SamBridge Posted January 24, 2013 Share Posted January 24, 2013 Ok I know what an irrational number is suppose to be, but how do we know if a number is actually irrational? It can't be expressed in a fraction that we know of, but there are infinite numbers, there could be a number that is so big we haven't counted to it yet, and not only that but we can't even generate them at will, in order to we'd have to spend an infinite amount of time writing out decimal places. So I know pi is suppose to be irrational, but how can it be proven that it isn't countable? I don't know if it has to do with cantor's proof or not, I don't fully understand it as I haven't taken set theory. Any explanations? Link to comment Share on other sites More sharing options...

John Posted January 24, 2013 Share Posted January 24, 2013 As an example (this is encountered early on in courses dealing with proof techniques, and isn't difficult to follow), we can prove that [math]\sqrt{2}[/math] is irrational using proof by contradiction, as follows: Suppose [math]\sqrt{2}[/math] is rational. Then it can be written as the ratio of two relatively prime integers, i.e. [math]\sqrt{2} = \frac{p}{q}[/math] where [math]p[/math] and [math]q[/math] are integers sharing no factor greater than 1. Since [math]\sqrt{2} = \frac{p}{q}[/math], then [math]2 = \frac{p^2}{q^2} \implies 2q^2 = p^2[/math]. Since [math]q[/math] is an integer, then [math]q^2[/math] must be an integer, which means [math]p^2[/math] is equal to 2 times an integer and is therefore even. Since [math]p^2[/math] is even, [math]p[/math] must be even, which means it's also equal to 2 times an integer, say [math]k[/math]. Now, [math]p = 2k \implies p^2 = 4k^2[/math], which, from our earlier calculations, means [math]4k^2 = 2q^2 \implies 2k^2 = q^2[/math]. By the same reasoning we used earlier, this means [math]q[/math] must be even. So we have that [math]p[/math] and [math]q[/math] are both even. But this contradicts our earlier assumption that they're relatively prime, and so [math]\sqrt{2}[/math] must be irrational. Proofs that pi is irrational are a bit harder to follow, but a variety of such proofs do exist. They can easily be found using Google if you'd like to take a gander. Link to comment Share on other sites More sharing options...

Bignose Posted January 24, 2013 Share Posted January 24, 2013 Several proofs pi is irrational: http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational It can also be proved e is irrational, as well as the square root of 2. But, there are some funny ones like we don't have a proof pi*e is irrational, or pi^e or e^pi or pi^pi, etc. Most likely they are, but we don't have a definitive proof. I guess, really, it depends on the number you're looking at. Again, some of them we know, but in general, we probably don't know for sure. Link to comment Share on other sites More sharing options...

imatfaal Posted January 24, 2013 Share Posted January 24, 2013 Several proofs pi is irrational: http://en.wikipedia.org/wiki/Proof_that_π_is_irrational It can also be proved e is irrational, as well as the square root of 2. But, there are some funny ones like we don't have a proof pi*e is irrational, or pi^e or e^pi or pi^pi, etc. Most likely they are, but we don't have a definitive proof. I guess, really, it depends on the number you're looking at. Again, some of them we know, but in general, we probably don't know for sure. "we don't have a proof pi*e is irrational, or pi^e or e^pi or pi^pi" Whereas we know that e^pi.i is as simple as you can get! You gotta love that mind-bending nature of maths. I have a feeling that there is a proof for gelfond's number e^pi being irrational but we do not know if it is transcendental. Link to comment Share on other sites More sharing options...

D H Posted January 24, 2013 Share Posted January 24, 2013 (edited) But, there are some funny ones like we don't have a proof pi*e is irrational, or pi^e or e^pi or pi^pi, etc. Most likely they are, but we don't have a definitive proof.The GelfondSchneider theorem says that e^pi is transcendental (so obviously irrational). http://www.wolframalpha.com/input/?i=%28-1%29%5E%28-i%29 Edited January 24, 2013 by D H Link to comment Share on other sites More sharing options...

Amaton Posted January 25, 2013 Share Posted January 25, 2013 (edited) It can't be expressed in a fraction that we know of, but there are infinite numbers, there could be a number that is so big we haven't counted to it yet, and not only that but we can't even generate them at will, in order to we'd have to spend an infinite amount of time writing out decimal places. But wouldn't that be painstaking trial-by-trial testing? A conjecture can be proven based on the very definitions and properties of the objects at hand, and I'm sure irrationality proofs are not exceptions to this. Edited January 25, 2013 by Amaton Link to comment Share on other sites More sharing options...

SamBridge Posted January 25, 2013 Author Share Posted January 25, 2013 (edited) But wouldn't that be painstaking trial-by-trial testing? A conjecture can be proven based on the very definitions and properties of the objects at hand, and I'm sure irrationality proofs are not exceptions to this. No math doesn't work like that because that mistake is made more often than you think, it has to be proven using logic, I can't say that just because we haven't found a place where it doesn't repeat so far means it's irrational, it has to be proven, just as any other theorem in mathematics has to be to be called "proven". There's many conjectures, but we do not know that they will work for an infinite number of numbers until they are proven to. I didn't look into it but I will assume that because it was mentioned by 2 experts/staff that there is a way to concretely prove if something's irrational. Edited January 25, 2013 by SamBridge Link to comment Share on other sites More sharing options...

Amaton Posted January 25, 2013 Share Posted January 25, 2013 No math doesn't work like that because that mistake is made more often than you think, it has to be proven using logic, I'm not sure where you're trying to get at. What mistakes are you referring to? Link to comment Share on other sites More sharing options...

D H Posted January 25, 2013 Share Posted January 25, 2013 So I know pi is suppose to be irrational, but how can it be proven that it isn't countable?Countable? You appear to be confusing the concept of sets and numbers. Countability is a property of sets, not individual elements of a set such as pi. No math doesn't work like that because that mistake is made more often than you think, it has to be proven using logic, I can't say that just because we haven't found a place where it doesn't repeat so far means it's irrational, it has to be proven, just as any other theorem in mathematics has to be to be called "proven".You can find many proofs on the net that pi is not only irrational, but also is transcendental. Link to comment Share on other sites More sharing options...

SamBridge Posted January 25, 2013 Author Share Posted January 25, 2013 Countable? You appear to be confusing the concept of sets and numbers. Countability is a property of sets, not individual elements of a set such as pi.You can find many proofs on the net that pi is not only irrational, but also is transcendental. The countability of a set containing all places of pi... Also I did not say that it couldn't be proven that pi was irrational or transcendental. Link to comment Share on other sites More sharing options...

Bignose Posted January 25, 2013 Share Posted January 25, 2013 The GelfondSchneider theorem says that e^pi is transcendental (so obviously irrational). http://www.wolframalpha.com/input/?i=(-1)^(-i) ah, my bad. I will admit I just threw a bunch of those together because I know in general we don't know about most of them. I shall be more careful in the future. Link to comment Share on other sites More sharing options...

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