Ytukyg

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 $sqrt{2}=frac{a}{b}cdotfrac{sqrt{2}-1}{sqrt{2}-1}=frac{sqrt{2}a-a}{sqrt{2}b-b}=frac{2b-a}{a-b}.,$ 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.

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 $,j.$

$begin{align}
[![1]!]qquadqquad, b sqrt k, &=, aqquad , Rightarrow,qquad text{$,b,$ is a denom of } sqrt k\
sqrt k,cdot, [![1]!] , Rightarrow,[![2]!]qquadqquad a sqrt k, &=, bkqquad Rightarrowqquad, text{ $a,$ is a denom of } sqrt k\
[![2]!] - [![1]!]q,Rightarrow,[![3]!] , (color{#c00}{a!-!bq})sqrt k, &=, bk!-!aq,Rightarrow, color{#c00}{abmod 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 $,sqrt k = a/b = nin Bbb Z,,$ 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.