HEADERS_END

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

*** 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

proof by contradiction

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!! Algebra, again

[[[>50

Exercises for the diligent reader:

* 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

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