# Some Irrational Numbers

## Who found the first irrational?

● The first irrational number, sqrt(2), is commonly credited to the Pythagoreans, and in particular to Hippasus. The Pythagoreans were strong believers that all numbers could be expressed by integers, or their ratios. It was also unpleasant to hold ideas that didn't fit into this mold. |

## Hippasus's Fate

Hippasus was so excited at his new finding, that he went off to enjoy his favorite activity, diving off a plank into shark infested waters wearing nothing but the finest cuts of meat. |

## Irrational Numbers

● What's an Irrational number?

● An Irrational number is not rational, and a

rational number is representable as a reduced

fraction .

●Rationals can also be said to be the set of numbers which
solve

● a linear polynomial with coefficients in Z .

## Some properties?

● Not representable as a terminating or repeating decimal .

● Do we have closure under addition or multiplication ?

● Are they bigger than the set of rational numbers? (Anyone know a 1-1 map from
irrationals to reals?)

## Closure?

● No sir, we've got no closure.

● Addition?

●Multiplication?

●Anti-Closed? Adding or multiplying by a rational will give an Irrational.

## How big are they?

● The rational numbers are big, but can be put into

a 1-1 correspondence with the integers, so we call

them countable.

● The real numbers are uncountable (Cantor's

diagonal argument), so the stuff left over after

we've taken out a countable set must be

uncountable, in this case that's the irrationals.

## Everyone's favorite unspeakable proof

● To get things started, let's show the irrationality

of a simple number. Assume sqrt(2) is rational

Now we take the prime factorization of each side. The power of 2 on the left hand side must be odd, and it is even on the right hand side. This is no good so perhaps we were a bit batty in assuming that sqrt(2) was rational, and it is irrational. |

## But Daddy! I wanted more irrationals!

● So we've gone and shown that one number is

irrational, and the proof works for the square root

of any prime number, such as 7, 13 or 31511.

● We can also get more by adding and multiplying

rationals to any previously found irrational

number.

● But that's not enough!

## Some more irrational proofs

● Let's see how far we can go by contradiction.

Now we go after a trickier number, e.

The left overs are slightly bigger than 0, and smaller than 1.

## E Squared?

● Can we user the same sort of trick to show e^2 is irrational? Almost...

We showed last slide that the left hand side is slightly

larger than an integer, maybe we can show the right

hand side is slightly smaller than an integer.

## The right hand side

● Now we play with the right hand side.

So we've found out that the right hand side is just a bit smaller

than an integer, but the left hand side was a tiny bit bigger than

an integer. Surely this can't be! So now we've found that e^2

is also an irrational number.

## Bigger Powers?

We will have to change our strategy for e^4. What happens if we multiply by n!?

The second summand, the non integer part now goes to something like

This doesn't get smaller as n goes to infinity, can we get rid of the extra powers of 2?

## Get rid of that power of 2!

Suppose n is a power of 2, then how many powers of 2 are there in n!?

n/2 | (even numbers) |

+n/4 | (numbers divisible by 4) |

+ n/8 | (numbers divisible by 8) |

+ .... | |

-------- | |

n-1 |

Using similar reasoning, we can see any other

factorial will have few factors of 2 in it. With

this in mind, we have another idea of what to

multiply by. Here n is a power of 2.

The first n tems in the sum look like

This is an integer since k! has at most k-1 powers of 2. Since n was even

the remainder is

Since the left hand side of the equation is an arbitrarily small amount

larger than an integer, and the right is an arbitrarily small amount

smaller than an integer, we've found a contradiction

Such a scheme works for showing numbers of the form e^{2p} are

irrational numbers. If we use some fancier machinery, we can show

e^{r} is irrational, for r a rational number. First we look at a particular

polynomial which will be of use.

## Nifty polynomial

● Let's look at the following polynomial

What else can we say about this polynomial? First it also has the form

The first 2 statements are easily seen, but the third is a bit trickier.

## Statement 3?

● The last bit?

## Put this polynomial to work

● So we've got a polynomial with some nice

properties, so let's try and show that e^{r}, for a

rational r must be irrational. It is enough to show

e^{p} is irrational for an integer p. Assume e^{s} is

rational

## Proof (cont)

● Make it an infinite sum (the rest of the terms are 0)

## Bounding?

● We also have a bound for the integral from our

second result in the lemma.

Since there aren't any integers between 0 and 1, so e^{s}
is not rational.

## Again, Again!

That seemed nice, let's try again with π^2.

So N is a positive integer (the integrand is positive), but...

But this can't be. We've found our contradiction

## And now for something completely diffrent

● Here is one more set of irrational numbers, with a different proof .

This is equivalent to saying that if we start at a point

on the unit circle and pick an orientation, and begin

circling around with chords of length 1 / √n, then we

will never hit a point twice.

## A helpful trig identity

We will make use of the following trig identity, to get a
recursion

formula.

## Churn, churn

By induction we have

## To the contradiction!

Assume that we do have a rational number

## Fin!

● Prizes (Irrational -> Real 1-1 Map?,

Transcendental -> Real 1-1 Map?)

● Thanks to Erdös, and the “Proofs from the Book”

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