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,
 $\sqrt{2}$ is never a ratio of whole numbers,
How can we show this? Here are four
proofs.
The usual way is as follows.
 Suppose $\sqrt{2}$ can be written as a fraction:
 Then $2=a^2/b^2$
 Then $2b^2=a^2$
 Then $a^2$ is even
 Hence $a$ is even
 (This takes a small proof of its own.)
 So we can write a=2k
 Hence $2b^2=(2k)^2=4k^2$
 So $b^2=2k^2$
 So now $b^2$ is even
 So now we can reduce $\frac{a}{b}$ to a smaller fraction.
 Lather, rinse, repeat, as they say.
 Following this process we can get smaller and smaller fractions, each of which is exactly the square root of two. Clearly this is nonsense.
 The argument is sound, so the premise must be false.
 Thus we can never write $\sqrt{2}$ as $\frac{a}{b}$
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.)
Repeat.
Eventually we run out of numbers and thus complete our
proof by contradiction
Exercises for the diligent reader:
 Show that $(2ab)^2=2(ba)^2$
 Show that ba < 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 $(2ab)^2=2(ba)^2$
so that $(ba,2ab)$ is a new pair of
integers with the same
property. But $ba\lt~a$ 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.
Contradiction.
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
 $T=1+\sqrt{2}=2+\epsilon$ where $\epsilon=\sqrt{2}1$
$\epsilon$ 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:
 $\LARGE{1+\sqrt{2}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$
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.
CategoryMaths