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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
|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.
● 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?)
● No sir, we've got no closure.
●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
● 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
● 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.
● 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.
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/4||(numbers divisible by 4)|
|+ n/8||(numbers divisible by 8)|
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 e2p are
irrational numbers. If we use some fancier machinery, we can show
er is irrational, for r a rational number. First we look at a particular
polynomial which will be of use.
● 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.
● 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 er, for a
rational r must be irrational. It is enough to show
ep is irrational for an integer p. Assume es is
● Make it an infinite sum (the rest of the terms are 0)
● 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 es is not rational.
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.
A helpful trig identity
We will make use of the following trig identity, to get a
By induction we have
To the contradiction!
Assume that we do have a rational number
● Prizes (Irrational -> Real 1-1 Map?,
Transcendental -> Real 1-1 Map?)
● Thanks to Erdös, and the “Proofs from the Book”