## Most recent change of RootTwoIsIrrational

Edit made on August 17, 2011 by ColinWright at 18:45:45

Deleted text in red / Inserted text in green

WW WM
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,
* EQN:\sqrt{2} is never a ratio of whole numbers,
** EQN:\sqrt{2} is not rational, a rational number,
*** EQN:\sqrt{2} is *irrational.*

How can we show this? Here are four proofs.

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!! Algebra

The usual way is as follows.
* Suppose EQN:\sqrt{2} can be written as a fraction:
** EQN:\sqrt{2}=a/b
* Then EQN:2=a^2/b^2
* Then EQN:2b^2=a^2
* Then EQN:a^2 is even
** Hence EQN:a is even
*** (This takes a small proof of its own.)
* So we can write /a=2k/
* Hence EQN:2b^2=(2k)^2=4k^2
* So EQN:b^2=2k^2
* So now EQN:b^2 is even
** Hence EQN:b is even
* So now we can reduce EQN:\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 EQN:\sqrt{2} as EQN:\frac{a}{b}

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!! Geometry
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 EQN:\sqrt{2}
is rational.

Consider the following diagram:
|>> COLUMN_START [[[ IMG:Sqrt2NotInQ.png ]]] COLUMN_END <<|
By Pythagoras we know that if the sides are EQN:a and the
diagonal is EQN:b then EQN:2a^2=b^2. This diagram shows that
given such a pair, EQN:a,b, we can find a smaller pair.
(Exercise for the reader: show that EQN:a and EQN:b in
the smaller pair are still integers.)

Repeat.

Eventually we run out of numbers and thus complete our

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!! Algebra, again

[[[>50
* Show that EQN:(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 EQN:b^2=2a^2 where EQN:a,b are integers and as small
as possible. Then a little algebra shows that EQN:(2a-b)^2=2(b-a)^2
so that EQN:(b-a,2a-b) is a new pair of integers with the same
property. But EQN:b-a\lt~a (exercise for the reader: why?) and so
we've got a smaller pair of integers whose ratio is EQN:\sqrt{2},
contrary to our original decision to take the /smallest/ pair.

(Exercise: what's the relationship between this proof and the
previous geometrical one?)

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!! Continued Fractions
[[[>50
Another exercise for the diligent reader:

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

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

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

* EQN:T=1+\sqrt{2}=2+\epsilon where EQN:\epsilon=\sqrt{2}-1

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

* EQN:\LARGE{1+\sqrt{2}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}

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

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CategoryMaths