Let's discuss Mathematics

[Atticus]

There is no "first" rational number. Also, there isn't only one irrational between any two irrationals, but infinitely many. There's also infinitely many rationals between any two irrationals.

The proof that there are more irrationals than rationals comes from Cantor's diagonal proof (via some easy maneuvering). The proof itself isn't that hard to understand, but typing it takes more time than I'm ready to invest at the moment.

EDIT: Missed a new page for a change...

EDIT2: If 0 is the first rational, what is the second? If it is m/n, then m/(2n) is smaller than it but greater than zero. The same logic applies to m/(2n) and so on. Thus you can't list rationals in order of magnitude. You can list them though.

 

[sanabas]

0 is a rational number, so we can use that for the first one... I fail to see your point?

What Atticus said. You can't find two consecutive rational numbers as you suggested, because between any two rational numbers there are infinitely many other rational numbers.

 

[tycoonist]

There is no "first" rational number. Also, there isn't only one irrational between any two irrationals, but infinitely many. There's also infinitely many rationals between any two irrationals.

The proof that there are more irrationals than rationals comes from Cantor's diagonal proof (via some easy maneuvering). The proof itself isn't that hard to understand, but typing it takes more time than I'm ready to invest at the moment.

EDIT: Missed a new page for a change...

EDIT2: If 0 is the first rational, what is the second? If it is m/n, then m/(2n) is smaller than it but greater than zero. The same logic applies to m/(2n) and so on. Thus you can't list rationals in order of magnitude. You can list them though.

I don't know if what paradigm posted is accurate, but if
#1 There is a rational between every two irrationals, and
#2 there is an irrational between every rationals

Then surely it follows that if we were able to write a list of all real numbers, any area of it would view:
I-R-I-R-I-R
where I represents an irrational and R represents a rational. Now surely I don't have to actually be able to write the list to conclude that I can pair them up?

 

[ParadigmShifter]

Just because there is one number inbetween doesn't mean there aren't infinitely many others inbetween as well

 

[tycoonist]

Just because there is one number inbetween doesn't mean there aren't infinitely many others inbetween as well

But if both are true, the pairing method still holds.

Let me try and explain my thinking:

Lets look at two rationals:
R-R
There must be an irrational inbetween right?
so it reads
R-I-R
and we can pair.

But what if there are two irrationals inbetween?
Then it reads
R-I-I-R
However, there must be a rational inbetween the two consecutive irrationals - So it now reads
R-I-R-I-R
and we can pair.

Iterate, we can pair infinitely. So there are the same number of rationals as irrationals if you take it to #1 and #2 to be true.

 

[ParadigmShifter]

That's what is weird about countable and uncountable infinities

 

[tycoonist]

That's what is weird about countable and uncountable infinities

Have I not just proved that there are the same number of irrationals as rationals? Or have I made an incorrect assumption?

 

[ParadigmShifter]

Let S be the set or rational numbers, and T be the set of irrationals. You can prove that there is an element of S that is zero distance from an element of T, I think that is what would cause the argument to fail.

EDIT: Actually make that "for each element of S there is an element of T at zero distance".

 

[Atticus]

It doesn't mean that there are as many of them because each of those pairings excludes both rationals and irrationals.

Also, that doesn't mean that rationals and irrationals alternate, because each of the lists is incomplete.

When you continue this process infinitely you might have all the rationals included, but not irrationals.

Here's how Cantor's diagonal proof goes: We're going to prove that there can't be infinitely long list of reals in ]0,1[. That means that if you pair up those reals with natural numbers, some of them will always be omitted.

Proof: Suppose there is such list. Let's write it in decimal form and write down the decimals after 0, as follows:
a₁₁a₁₂a₁₃a₁₄...
a₂₁a₂₂a₂₃a₂₄...
a₃₁a₃₂a₃₃a₃₄...
a₄₁a₄₂a₄₃a₄₄...
.
.
.
So a₁₁ is the first decimal of the first number, a₁₂ the second decimal of first number, a₇₅ the fifth decimal of seventh number and so on.

Now look at the diagonal of this matrix, the numbers a_kk, and form a new number, whose decimals are defined as follows b_k= 4 if a_kk is 7 and b_k=7 if a_kk is not 7 (these specific numbers 4 and 7 are pretty much arbitary choices*).

Now this new number whose decimals are b_k is not on our original list, because it's k:th decimal differs from the k:th decimal of any number on that list! So the list didn't contain all the reals in ]0,1[.

Notice also, that putting this number on the list doesn't help, because the same argument applies to the new list too.

It's pretty easy from this to go on to prove that there are more irrationals than rationals.

 

[Souron]

But if both are true, the pairing method still holds.

Let me try and explain my thinking:

Lets look at two rationals:
R-R
There must be an irrational inbetween right?
so it reads
R-I-R
and we can pair.

But what if there are two irrationals inbetween?
Then it reads
R-I-I-R
However, there must be a rational inbetween the two consecutive irrationals - So it now reads
R-I-R-I-R
and we can pair.

Iterate, we can pair infinitely. So there are the same number of rationals as irrationals if you take it to #1 and #2 to be true.

You try to list consecutive real numbers. This cannot be done. That's where this logic falls apart.

 

[Petek]

Here's a question with a nifty answer (if you haven't seen it before):

Do there exist irrational numbers a and b such that a^b is rational? No fair using high-powered theorems such as Gelfond-Schneider.

 

[SS-18 ICBM]

A set containing all sets also contains itself, right? I don't see any reason for that statement to be paradoxical.

 

[nc-1701]

Here's a question with a nifty answer (if you haven't seen it before):

Do there exist irrational numbers a and b such that a^b is rational? No fair using high-powered theorems such as Gelfond-Schneider.

Unless I misunderstand the question...


Of course! let a=b=2 => a^b=2^2=4 which is rational...

 

[Petek]

Well, I specified that a and b be irrational, and your a and b are about as rational as they could be. That's OK. It's Friday, so you get a free pass!

 

[Atticus]

A set containing all sets also contains itself, right? I don't see any reason for that statement to be paradoxical.

Let A be the set of all sets and P(A) it's power set. Then every set in P(A) is contained also in A, and therefore P(A)\subset A. It's a contradiction since P(A) is of greater cardinality than A.

Have to think Petek's question a while. Is it a bad spoiler if you give the answer but not the proof?

 

[IdiotsOpposite]

Here's a question for y'all. Say I want to find all functions that obey this property:

Re(f(z)) = ((1+i)/2) f(z)

Obviously, f(z)=0 is a solution. How would I find other solutions, or prove that there aren't any?

 

[Atticus]

Break up f(z) into real and imaginary parts.

 

[ParadigmShifter]

e^i*pi is rational, and e and i*pi aren't. Unless you insist both irrational numbers are real...

 

[IdiotsOpposite]

e^i*pi is rational, and e and i*pi aren't. Unless you insist both irrational numbers are real...

Considering that rational numbers are defined as "the set of numbers you can get by dividing one integer (integers are real) by another, nonzero integer", and irrational numbers are defined as "not belonging to set Q", I believe that there is no requirement that an irrational number be real.

 

[ParadigmShifter]

2^1/2 is irrational as well. The proof is easy but I can't be bothered typing it up I can also come up with infinitely many more irrational numbers of this form...

EDIT: Maybe petek is thinking if transcendental rather than merely irrational numbers.b (e and i*pi are both transcendental as well though).

EDIT2: Whoops, turned the question the wrong way round Well at least I proved the converse false