can rewrite the above proof more conceptually as below, where "$\,n\,$ is a denom of $\,r$" means that the rational $\,r\,$ can be written with denominator $\,n,\,$ i.e. $\,n\,r = j\,$ for some integer $\,j.$

$\begin{align} [\![1]\!]\qquad\qquad\, b \sqrt k\, &=\, a\qquad\ \, \Rightarrow\,\qquad\ \ \ \text{$\,b\,$ is a denom of }\ \sqrt k\\ \sqrt k\,\cdot\, [\![1]\!]\ \ \, \Rightarrow\,[\![2]\!]\qquad\qquad a \sqrt k\, &=\, bk\qquad \Rightarrow\qquad\,\ \ \ \text{ $a\,$ is a denom of }\ \sqrt k\\ [\![2]\!] - [\![1]\!]q\,\Rightarrow\,[\![3]\!]\ \ \ \ \ \, (\color{#c00}{a\!-\!bq})\sqrt k\, &=\, bk\!-\!aq\,\Rightarrow\, \color{#c00}{a\bmod b} \, \text{ is a denom of }\ \sqrt k\\ \end{align}$

If $\,b\,$ doesn't divide $\,a\,$ we get a smaller denom $\, 0 < \color{#c00}{a \bmod b} < b\,$ so infinite descent (on denoms), contra $\Bbb N\,$ is well-ordered. Hence $\,b\,$ divides $\,a,\,$ so $\,\sqrt k = a/b = n\in \Bbb Z,\,$ so $\,k = n^2$.

Alternatively we can initially assume that $\,b\,$ is the least denominator then deduce a contradiction that a smaller denominator exists if $\,b\,$ doesn't divide $\,a.$

This method generalizes to show the $\,\Bbb Z\,$ (or any PID) is integrally-closed, i.e. no proper fraction is a root of a polynomial that is monic (lead coef $= 1),\,$ i.e. the monic case of the Rational Root Test. You can find much further discussion of this and related ideas in my posts on denominator ideals.

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