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 $(2a-b)^2=2(b-a)^2$
- 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.
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
