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 2 years ago
Prove OR Disprove
Every irrational number can be expressed as (x)^y where x and y are rational numbers.
 2 years ago
Prove OR Disprove Every irrational number can be expressed as (x)^y where x and y are rational numbers.

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terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.0Am I allowed to assume that pi is transcendental?

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0Any transcendental number disproves this.

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.0Suppose pi (which is irrational) can be expressed as (x)^y where x and y are rational. Then (x)^y can be expressed as \[\huge b^{\frac{p}{q}}\] Where p and q are integers and b is rational Then pi is a solution of \[\huge x  b^{\frac{p}{q}}=0\] \[\huge x = b^{\frac{p}{q}}\] \[\huge x^{q} = b^{p}\] \[\huge x^{q}  b^{p} = 0\]effectively contradicting the transcendentalness of pi XD

wio
 2 years ago
Best ResponseYou've already chosen the best response.2I'm guessing the point is to prove the existence transcendental numbers.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Or rather, to do the proof without assuming they exist.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Let's say \(x = m/n\) and \(y=a/b\) \[ x^{y} = \left( \frac{m}{n}\right)^{a/b} = \left( \frac{m^a}{n^a}\right)^{1/b} = \sqrt[b]{x^a} \]We know \(x^a\) is still rational so the question then just becomes "every irrational number can be expressed as an integer root of a rational number"

wio
 2 years ago
Best ResponseYou've already chosen the best response.2But this also means "Every irrational number q, when taken to some integer number n, results in a rational number"

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Probably should have noted earlier that if \(y\) is negative, then we can take the reciprocal of \(x\) which is still rational.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Anyway, I'm not sure where to continue from here, but I think: "Every irrational number q, when taken to some integer number n, results in a rational number" Is a bit easier of a thing to disprove. That's what my gut says.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2We could note that: \[ \Large \sum_{n=1}^{\infty} \frac{1}{n^{2}} \]Converges, and thus is an irrational number...and try to show that there is no way putting it to any integer power would give you a rational number?? I think...

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Oh wait! I got an idea, how about \[ \sqrt[\sqrt{2}]{2} \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2This is obviously an irrational number

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0I don't see how that is relevant.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Anyway, if we take it to the \(n\)th power: \[ \sqrt[\sqrt{2}]{2}^n= \sqrt{2}^{n/\sqrt{2}} \]No matter what we set \(n\) to (it must be an integer), we are going to get and irational number to an irrational power. hmmmm

wio
 2 years ago
Best ResponseYou've already chosen the best response.2@oldrin.bataku What do you think then?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Without the for sight of transcendental numbers, how would you disprove it?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Hmmm, I think maybe a good way is to do a proof by contradiction. Suppose \[ \sqrt{2}^\sqrt{2} = x^y \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2We know the inverse of \(y\) is also rational, let's say \(z = 1/y\)

wio
 2 years ago
Best ResponseYou've already chosen the best response.2\[ \Large x = \sqrt{2}^{ \sqrt{2}z} \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2\[ \Large x = 2^{\sqrt{2}z/2} \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2\[ \Large x^2 = 2^{\sqrt{2}z} \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Now if only I can show that \[ \Large 2^{\sqrt{2}z} \]is not rational....

wio
 2 years ago
Best ResponseYou've already chosen the best response.2I know that a rational number times an irrational number is irrational, but an integer to an irrational power...?

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0@wio the question states \(x,y\) are *rational*, so \(\sqrt2\) is no good here.

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0@wio ... that doesn't disprove the claim however, at all. The only relevance \(\sqrt2\) has is that it's an algebraic irrational number that works for our claim, i.e. \(2^\frac12\) where \(2,\frac12\) are rational.

wio
 2 years ago
Best ResponseYou've already chosen the best response.2We know that \(x^2\) is rational. If \(2^{\sqrt{2}z}\) isn't rational then it's a contradiction, that's where I was going.
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