# The Very Last, for 2014, proof of the Irrationality of sqrt(2)

### Amrik Singh Nimbran

*American Mathematical Monthly*, Vol. 121, No. 10, 964

$\sqrt{2}$ is irrational.

### Proof

Clearly, $1^{2} \lt 2 \lt 2^{2}.$ Since $2$ lies between two consecutive integral squares, $\sqrt{2}$ cannot be an integer. Suppose $\sqrt{2}$ is rational, i.e., $\sqrt{2} =\displaystyle\frac{m}{n},$ $1 \lt n \lt m \lt 2n;$ $m, n\in\mathbb{N}.$ Since $m \gt n,$ we may write: $m = n + r,$ $n \gt r,$ $r\in\mathbb{N}.$ Hence, $\sqrt{2} = \displaystyle\frac{n+r}{n}.$ Let $g = \gcd(n, r).$ So, $n = gs$ and $r = gt,$ $s \gt t;$ $\gcd(s, t) =1;$ $g, s, t\in\mathbb{N}.$ Then, $\sqrt{2} = \displaystyle\frac{s+t}{s}.$ In other words, $2s^{2} = s^{2} + 2st + t^{2}.$ So,

(1)

$s^2 = t (2s + t).$

Since $t$ divides the right-hand side of (1) and $\gcd(s, t) = 1,$ so $t = 1$ if $t$ is to divide the left-hand side of (1). Hence, $s^2 = 2s + 1,$ i.e., $s(s-2)=1,$ implying $s|1,$ i.e., $s=1.$ This contradicts $s\gt t\ge 1.$ Hence, our supposition was wrong and so $\sqrt{2}$ is irrational.

The above argument admits a generalization for $a$ which are not a complete square.

Let $a$ be such an integer and $\lfloor\sqrt a\rfloor=b,$ where $\lfloor z\rfloor$ denotes the floor function. Then $b^{2} \lt (\sqrt{a})^{2} \lt (b+1)^{2}.$ Thus, $a$ lies between two consecutive integral squares; hence, $\sqrt{a}$ is not an integer. Suppose $\sqrt{a}$ is rational, i.e., $\sqrt{a} =\displaystyle\frac{m}{n},$ $1 \lt n \lt m \lt (b+1)n;$ $m, n\in\mathbb{N}.$ Using the division algorithm, we may write: $m = bn + r,$ $n \gt r,$ $r\in\mathbb{N}.$ Hence, $\sqrt{a} = \displaystyle\frac{bn+r}{n}.$ Let $g = \gcd(n, r).$ So, $n = gs$ and $r = gt,$ $s \gt t;$ $\gcd(s, t) =1;$ $g, s, t\in\mathbb{N}.$ Then, $\sqrt{a} = \displaystyle\frac{bs+t}{s}.$ In other words,

(2)

$(a-b^{2})s^{2} = t(2bs+t).$

Since $t$ divides the right-hand side of (2) and $\gcd(s, t) = 1,$ either $(a-b^{2})=tu,$ $u\in\mathbb{N},$ or $t=1.$ In the first case, we would have $s(us-2b)=t,$ which would imply $s|t,$ in contradiction to $s\gt t.$ In the second case $(t=1)$ we obtain

$s((a - b^{2})s - 2b) = 1$

which implies $s|1,$ i.e., $s=1,$ in contradiction with $s\gt t\ge 1.$ Further, it leads to $a=(b+1)^2,$ contradicting our supposition $a\lt (b+1)^2.$ Hence, $\sqrt{a}$ is irrational.

|Contact| |Front page| |Contents| |Algebra| |Up|

Copyright © 1996-2018 Alexander Bogomolny64660265 |