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sauravshakya
Group Title
Prove OR Disprove
Every irrational number can be expressed as (x)^y where x and y are rational numbers.
 one year ago
 one year ago
sauravshakya Group Title
Prove OR Disprove Every irrational number can be expressed as (x)^y where x and y are rational numbers.
 one year ago
 one year ago

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terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Am I allowed to assume that pi is transcendental?
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
Any transcendental number disproves this.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Suppose 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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
I'm guessing the point is to prove the existence transcendental numbers.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Or rather, to do the proof without assuming they exist.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Let'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"
 one year ago

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

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

wio Group TitleBest ResponseYou've already chosen the best response.2
Anyway, 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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
We 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...
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.2
This is obviously an irrational number
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
I don't see how that is relevant.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Anyway, 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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
@oldrin.bataku What do you think then?
 one year ago

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

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

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

wio Group TitleBest ResponseYou've already chosen the best response.2
\[ \Large x = \sqrt{2}^{ \sqrt{2}z} \]
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.2
\[ \Large x^2 = 2^{\sqrt{2}z} \]
 one year ago

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

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

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

wio Group TitleBest ResponseYou've already chosen the best response.2
'proof by contradiction'
 one year ago

oldrin.bataku Group TitleBest 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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
We 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.
 one year ago
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