# Irrationality of $\sqrt{2}$ via Bezout's Lemma

$\sqrt{2}$ is irrational.

Assuming Pythagoras understood Euclid's algorithm, the following proofs show how he could have demonstrated that any integer root of an integer is an irrational or an integer, and even that the cube root of an integer either is not the root of a quadratic (i.e., not in the form $(a + b\sqrt{N}) / c$ or is an integer.

In the following, all variables but $r$ are restricted to integer values.

The (shorter) proof of Sagher's special case comes first, to motivate the others. If $r = \sqrt{k} = m/n$ (in lowest terms), Euclid's algorithm (this is also known as Bézout's Lemma) gives $\alpha$ and $\beta$ for which $\alpha m + \beta n = 1.$ Then $rn = m,$ $rm = r^{2}n = kn,$ and

$r = r(\alpha m + \beta n) = \alpha rm + \beta rn = \alpha kn + \beta m,$

an integer.

Now take $r = m/n = k^{1/3},$ $\alpha m + \beta n = 1.$ Then

$m = rn,$ $rn^{2} = mn,$ $rmn = m^{2},$ $rm^{2} = r^{3}n^{2} = kn^{2}.$

$r = r(\alpha m + \beta n)^{2} = \alpha ^{2}rm^{2} + 2\alpha \beta rmn + \beta ^{2}rn^{2} = \alpha ^{2}kn^{2} + 2\alpha \beta m^{2} + \beta ^{2}mn,$

an integer.

Now make the weaker assumption that $r = k^{1/3},$ and that $r$ satisfies a proper quadratic equation $ar^{2} + br + c = 0,$ $a \ne 0.$ then

$0 = (ar^{2} + br + c)(ar - b) = a^{2}r^{3} + (ac - b^{2})r - bc.$

If $ac = b^{2},$ divide the equation by $a^{2}$ to find $r^{3} = (b/a)^{3}$ and $r = b/a.$ Otherwise, put $k$ for $r^{3}$ and find $r = (bc - a^{2}k)/(ac - b^{2}).$ Either way, $r$ is rational, and consequently an integer.

### References

- Robert W. Floyd,
*Am Math Monthly*, Vol. 96, No. 1 (Jan., 1989), p. 67

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

Copyright © 1996-2017 Alexander Bogomolny61242819 |