Irrational numbers to irrational powers being rational?
So some of you may be familiar with the proof that some irrational numbers to irrational powers are rational, that is: if $A = sqrt2^sqrt{2}$ then it follows that $A^sqrt{2} = 2$. So, I've found a fun brain teaser, so to speak, for people (probably everyone on here) who like this type of math problem.
Given an irrational number $q$, find the set of all
irrational numbers $p$ such that: $q^p$ is rational.
The solution that I found is the set of all $p = log_q(x):xin Q $ where Q is the field of rational numbers. The example I gave at first does satisfy this. As you can see, this problem will be more fun for the presenter than the solver since it's such an annoyingly simple solution, which is why I wanted to ask if there is an example that falls outside of the solution set that I gave for some bizarre reason, and would I need a more rigorous proof that no solution can fall outside the set? I mean all I really did was set $q^p = x$, say x was rational, and solver for p in terms of q.
number-theory logarithms exponentiation rationality-testing
edited 19 hours ago
Klangen
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
add a comment |
So some of you may be familiar with the proof that some irrational numbers to irrational powers are rational, that is: if $A = sqrt2^sqrt{2}$ then it follows that $A^sqrt{2} = 2$. So, I've found a fun brain teaser, so to speak, for people (probably everyone on here) who like this type of math problem.
Given an irrational number $q$, find the set of all
irrational numbers $p$ such that: $q^p$ is rational.
The solution that I found is the set of all $p = log_q(x):xin Q $ where Q is the field of rational numbers. The example I gave at first does satisfy this. As you can see, this problem will be more fun for the presenter than the solver since it's such an annoyingly simple solution, which is why I wanted to ask if there is an example that falls outside of the solution set that I gave for some bizarre reason, and would I need a more rigorous proof that no solution can fall outside the set? I mean all I really did was set $q^p = x$, say x was rational, and solver for p in terms of q.
number-theory logarithms exponentiation rationality-testing
edited 19 hours ago
Klangen
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
8
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
You should mention $q ne 1$
– harshit54
19 hours ago
add a comment |
So some of you may be familiar with the proof that some irrational numbers to irrational powers are rational, that is: if $A = sqrt2^sqrt{2}$ then it follows that $A^sqrt{2} = 2$. So, I've found a fun brain teaser, so to speak, for people (probably everyone on here) who like this type of math problem.
Given an irrational number $q$, find the set of all
irrational numbers $p$ such that: $q^p$ is rational.
The solution that I found is the set of all $p = log_q(x):xin Q $ where Q is the field of rational numbers. The example I gave at first does satisfy this. As you can see, this problem will be more fun for the presenter than the solver since it's such an annoyingly simple solution, which is why I wanted to ask if there is an example that falls outside of the solution set that I gave for some bizarre reason, and would I need a more rigorous proof that no solution can fall outside the set? I mean all I really did was set $q^p = x$, say x was rational, and solver for p in terms of q.
number-theory logarithms exponentiation rationality-testing
edited 19 hours ago
Klangen
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
So some of you may be familiar with the proof that some irrational numbers to irrational powers are rational, that is: if $A = sqrt2^sqrt{2}$ then it follows that $A^sqrt{2} = 2$. So, I've found a fun brain teaser, so to speak, for people (probably everyone on here) who like this type of math problem.
Given an irrational number $q$, find the set of all
irrational numbers $p$ such that: $q^p$ is rational.
The solution that I found is the set of all $p = log_q(x):xin Q $ where Q is the field of rational numbers. The example I gave at first does satisfy this. As you can see, this problem will be more fun for the presenter than the solver since it's such an annoyingly simple solution, which is why I wanted to ask if there is an example that falls outside of the solution set that I gave for some bizarre reason, and would I need a more rigorous proof that no solution can fall outside the set? I mean all I really did was set $q^p = x$, say x was rational, and solver for p in terms of q.
number-theory logarithms exponentiation rationality-testing
number-theory logarithms exponentiation rationality-testing
edited 19 hours ago
Klangen
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
edited 19 hours ago
Klangen
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
edited 19 hours ago
Klangen
1,65411334
edited 19 hours ago
Klangen
1,65411334
edited 19 hours ago
Klangen
1,65411334
1,65411334
asked Mar 22 '15 at 2:04
Eben Cowley
32718
asked Mar 22 '15 at 2:04
Eben Cowley
32718
asked Mar 22 '15 at 2:04
Eben Cowley
32718
32718
8
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
You should mention $q ne 1$
– harshit54
19 hours ago
add a comment |
8
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
You should mention $q ne 1$
– harshit54
19 hours ago
8
8
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
You should mention $q ne 1$
– harshit54
19 hours ago
You should mention $q ne 1$
– harshit54
19 hours ago
add a comment |
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8
Your proposed solution pretty much says "the set of solutions is the set of solutions". It isn't humanly (I think) more comprehensible to find $log_q(x)$ than the initial problem.
– Git Gud
Mar 22 '15 at 2:07
Given that we don't even know if $e^e$ is rational, I think anybody trying to solve your problem in general is going to have a sad time.
– William Stagner
Mar 22 '15 at 2:33
Related: For which $x$ is $e^x$ rational? Transcendental?
– barto
Aug 17 '15 at 14:59
You should mention $q ne 1$
– harshit54
19 hours ago