There are the same amount of fractions as whole numbers

Syth

Most people on reading the subject line will probably disagree, caiming that it's intuitivly obvious that there are more fractions than whole numbers.

If you have two groups of numbers (ie two sets), then there are the same amount of numbers (aka elements) if you can match up each number in one group to another, and vice versa. If you can do this then there will be no number left over, no number unmatched. So there must be equal numbers of numbers (or elements) in each group.

So how does this apply to fractions and whole numbers? In other words how can you match up the fractions and whole numbers?

Well you draw a grid like this :
1/1 2/1 3/1 4/1 ...
1/2 2/2 3/2 4/2 ...
1/3 2/3 2/3 4/3 ...
... ... ... ....

Going on forever. This grid holds all fractions since it goes on forever. Then you draw a line like this(where X's represent fractions)
X-X X-X
/ / /
X X X
| / /
X X X

and so on. The you wind up with a list of fractions by starting at the start of the line (the top left), and following the line. ie you get this list:
1/1, 2/1, 1/2, 1/3, 2/2, 3/1, ...
and so on forever. ie the list has every fraction.
You then write the whole numbers underneth and match em up.
ie
1/1, 2/1, 1/2, 1/3, 2/2, 3/1, ...
1, 2, 3, 4, 5, 6, ...

Thus you can match up the whole numbers and the fractions, so there will never be a fraction that is not matched to a whole number and vice versa, so there must be one fraction for every whole number and one whole numbers for every fraction.
SO THERE MUST BE EQUAL NUMBERS OF FRACTIONS AS WHOLE NUMBERS.

Cool, eh?

BTW don't reply calling me an idot with a crackpot theory. This was proved in 1845 by George Cantor, I'm just speading it around.

davej

BTW don't reply calling me an idot

Ok I'll call you and idiot instead!

No just kidding, but what is the point of your post ?
Do you want to discuss something or are you just expending electrons in an extravagant manner ?

davej

hussey

1/1, 2/1, 1/2, 1/3, 2/2, 3/1, ...
1, 2, 3, 4, 5, 6, ...


but 1/1 and 2/2 are the same ??

Yes you are correct though .. once you think of a fraction as a rational number ie p/q where p,q are integers.

and number like sqrt(2) is not a fraction

Syth

What's the point? The fact that it's totally counter intuitive and cool.

Ok then for fractio that are identical then you just skip over them. It won't matter, you still prove that there are same amount of fractions as whole numbers. It just goes to show that infinity is a tricky thing and can lead to freakky results.

smiles

No, you're not an idiot, you just havent really thought about it.

What you are talking about it pairing up the set of natural numbers N and the set of fractions Q, well similar to that anyway, and that is how you determine if a set is countable (it you can put Q into one to one correspondance with N)

this does not mean they have the same number of elements.

In fact N and Q have a different amount of infinite elements, called alephs.

<< Fio >>

rob1891

you kids ought to try counting the number of infinite fractions and then the number of infinite natural numbers, and compare them ... that would sort it out.

That is how stoopid this thread sounds! There is no point in comparing the number of elements in an infinite series, yeah, they have a proof for it, but those mathematicians have a proof for everything, it doesn't make worth knowing!! Adding them up is interesting sometimes tho.

sum n^-1 as n goes to infinity = 1/1+1/2+1/3+...1/n = ??


Has anyone done the root 2 is irrational proof yet, love that one, irrational!! teehee!!

edit: lol, smiles make me sound like an idiot, tell us about alephs please

smiles

There is no formula/value for the sum of the sequence n^-1, it diverges. Sorry!

As for alephs:
Aleph nought (0) is used to describe sets of countable infinity, ie they can be put into correspondance with N (the natural numbers), alepth 1 is for Real numbers, as they are uncountable, and it goes on.... so technically the guy who started the thread is correct, both rational numbers and natural numbers have aleph 0 members, but not at the same time....

http://whatis.techtarget.com/definition/0,,sid9_gci852621,00.html
He probably explains better than I do.

Root 2 being irrational?
The proof is to say that it is rational and then prove that that is wrong (Proof by Contradition)

Ok, say root 2 = p/q, where p and q are integers and are relatively prime (ie you cant cancel them anymore)

then you say 2 = p^2/q^2

so 2 * q^2 = p^2
Therefore p^2 is a multiple of 2, this p is also a multiple of 2, say 2s, so p^2 is a multiple of 4, ie 4 * s^2.

So q^2 = 2 * s^2 (dividing by 2 from the above equation).
This says that q^2 is a multiple of 2, so q is a multiple of 2.

But this is a contradiction, because if p and q are multiples of 2 then they cannot be relatively prime.
Therefore root 2 is irrational.

Make sense? you can use the same proof for proving the root of any prime is irrational, just replace 2 with whatever.

<< Fio >>

Gordon

Originally posted by rob1891
those mathematicians have a proof for everything,

Is there not a number called the omega number that mathematicians cannot calculate?

Cyan

Although both proofs (that |Q| = |N|, and that sqrt(2) is irrational) are perfectly valid, there is something nice about getting a fundamental insight into why these things are true. It's interesting to note that much more elegant (in my opinion) proofs are available for both of these claims, and furthermore they are corollaries of the same principle: The Fundamental Theorem of Algebra.

For those of you who never did mathematics in college, this states:

Any natural number greater than 1 has a unique prime factorisation.

Accepting this as true, here are the proofs.

Sqrt(2) is irrational:

Suppose the opposite. Then

p^2 = 2*q^2 as before.

Now p and q, being natural numbers, have unique prime factorisations, being expressible as p = p1^a1 * p2^a2 * ... * pr^ar, and q = q1^b1 * ... * qs^bs.

Note that p^2 = p1^2a1 * p2^2a2 * ... * pr^2ar. Thus the exponent of every prime in a square number is even . This is also true for q^2.

So, if p^2 = 2*q^2 ...

• there is an even number of 2's in the prime factorisation of the left hand side.
• there is an odd number of 2's in the right hand side (the 2, and the even number from q^2)

This is a contradiction.

Another nice feature of this proof is that it is easily generalized to anything that isn't a square number, unlike the plodding methodical proof seen above.

As for |R| = |N| = Aleph-0 (yes, Fio, they both are Aleph-0: that's what we're trying to prove .. :-))

Each rational number can be represented as m/n, with m and n relatively prime (i.e. no factors in common -- if they do have factors in common, cancel them out and you're left with the same number.)

m and n both have unique prime factorisations. Let the same notation used above be assumed here. (m = p1..., n = q1...).

Then consider the function K(m, n) := p1^2a1 * p2^2a2 * ... * pr^2ar * q1^(2b1 - 1) * ... * qs^(2bs - 1).

Note:

• K(m, n) is unique for unique m, n.
• K(m, n) is always a unique natural number.
• Furthermore, given a natural number, you can construct unambiguously m and n again.

(This is because all the prime powers that came from m are odd, all the prime powers that came from n are even.)

So a function mapping a unique rational to a unique natural exists, and it has an inverse.

Thus there is a 1-to-1 relationship between Q and N.

Thus |Q| = |N|, and thus Q is countable.

Quod Erat Demonstrandum. :-)