# Rationals between irrationals

1. Sep 21, 2008

### Bob3141592

I understand that there is an irrational number between any (and every) two rationals. And I can see that there is a rational number between any (and every) two algebraic irrational numbers. But I'm having a hard time convincing myself that there must be a rational between any (and every) two transcendental numbers. Can this be proven, and is the proof accessible?

It seems to me that if every two transcendental numbers are separated by a rational, then there has to be as many of one as of the other. But the rationals and algebraics are countable, and the transcendentals are uncountable. This is where I run into trouble.

Now, I can see that for any two specifically named transcendentals there must be a rational between them. But since the transcendentals are uncountable, there must be transcendentals that cannot be so easily named (except perhaps to say that one must be very, very near the other).

I've no idea how they prove specific numbers are transcendental, but I gather it's hard. Is it impossible in the general case? And I know I was playing fast and loose with the description "very, very close" but the specification of a distance $$\delta$$ makes the second point separated by a nameable distance from the first, and I'm thinking of something much, much closer. Is there a way to express this more formally?

Thanks in advance for any (and every) responses.

2. Sep 21, 2008

### CRGreathouse

If the irrationals are different, there is some decimal place at which they differ. Say this is 10^-d * a for the first and 10^-d * b for the second. Then take the initial decimals where the two agree ending with min(a, b) * 10^-d. This is clearly rational and also clearly between the two irrationals.

Last edited: Sep 22, 2008
3. Sep 22, 2008

### ssd

Perfect.

4. Sep 22, 2008

### ssd

Perfect.

5. Sep 22, 2008

### CRGreathouse

I edited the post for correctness.

6. Sep 22, 2008

### Moo Of Doom

Hmm... not quite. That rational has to be less than both of the irrationals. Maybe if you take max(a,b) instead of min, then you're somewhere. Clearly

x.xxxxxxxaxxxxx... > x.xxxxxxxa

EDIT: And what was wrong with your original argument? (a + b)/2*10^-d is clearly rational, with a < (a + b)/2*10^-d < b (WLOG a < b), and since the number consisting of only the decimals up to 10^-(d-1) is rational, the resulting sum is rational, and clearly between the two irrationals. "Perfect" is right.

7. Sep 22, 2008

### CRGreathouse

Yes, max is what I meant.

If the numbers first differ at
...4
...5
then choosing half the two gives
...45
which is indeed rational. But if the numbers continued
...48...
...5...
then ...45 is not between the two.

8. Sep 22, 2008

### Bob3141592

I still don't feel good about this, since the rationals are countable and the transcendentals are uncountable. If this method is universally correct, I suspect it could be specified in a way that allows every transcendental to have at least one rational correspond to it. Consider the diagonal argument. I'll grant you that any transcendental on the list has a rational that separates it from every other transcendental on the list, but what about all the transcendentals that aren't on the list? Those are the ones I'm most interested in, and there's a whole lot of them.

I have an idea, but I should express it in more formal notation for clarity. It's too late tonight, and tomorrow is committed. Hopefully I'll get to write it on Wednesday.

9. Sep 23, 2008

### CRGreathouse

You're having trouble reconciling two different notions of size: density in R (making Q "large") and countability (making Q "small").

10. Sep 24, 2008

### Bob3141592

Sorry I haven’t put this into proper formyet, but it’s gotten surprisingly late, and my wife worries about my health when I don’t get enough sleep. But I’d like your thoughts, so I’m posting this crude write up now, and will refine the content and presentation with LaTex tomorrow

I think I know how to write what I want. I hope I get the notation right.

Let’s use a Universal set U from the closed interval of the reals [0,1]

Well define a countable set of rationals S from that interval. We can reference these points as x sub i where i is an element of N and s sub i is an element of Q. The integers of N are our index set.

Well make a set of only irrationals T by taking our interval and removing the points from S. Can I write that as T = U – S, or does it have to be written as U intersection with complement of S? The set T is uncountably infinite, so we must index them with an uncountably infinite index set, called alpha, which we can equate with the positive reals. We can make two sets from this, T1 containing t (is an element of T) sub x (element of alpha) if x is integer, and T2 containing t sub x if x is noninteger.

By the way, I’m following the Fortran convention of representing integers with the letters i through n and floats with x, y, z, t, etc. I told you I was an old geezer. :-)

Now we can make an infinite list of decimal representations of numbers between zero and 1. This list is the set L formed by the union of all the rational points in S with indexes from 1 to infinity (which is all the points in S) with all of the irrational points in T1 whose index is also an integer. We can interleave these points, forming l sub i = s sub 2 (i+1) / 2 if i is odd and l sub i = t1 sub i/2 for even i. (I believe that this can automatically order the elements of L in a monotonically increasing progression, but that should be proven. My intuition tells me it should be provable, but I haven’t thought about how to do that just yet. In any case we can impose such order). Thus between every two rational numbers we have an irrational, and between every two irrationals we have a rational.

The points I’m most interested in, the points not on this list, are the irrationals which, by Cantor’s Daigonalization Argument, have at least one digit different from any and all of the numbers on the list, and so are not on the list. These are the points in T2. Note that the union of L and T2 is U, the universal set, and the intersection of L and T2 is the empty set.

We can order these sets to match the ordering of their index sets, however, the values will clearly not match, since the values of the elements in L and T2 range from 0 to 1 while the values of the indicis are all the positive numbers.

Ordered this way, T2 is not a continuous set, since it excludes points with an integer index. It is the union of a countably infinite number of continuous subsets (the continuity of these subsets seems obvious to me but this point might deserve a better confirmation). We can label these sets as V sub 1 for all the irrationals between T2 sub x where 0 < x < 1, and V sub 2 for all the irrationals between T2 sub x where 1 < x < 2, etc. Therefore V sub i is a continuous subset of the reals containing no rationals between them.

Is there anything wrong with this construction? Am I on to something, or am I missing something fundamental?

11. Sep 25, 2008

### CRGreathouse

All of this is fine. You can write T = U - S or T = U \ S for T = U ∩ ~S. Note that partitioning T into T1 and T2 requires the Axiom of Choice.

You certainly can't prove that the elements can be put in order! For all you know the elements of T1 are all between 0.001 and 0.002.

But you don't know anything about them; you chose them arbitrarily.

It's not clear what this means, since you don't know where the elements of T1 are. It's not even clear to me that you can do this given some placement for the elements of T1 (say, T1 = {q + sqrt 2 mod 1, q rational}).

12. Sep 27, 2008

### Bob3141592

Thanks for your comments, Charles. My notion wasn't clear enough, either in my writing or in my mind. Dealing at the edges of infinity at either end makes my head hurt, but its wrth the pain to try and understand this stuff.

From your second comment, it seems I need to work out a mapping function from T to T1, which will be T1's index set, so that it can demonstrate a suitably equitable allocation of points from T. I'd like this function to preserve the ordering of the original interval but otherwise be pretty arbitrary. I should be able to do that, and for at least one case in such a way that I hit all the algebraic irrationals. May be some details have to be filled in, but I'd think it should be doable. Can I simply specify that T1 is make up of the algebraic irrationals between zero and 1 in order?

13. Sep 27, 2008

### CRGreathouse

Hmm... certainly their cardinality is small enough, but I worry about their ordinality when you put them in order. Before you can get to 1/sqrt(2) you must pass 1/sqrt(3), 1/sqrt(4), ... and so you never get to 1/sqrt(n) in $\omega$ steps. At least that's how it seems to go. I don't do much with infinite ordinals, myself.

14. Sep 27, 2008

### Bob3141592

Is there a problem with mapping between open sets and closed sets in general? The interval I'm mapping from (T1) is open but the index set, being the natural numbers, is half closed. Are there ways of dealing with this, or is it a show stopper?

15. Sep 28, 2008

### CRGreathouse

I don't think that's the problem -- adding an endpoint to T1 doesn't make it any easier.

16. Sep 28, 2008

### HallsofIvy

This would be true if any two specifically named transcendentals there were one rational number between them. In fact, for any two real number, rational, irrational, or transcendental, there are (countably) infinite rationals between them and (uncountably) infinite irrationals between them.

I have no idea what "a rational number between any two transcendentals" has to be with being "easily named" or even what that could mean.

You seem to be trying for "infinitesmals"- and that way leads to madness! (for me anyway)

It is true that, given any positive integer, n, and given any irrational (or transcendental) number, x, there exist another irrational (or transcendental) number, y, such that |x- y|< 1/n.

You cannot, however, find two irrationals, x and y, such that |x-y|< 1/n for all n.

17. Sep 28, 2008

### Bob3141592

That sounds perfectly right and reasonable, but I'm still having a very hard time getting a handle on this. Given an interval of reals, when I divide it into two complementary sets, one with only the rationals and the other without any rationals, I keep thinking this gives me a countably infinite partitioning of the irrationals (not sure partitioning is the right word there). Apparently it doesn't, though it seems like it would have to. And it also seems necessary that the irrationals are all disconnected single points, just like the rationals are, but this "just like" isn't any sort of a relationship.

This stuff is driving me nuts, and it won't take much gas to get there. :uhh:

18. Sep 29, 2008

### HallsofIvy

"partition" is correct here, although this is a partition of the interval of the reals, not the irrationals, into two sets, one containing the rationals, the other the irrationals. But why reason do you have to think that the two sets will be the same size or both countable?

19. Sep 29, 2008

### CRGreathouse

Bob is suggesting this:

...QiiiQiQiiiiiiiiiiiiiQ...

where the 'Q's are rational numbers and the 'i's irrational, with a countable number of 'Q's and an uncountable number of 'i's. Finitary reasoning would suggest that there are now countably many blocks of irrational numbers between the rational numbers.

This fails because there are no such spans with only irrationals between two (infinitesimally close) rationals.

You can get a partition of the irrationals into countably many (uncountable) subsets, though. Instead of all the rationals, use the rationals 1 - 1/n. Then the irrationals in (0, 1/2), (1/2, 2/3), (2/3, 3/4), ... is a partition of the irrationals into countably many uncountable subsets.