Most, but perhaps not all, of us know that the square root of 2 can never be written exactly as a fraction. In other words,

How can we show this? Here are four proofs.


The usual way is as follows.


Suppose we can draw a square that has integral sides and an integral diagonal. Using Pythagoras' theorem we can see that this is equivalent to saying that $\sqrt{2}$ is rational.

Consider the following diagram:

By Pythagoras we know that if the sides are $a$ and the diagonal is $b$ then $2a^2=b^2.$ This diagram shows that given such a pair, $a,b,$ we can find a smaller pair. (Exercise for the reader: show that $a$ and $b$ in the smaller pair are still integers.)


Eventually we run out of numbers and thus complete our proof by contradiction

Algebra, again

Exercises for the diligent reader:
  • Show that $(2a-b)^2=2(b-a)^2$
  • Show that b-a < a
  • what's the relationship between this proof and the previous geometrical one?
Suppose $b^2=2a^2$ where $a,b$ are integers and as small as possible. Then a little algebra shows that $(2a-b)^2=2(b-a)^2$ so that $(b-a,2a-b)$ is a new pair of integers with the same property. But $b-a\lt~a$ and so we've got a smaller pair of integers whose ratio is $\sqrt{2},$ contrary to our original decision to take the smallest pair.


Continued Fractions

Another exercise for the diligent reader:

Show that for every number of the form $a/b,$ its continued fraction will terminate.

Every rational number has a finite continued fraction. We show now that the continued fraction for $1+\sqrt{2}$ does not terminate. That means $1+\sqrt{2}$ is irrational, and hence $\sqrt{2}$ is irrational.

To compute the continued fraction we separate the target number into its integer part and the fractional part

$\epsilon$ is between 0 and 1 as required, so we can take its reciprocal, giving $\frac{1}{\sqrt{2}-1}$ which after rationalising the denominator is $1+\sqrt{2}.$ That's what we started with, so the continued fraction for $1+\sqrt{2}$ is:

Thus the continued fraction for $1+\sqrt{2}$ is [2;2,2,2,2,...] which does not terminate. Hence $1+\sqrt{2}$ is irrational, and so $\sqrt{2}$ is irrational.

Last change to this page
Full Page history
Links to this page
Edit this page
  (with sufficient authority)
Change password
Recent changes
All pages