An irrational power of an irrational number

[Genady]

Can the thing in the title be rational?

[studiot]

1 hour ago, Genady said:

Can the thing in the title be rational?

Are you referring to euler's identity ?

[math]{e^{\pi i}} + 1 \equiv 0[/math]

 

[Genady]

2 minutes ago, studiot said:

Are you referring to euler's identity ?

eπi+1≡0

 

I thought about it, but I don't know if πi is rational or irrational. To make it well-defined, let's stay in the real numbers.

[studiot]

59 minutes ago, Genady said:

I thought about it, but I don't know if πi is rational or irrational. To make it well-defined, let's stay in the real numbers.

Can't see why.

e and pi are irrational (and also trancendental but that is irrelevant here).
i and 1 and 0 are all rational.
So proceed as follows

[math]{e^{\pi i}} + 1 = 0[/math]           Euler

Rearrange

[math]{e^{\pi i}} =  - 1[/math]

Raise each side to the power -i

[math]{\left( {{e^{\pi i}}} \right)^{ - i}} = {\left( { - 1} \right)^{ - i}}[/math]

Which is

[math]\left( {{e^{ - i\pi i}}} \right) = {\left( { - 1} \right)^{ - i}}[/math]

Which is

[math]{e^\pi } = {\left( { - 1} \right)^{ - i}}[/math]

Which is an irrational number raised to an irrational power expressed as a rational fraction of two rational numbers.

 

[Genady]

OK. I just don't see yet that (-1)^-i is a rational number as per definition "a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q."   

[joigus]

You said an irrational raised to an irrational. (-1)^-i is a negative integer number raised to an imaginary (complex) number. Both can be done. One is more sophisticated. Which one is your question?

For example, \( \left(-1\right)^{\pi} \) presents its own challenge. Which one are you interested in?

It's better perhaps to tackle directly \( z^{w} \) with \( z,w\in\mathbb{C} \).

[Genady]

I am interested to know if there are two real irrational numbers r and s such that r^s is rational.

45 minutes ago, joigus said:

You said an irrational raised to an irrational. (-1)^-i is a negative integer number raised to an imaginary (complex) number. Both can be done. One is more sophisticated. Which one is your question?

For example, (−1)π presents its own challenge. Which one are you interested in?

It's better perhaps to tackle directly zw with z,w∈C .

The comment you refer to is a reply to an attempted proof above that:

1 hour ago, studiot said:

Can't see why.

e and pi are irrational (and also trancendental but that is irrelevant here).
i and 1 and 0 are all rational.
So proceed as follows

eπi+1=0            Euler

Rearrange

eπi=−1

Raise each side to the power -i

(eπi)−i=(−1)−i

Which is

(e−iπi)=(−1)−i

Which is

eπ=(−1)−i

Which is an irrational number raised to an irrational power expressed as a rational fraction of two rational numbers.

 

 

Edited January 7, 2022 by Genady

[joigus]

39 minutes ago, Genady said:

I am interested to know if there are two real irrational numbers r and s such that r^s is rational.

Ah, OK. Yes, that can happen. Studiot gave an important clue, I think. Think Euler.

I think you will agree that \( \log_{\pi}2 \) is irrational. Take \( r=\pi \) and \( s=\log_{\pi}2 \). Then,

\[ r^{s}=\pi^{\log_{\pi}2}=2 \]

The fact that \( \pi^{x} =2\) cannot be solved with x rational should be easy to prove by contradiction.

Edit: Actually, I don't think it's 'easy', it's a somewhat elaborate result of number theory.

Edited January 7, 2022 by joigus

[studiot]

42 minutes ago, Genady said:

I am interested to know if there are two real irrational numbers r and s such that r^s is rational.

The comment you refer to is a reply to an attempted proof above that:

 

 

5 minutes ago, joigus said:

Ah, OK. Yes, that can happen. Studiot gave an important clue, I think. Think Euler.

I think you will agree that logπ2 is irrational. Take r=π and s=logπ2 . Then,

 

rs=πlogπ2=2

 

I didn't think of that.  +1

 

There is a more detailed discussion here.

I am not an expert on number theory, I find the minutiae rather boring after a while.

https://math.stackexchange.com/questions/823970/is-i-irrational

Edited January 7, 2022 by studiot

[Genady]

7 minutes ago, joigus said:

Ah, OK. Yes, that can happen. Studiot gave an important clue, I think. Think Euler.

I think you will agree that logπ2 is irrational. Take r=π and s=logπ2 . Then,

 

rs=πlogπ2=2

 

The fact that πx=2 cannot be solved with x rational should be easy to prove by contradiction.

Edit: Actually, I don't think it's 'easy', it's a somewhat elaborate result of number theory.

Yes, it answers it. If we know that logπ2 is irrational.

I got a simpler proof, without that knowledge:

r=sqrt(2)^sqrt(2) , s=sqrt(2).

 

[joigus]

Let me add something that involves probabilities. I don't think it's very likely that picking r and s irrational, at random, you can get r^s to be rational. The simple 'probabilistic' argument being that the cardinality of irrationals is aleph 1, while that of rationals is aleph naught, which means that there are incommensurably more irrationals than rationals. So, what are the chances. But I do think there are infinitely many occurrences of (irrational)^irrational that are.

11 minutes ago, Genady said:

Yes, it answers it. If we know that logπ2 is irrational.

I got a simpler proof, without that knowledge:

r=sqrt(2)^sqrt(2) , s=sqrt(2).

 

LOL. Good one! I do think it's equally simple, tho.  

No serious. Good one. Now try to do something that simple with \( \left( -1 \right)^\pi \)! 

Edited January 7, 2022 by joigus

[Genady]

25 minutes ago, studiot said:

 

I didn't think of that.  +1

 

There is a more detailed discussion here.

I am not an expert on number theory, I find the minutiae rather boring after a while.

https://math.stackexchange.com/questions/823970/is-i-irrational

The following answer doesn't require number theory, Euler, or complex powers, just algebra:

r=sqrt(2)^sqrt(2) , s=sqrt(2)

Either r is rational or r^s is rational.

So, the answer to the OP is, Yes.

[joigus]

5 minutes ago, Genady said:

The following answer doesn't require number theory, Euler, or complex powers, just algebra:

r=sqrt(2)^sqrt(2) , s=sqrt(2)

Either r is rational or r^s is rational.

So, the answer to the OP is, Yes.

How do you know \( \sqrt{2}^{\sqrt{2}} \) is irrational?

[Genady]

Just now, joigus said:

How do you know 2–√2√ is irrational?

I don't. But if it is not, then we take r=sqrt(2)=s, and r^s is rational. Which answer the OP.

One of the two has to be rational, and this answers the question.

[joigus]

2 minutes ago, Genady said:

I don't. But if it is not, then we take r=sqrt(2)=s, and r^s is rational. Which answer the OP.

One of the two has to be rational, and this answers the question.

Not quite: If it's not, then it's rational. Then you use your algebra and you prove that some 'rational' you've found, raised to sqrt(2), gives a rational.

You need to prove that \( \sqrt{2}^{\sqrt{2}} \) is irrational in order to make that claim. See my point?

So you're back to square one, which is what I tried to tell you: Number theory.

6 minutes ago, dimreepr said:

 

??

[Genady]

Just now, joigus said:

Not quite: If it's not, then it's rational. Then you use your algebra and you prove that some 'rational' you've found, raised to sqrt(2), gives a rational.

You need to prove that 2–√2√ is irrational in order to make that claim. See my point?

No, I don't. Here is why.

The question is: are there such irrational r and s that r^s is rational?

Consider two possibilities.

1. r=sqrt(2) s=sqrt(2) If r^s = sqrt(2)^sqrt(2)  is rational then this answers the question.

2. If sqrt(2)^sqrt(2) is irrational, then r=sqrt(2)^sqrt(2) s=sqrt(2), and r^s is rational, answering the question.

[joigus]

OK. I see. Why didn't you just say that was your answer from the beginning, instead of giving it piecemeal?

You had this 'question' all ready with the answer and all. Then you ask a question pretending not to know the answer.

The answer is kinda obvious TBH. You can find about infinitely many possibilities to do that. I gave you one that's pretty obvious too. About as obvious as the fact that \( \sqrt{2} \) is irrational, which you haven't proved either. Yes, that's a result in number theory too, and you're resting your answer on shoulders of giants.

(pi)^rational cannot give you a rational is pretty obvious to me too. Proving it rigorously is another matter.

You, though, for some reason, don't like the argument. You prefer yours, (which is to come pretty soon.)

You reappear then in intervals of less than one minute declaring that your answer is the answer, and it's simpler than everybody else's. Your answer in every step is, of course, flawed unless you provide the looping argument which is your final effect. Then you pull the rabbit out of the hat that you've been silent about for the whole conversation. 

Voilá! --Applause. 👏👏👏

To me, it's been a considerable amount of time down the drain. I have better things to do.

Cute.

Thank you.

[Genady]

1 hour ago, joigus said:

OK. I see. Why didn't you just say that was your answer from the beginning, instead of giving it piecemeal?

You had this 'question' all ready with the answer and all. Then you ask a question pretending not to know the answer.

The answer is kinda obvious TBH. You can find about infinitely many possibilities to do that. I gave you one that's pretty obvious too. About as obvious as the fact that 2–√ is irrational, which you haven't proved either. Yes, that's a result in number theory too, and you're resting your answer on shoulders of giants.

(pi)^rational cannot give you a rational is pretty obvious to me too. Proving it rigorously is another matter.

You, though, for some reason, don't like the argument. You prefer yours, (which is to come pretty soon.)

You reappear then in intervals of less than one minute declaring that your answer is the answer, and it's simpler than everybody else's. Your answer in every step is, of course, flawed unless you provide the looping argument which is your final effect. Then you pull the rabbit out of the hat that you've been silent about for the whole conversation. 

Voilá! --Applause. 👏👏👏

To me, it's been a considerable amount of time down the drain. I have better things to do.

Cute.

Thank you.

It was not my answer, unfortunately. Somebody has shown it to me, and I was curious to know, how easy it is to find it.

Edited January 7, 2022 by Genady
minor grammar correction

[joigus]

21 hours ago, Genady said:

It was not my answer, unfortunately. Somebody has shown it to me, and I was curious to know, how easy it is to find it.

Sorry, I misinterpreted the whole sequence of events. I suppose I'm just tired. It's a very clever solution anyway. It still rests on sqrt(2) being irrational, but that's easier than pi.