Let's discuss Mathematics

[Atticus]

I took it to mean he's talking about only real numbers. Also, I don't like IO's definition, because it would make me for example irrational (since I'm not in Q).

I also thought that e^{pi*i} thingie or similar could do the job with some manipulation.

 

[IdiotsOpposite]

I took it to mean he's talking about only real numbers. Also, I don't like IO's definition, because it would make me for example irrational (since I'm not in Q).

I also thought that e^{pi*i} thingie or similar could do the job with some manipulation.

Can you be expressed as a ratio?

 

[nc-1701]

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!

Erm yeah...:facepalm: Wish I could claim to have been drunk or something...

 

[Petek]

To clarify, in my question a and b should be real and irrational.

 

[SS-18 ICBM]

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.

Alright, thanks. Mostly makes sense. Incidentally, where do they start teaching set theory in your country?

 

[Atticus]

I learnt it at the first year in university (which here corresponds partly to your colleges, I've understood).

Can you be expressed as a ratio?

Not per se.

 

[ParadigmShifter]

You certainly got one over him!

 

[IdiotsOpposite]

I learnt it at the first year in university (which here corresponds partly to your colleges, I've understood).



Not per se.

Exactly. Therefore, you are an irrational number.

 

[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.

I'll post my solution in a spoiler. Recall that a and b are both irrational and real.

Spoiler :

Yes, such irrationals exist. Let a = sqrt(2). We take it as known that a is irrational. Consider a^a. This number is either rational or irrational. If it is rational, then we are done. If it is irrational, then consider (a^a)^a, which equals 2 and so is rational.

This solution is nonconstructive, but elementary. You can invoke Gelfond-Schneider (cited above) to conclude that a^a is trancendental to get explicit a and b.

 

[IdiotsOpposite]

The set containing Bin Laden's life is now empty. YES!

 

[Atticus]

That was clever, Petek!

 

[Petek]

Thanks. Not my proof, though. I saw it somewhere else.

Since I used the result that sqrt(2) is irrational, here's a quick proof that's not as well-known as the standard one:

Suppose that sqrt(2) is rational. Let k be the least positive integer such that k*sqrt(2) is an integer. Then n = k*sqrt(2) - k is a positive integer less than k and n*sqrt(2) is an integer, a contradiction.

 

[ParadigmShifter]

I'm too drunk too understand that tonight

I'm just wondering if we will get a log₁₀ = 3 thread and we lose all our quality posts in this thread

 

[Atticus]

You mean when log₁₀ <n<log₂, where n is the number of posts in this thread?

We can't meaningfully discuss it in this thread of course. If there's gonna be a new thread, this shouldn't be deleted, because people might want to take look at this afterwards too. (I know they delete spammier threads right away).

 

[sanabas]

Thanks. Not my proof, though. I saw it somewhere else.

Since I used the result that sqrt(2) is irrational, here's a quick proof that's not as well-known as the standard one:

Suppose that sqrt(2) is rational. Let k be the least positive integer such that k*sqrt(2) is an integer. Then n = k*sqrt(2) - k is a positive integer less than k and n*sqrt(2) is an integer, a contradiction.

Shouldn't you prove that n is less than k? Otherwise I could 'prove' sqrt (4) or (9) is irrational.

Do the same thing for sqrt (4), and k = 1, n =1. Do it for sqrt (9), and k = 1, n = 2.

 

[Petek]

Shouldn't you prove that n is less than k? Otherwise I could 'prove' sqrt (4) or (9) is irrational.

Do the same thing for sqrt (4), and k = 1, n =1. Do it for sqrt (9), and k = 1, n = 2.

Yes, n < k requires proof, and I made that claim in the post. I omitted the proof because it's simple:

n = k*sqrt(2) - k < k if and only if

k*sqrt(2) < 2k if and only if

sqrt(2) < 2, which is true.

 

[sanabas]

Makes sense. Though that wouldn't work for proving that sqrt (5) is irrational, for instance. Easy to change though, you simply change the (-k) part to be (-4k), and then you get n<k iff sqrt (5) < 5.

Same deal for any other root.

Except that still gives a seemingly valid proof to prove that sqrt (4) is irrational.

Suppose that sqrt(4) is rational. Let k be the least positive integer such that k*sqrt(4) is an integer. Then n = k*sqrt(4) - 3k is a positive integer less than k and n*sqrt(4) is an integer, a contradiction.

The problem there is that n is a negative integer.

In the first one, n = k*sqrt(2) - k is clearly positive.

But n = k*sqrt(5) - 4k isn't.

Should be able to use a very similar proof to prove sqrt (5) is irrational, but how do you prove that n<k and n>0? I'll have to think about that one when more awake.

 

[Petek]

Yes, you can modify the original argument to prove that the square root of any positive, non-square integer is irrational. I'll let you think about it.

 

[Souron]

n>0 as long as sqrt(2)>1.

If sqrt(2) is rational, then it can be expressed in the form l/k, and l > k iff l/k>1. n = l - k, which is positive if l > k.

But I don't get why we conclude that n*sqrt(2) is an integer.

 

[Petek]

n>0 as long as sqrt(2)>1.

If sqrt(2) is rational, then it can be expressed in the form l/k, and l > k iff l/k>1. n = l - k, which is positive if l > k.

But I don't get why we conclude that n*sqrt(2) is an integer.

By definition, n = k*sqrt(2) - k. So

n*sqrt(2)
= (k*sqrt(2) - k)*sqrt(2)
= 2k - k*sqrt(2)

and k was chosen to be such that k*sqrt(2) is an integer.