- #1
kenewbie
- 239
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Can we "reach" all the irrationals?
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
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