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Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is...
$begingroup$
In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = sqrt 2$, so that $x^y = 2^sqrt 2$.
Why is this not enough? How come I have to go through the case whether $x^y = 2^sqrt 2$ is rational?
If $2^sqrt 2$ is rational than let $x = 2^sqrt 2$, and $y = sqrt 2 / 4$
$x^y = (2^sqrt 2)^{sqrt 2 /4} = 2^{(sqrt 2*sqrt 2) /4} = 2^{2/4} = 2^{1/2} = sqrt 2$ (previous value for y that was established as irrational.
proof-writing exponentiation irrational-numbers rational-numbers
asked Jan 20 at 21:47
ElliottElliott
596
$endgroup$
|
show 8 more comments
$begingroup$
In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = sqrt 2$, so that $x^y = 2^sqrt 2$.
Why is this not enough? How come I have to go through the case whether $x^y = 2^sqrt 2$ is rational?
If $2^sqrt 2$ is rational than let $x = 2^sqrt 2$, and $y = sqrt 2 / 4$
$x^y = (2^sqrt 2)^{sqrt 2 /4} = 2^{(sqrt 2*sqrt 2) /4} = 2^{2/4} = 2^{1/2} = sqrt 2$ (previous value for y that was established as irrational.
proof-writing exponentiation irrational-numbers rational-numbers
asked Jan 20 at 21:47
ElliottElliott
596
$endgroup$
6
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
2
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
4
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03
|
show 8 more comments
$begingroup$
In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = sqrt 2$, so that $x^y = 2^sqrt 2$.
Why is this not enough? How come I have to go through the case whether $x^y = 2^sqrt 2$ is rational?
If $2^sqrt 2$ is rational than let $x = 2^sqrt 2$, and $y = sqrt 2 / 4$
$x^y = (2^sqrt 2)^{sqrt 2 /4} = 2^{(sqrt 2*sqrt 2) /4} = 2^{2/4} = 2^{1/2} = sqrt 2$ (previous value for y that was established as irrational.
proof-writing exponentiation irrational-numbers rational-numbers
asked Jan 20 at 21:47
ElliottElliott
596
$endgroup$
In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = sqrt 2$, so that $x^y = 2^sqrt 2$.
Why is this not enough? How come I have to go through the case whether $x^y = 2^sqrt 2$ is rational?
If $2^sqrt 2$ is rational than let $x = 2^sqrt 2$, and $y = sqrt 2 / 4$
$x^y = (2^sqrt 2)^{sqrt 2 /4} = 2^{(sqrt 2*sqrt 2) /4} = 2^{2/4} = 2^{1/2} = sqrt 2$ (previous value for y that was established as irrational.
proof-writing exponentiation irrational-numbers rational-numbers
proof-writing exponentiation irrational-numbers rational-numbers
asked Jan 20 at 21:47
ElliottElliott
596
asked Jan 20 at 21:47
ElliottElliott
596
edited Jan 22 at 4:32
Elliott
asked Jan 20 at 21:47
ElliottElliott
596
asked Jan 20 at 21:47
ElliottElliott
596
asked Jan 20 at 21:47
ElliottElliott
596
596
6
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
2
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
4
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03
|
show 8 more comments
6
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
2
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
4
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03
6
6
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
2
2
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
4
4
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03
|
show 8 more comments
2 Answers
2
active
oldest
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$begingroup$
Let $mathbb{I}$ denote the set of irrational numbers.
Function $f : mathbb{I} rightarrow mathbb{R} : x mapsto 2^x$ is an injection.
$mathbb{I}$ is uncountable $Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.
answered Jan 20 at 22:42
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $log x = b log a$ or $b =frac {log x}{log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).
Now the expressions $a^b$ where $ain mathbb Q^+; bin mathbb Q$ are countable and we can choose $xin mathbb R^+$ which is neither in the set of such expressions nor in $mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $aneq 1$ and the corresponding $b$ must be irrational.
This is a non-constructive proof, which shows that there will be lots of examples.
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
Let $mathbb{I}$ denote the set of irrational numbers.
Function $f : mathbb{I} rightarrow mathbb{R} : x mapsto 2^x$ is an injection.
$mathbb{I}$ is uncountable $Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.
answered Jan 20 at 22:42
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Let $mathbb{I}$ denote the set of irrational numbers.
Function $f : mathbb{I} rightarrow mathbb{R} : x mapsto 2^x$ is an injection.
$mathbb{I}$ is uncountable $Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.
answered Jan 20 at 22:42
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Let $mathbb{I}$ denote the set of irrational numbers.
Function $f : mathbb{I} rightarrow mathbb{R} : x mapsto 2^x$ is an injection.
$mathbb{I}$ is uncountable $Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.
answered Jan 20 at 22:42
ZeeklessZeekless
577111
$endgroup$
Let $mathbb{I}$ denote the set of irrational numbers.
Function $f : mathbb{I} rightarrow mathbb{R} : x mapsto 2^x$ is an injection.
$mathbb{I}$ is uncountable $Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.
answered Jan 20 at 22:42
ZeeklessZeekless
577111
answered Jan 20 at 22:42
ZeeklessZeekless
577111
answered Jan 20 at 22:42
ZeeklessZeekless
577111
answered Jan 20 at 22:42
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
$begingroup$
If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $log x = b log a$ or $b =frac {log x}{log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).
Now the expressions $a^b$ where $ain mathbb Q^+; bin mathbb Q$ are countable and we can choose $xin mathbb R^+$ which is neither in the set of such expressions nor in $mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $aneq 1$ and the corresponding $b$ must be irrational.
This is a non-constructive proof, which shows that there will be lots of examples.
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
$endgroup$
add a comment |
$begingroup$
If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $log x = b log a$ or $b =frac {log x}{log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).
Now the expressions $a^b$ where $ain mathbb Q^+; bin mathbb Q$ are countable and we can choose $xin mathbb R^+$ which is neither in the set of such expressions nor in $mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $aneq 1$ and the corresponding $b$ must be irrational.
This is a non-constructive proof, which shows that there will be lots of examples.
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
$endgroup$
add a comment |
$begingroup$
If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $log x = b log a$ or $b =frac {log x}{log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).
Now the expressions $a^b$ where $ain mathbb Q^+; bin mathbb Q$ are countable and we can choose $xin mathbb R^+$ which is neither in the set of such expressions nor in $mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $aneq 1$ and the corresponding $b$ must be irrational.
This is a non-constructive proof, which shows that there will be lots of examples.
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
$endgroup$
If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $log x = b log a$ or $b =frac {log x}{log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).
Now the expressions $a^b$ where $ain mathbb Q^+; bin mathbb Q$ are countable and we can choose $xin mathbb R^+$ which is neither in the set of such expressions nor in $mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $aneq 1$ and the corresponding $b$ must be irrational.
This is a non-constructive proof, which shows that there will be lots of examples.
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
answered Jan 20 at 22:42
Mark BennetMark Bennet
81.4k983180
81.4k983180
add a comment |
add a comment |
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6
$begingroup$
And how do you know that $2^{sqrt{2}}$ is irrational?
$endgroup$
– Zeekless
Jan 20 at 21:50
$begingroup$
Elliott and @Zeekless: Oh... it isn't rational.
$endgroup$
– David G. Stork
Jan 20 at 21:50
$begingroup$
If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $left(2^sqrt2right)^sqrt2$
$endgroup$
– Ivan Neretin
Jan 20 at 21:54
2
$begingroup$
@David G. Stork, you misunderstand the question. Your last edit of the title is wrong.
$endgroup$
– Zeekless
Jan 20 at 21:55
4
$begingroup$
The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof.
$endgroup$
– Ross Millikan
Jan 20 at 22:03