Can we "reach" all the irrationals?

**kenewbie** · Mar23-09, 04:37 AM

Pardon my lack of terminology, I'll do my best to explain myself.

Imagine that the only integers we have are 1, 2 and 3.

We can only describe a very small subset of the irrational numbers using these, namely
Sqrt(2) and Sqrt(3).

Oh, and perhaps Sqrt(2) / Sqrt(3), ((Sqrt(2) * Sqrt(3)) / ((Sqrt(2) * Sqrt(3)), and so on. If this works then you can reach an infinite number of irrationals using just the numbers 2 and 3 and the arithmetic operators as a basis. How frickin' cool is that!

Sorry, I just discovered the above paragraph while typing the post. I'll get back to my original question now.

Now given the fact that the infinite number of irrationals are larger than the infinite number of integers, it would seem to be that we are forever destined to be unable to "reach" all the irrationals? By reaching them I mean describing them by some finite operation of integers under the arithmetic operators.

I was basically wondering if this is true or not, and if proofs or more information exists on this. Do we know how large the infinite of the irrationals are compared to the infinite of integers? (ie, the infinite number of odds should be half the size of the infinite number of integers?)

Are there infinities that we can define which are larger than the irrationals? Of the top of my head I would guess one could say something like the complex irrationals, but that goes against the spirit of what I am looking for. I guess I mean a series of numbers using just the basic numerical numbers and arithmetic operators.

If this post is just filled with drivel, then feel free to ignore it. I realize that I am out of my depth here but this is a seriously entertaining topic.

k

**matt grime** · Mar23-09, 05:03 AM

it would seem to be that we are forever destined to be unable to "reach" all the irrationals? By reaching them I mean describing them by some finite operation of integers under the arithmetic operators.

Correct. Irrationals that satisfy a polynomial relation (essentially what you're describing) are called algebraic numbers. There are only countably many of these. Countable means 'is in one-to-one correspondence with the natural numbers'. Sets which are countable are the natural number, the rational numbers, the algebraic integers, and the algebraic numbers, in fact each of these sets is a subset of the later ones in the list.

They constitute only a very small proportion of the real numbers. For instance neither pi nor e are algebraic (in the algebraic integers).

I was basically wondering if this is true or not, and if proofs or more information exists on this. Do we know how large the infinite of the irrationals are compared to the infinite of integers?

Yes and no. The real numbers have cardinality c (for continuum). Precisely what this is is the subject of the continuum hypothesis and the answer depends on the model of set theory that you use.

(ie, the infinite number of odds should be half the size of the infinite number of integers?)

For infinite sets size (or cardinality) is defined in terms of functions. There is a bijection between odd and even integers, thus they have the same cardinality. A bijection is a one-to-one correspondence (the maps x to x+1 and x to x-1 are bijections between odd and even integers).

Are there infinities that we can define which are larger than the irrationals?

Yes. Lots...

Of the top of my head I would guess one could say something like the complex irrationals, but that goes against the spirit of what I am looking for.

The real and complex numbers have the same cardinality

I guess I mean a series of numbers using just the basic numerical numbers and arithmetic operators.

By definition these are countable - assuming you mean a finite number of operations.

**kenewbie** · Mar23-09, 05:26 AM

Originally Posted by matt grime

Countable means 'is in one-to-one correspondence with the natural numbers'.

This confused me a little. Given the integers 2 and 3 and the arithmetic operators, I can construct more than 2 irrationals. So I guess the term "one-to-one correspondence" means something other than what I intuitively assign to the phrase?

Thanks for the reply, you gave me heaps of keywords to use for finding reading material.

Is this generally part of set theory? Do you know of any books which have meaningful discussions on these topics at my level? (Pre-calc)

Thanks again,

k

**matt grime** · Mar23-09, 05:36 AM

A one-to-one correspondence between two sets A and B is a rule that assigns one and exactly one element of B to A and vice versa. This is how you define size for infinite sets. Any other method, such as saying 'the integers are larger than the odd numbers because the latter is a subset of the former' is unacceptable. Not least because it only provides a way of defining relative size of sets that are nested. It does not allow you to compare the odd and even integers for size, does it? So you need some way of defining size that is independent of what the sets in question are.

I showed you a way to do this if A is the even integers, and B the odd integers:

n goes to n+1 from even to odd, and m goes to m-1 for odd to even.

It should also be clear that the integers are in one-to-one correspondence with the even numbers: n maps to 2n, integer to even, and m to m/2 from even to integer.

A one-to-one correspondence is what you think it is. It is just that I am not relating it at all to your method of construction of more numbers from algebraic operations on 2 and 3.

**kenewbie** · Mar23-09, 06:16 AM

Originally Posted by matt grime

I showed you a way to do this if A is the even integers, and B the odd integers:

n goes to n+1 from even to odd, and m goes to m-1 for odd to even.

It should also be clear that the integers are in one-to-one correspondence with the even numbers: n maps to 2n, integer to even, and m to m/2 from even to integer.

Ahh, of course. I had this mental image of an infinite line, yet I had a beginning and an end on it. If the line never stops, mapping n to 2n makes perfect sense, you never run out of numbers anyway to map towards anyway.

Thanks,

k

**MathematicalPhysicist** · Mar24-09, 02:05 AM

Originally Posted by kenewbie

Pardon my lack of terminology, I'll do my best to explain myself.

Imagine that the only integers we have are 1, 2 and 3.

We can only describe a very small subset of the irrational numbers using these, namely
Sqrt(2) and Sqrt(3).

Oh, and perhaps Sqrt(2) / Sqrt(3), ((Sqrt(2) * Sqrt(3)) / ((Sqrt(2) * Sqrt(3)), and so on. If this works then you can reach an infinite number of irrationals using just the numbers 2 and 3 and the arithmetic operators as a basis. How frickin' cool is that!

((Sqrt(2) * Sqrt(3)) / ((Sqrt(2) * Sqrt(3))=1.

**kenewbie** · Mar24-09, 03:29 AM

Originally Posted by MathematicalPhysicist

((Sqrt(2) * Sqrt(3)) / ((Sqrt(2) * Sqrt(3))=1.

Ugh yeah, the second multipliers are supposed to be addition.

k