1. ## Existence proofs

Couple of problems i can't think through looking for some help...

Prove there exists a rational number a and an irrational number b such that a^b is irrational.

I was thinking that I could prove they were both irrational, taking square root two^square root two as A and square root two as B and it works. can't put my finger on this one.

Last one disprove statement there is a real number x such that x^6+x^4+1=2x^2.

Thank you for any help

2. Originally Posted by j5sawicki
Couple of problems i can't think through looking for some help...

Prove there exists a rational number a and an irrational number b such that a^b is irrational.

I was thinking that I could prove they were both irrational, taking square root two^square root two as A and square root two as B and it works. can't put my finger on this one.

Last one disprove statement there is a real number x such that x^6+x^4+1=2x^2.

Thank you for any help

$\sqrt{2}\mbox{ is irrational and } w=\sqrt{2}^{\sqrt{2}}$ $\mbox { is either rational or irrational, but } w^{\sqrt{2}} \mbox{ de{}finitely is rational}$

Tonio

3. Originally Posted by j5sawicki
Couple of problems i can't think through looking for some help...

Prove there exists a rational number a and an irrational number b such that a^b is irrational.

I was thinking that I could prove they were both irrational, taking square root two^square root two as A and square root two as B and it works. can't put my finger on this one.

Last one disprove statement there is a real number x such that x^6+x^4+1=2x^2.

Thank you for any help
For the first part, read what tonio wrote he is right; you are probably confusing yourself immensely.

For the second part take a look at the graph of f(x) = x^6 + x^4 -2x^2 + 1

notice that f(x) is always > 0. You can prove this by looking at f'(x) and f''(x). After you find f''(x) find the points where it is 0, then the function is decreasing or increasing in the intervals between the critical points, based on the sign of f'(x) which you can prove from f''(x). This problem is pretty trivial in the way you prove it, but its very ugly.

4. Hello tonio
Originally Posted by tonio
$\sqrt{2}\mbox{ is irrational and } w=\sqrt{2}^{\sqrt{2}}$ $\mbox { is either rational or irrational, but } w^{\sqrt{2}} \mbox{ de{}finitely is rational}$

Tonio
Sorry, I'm perhaps being very slow here. I can't really see how this helps. Are you suggesting a value for $a$ and a value for $b$? If so, what values?

5. For the first problem, Tonio had the right idea, and so do you

Consider $\sqrt{2}^{\sqrt{2}}$. If it was rational, what would that imply?

If it was irrational, what would that imply?

For the second one, gmatt's solution will work. However, there is a better way: (answer in spoiler)

Spoiler:
$x^6 + (x^4 - 2x^2 + 1) = x^6 + (x^2 - 1)^2$
As the sum of two squares, this will always be positive, meaning the function has no real roots.

6. Originally Posted by Turiski
For the first problem, Tonio had the right idea, and so do you

Consider $\sqrt{2}^{\sqrt{2}}$. If it was rational, what would that imply?

If it was irrational, what would that imply?

For the second one, gmatt's solution will work. However, there is a better way: (answer in spoiler)

Spoiler:
$x^6 + (x^4 - 2x^2 + 1) = x^6 + (x^2 - 1)^2$
As the sum of two squares, this will always be positive, meaning the function has no real roots.
Really nice approach to problem 1. Thanks very much.

Infact from the same approach - we can show that
there exist a - rational and b - irrational, such that a^b is rational.

Am I correct?

Must admit, this was very conterintutive (to me at least)

7. Originally Posted by Turiski
For the first problem, Tonio had the right idea, and so do you

Consider $\sqrt{2}^{\sqrt{2}}$. If it was rational, what would that imply?

If it was irrational, what would that imply?
...
OK, but this is incomplete. If $\sqrt2^{\sqrt2}$ is irrational (and, although I assume it is, how do we prove it?), all it proves is that there exist $a\, (=\sqrt2^{\sqrt2})$ and $b\,(=\sqrt2)$, both irrational, for which $a^b$ is rational. Which is not what the question asks for. We need to establish the existence of a rational, $a$, and an irrational, $b$, for which $a^b$ is irrational.

However, even if we can establish that $\sqrt2^{\sqrt2}$ is irrational, we would still need to go on to say:
We have found $a,\, b$ such that $a^b = c$, where $a,b \notin \mathbb{Q}$ and $c \in \mathbb{Q}$

Now $a^b=c\Rightarrow a = c^{\frac1b}$
But if $b \notin \mathbb{Q}$, then $\frac1b \notin \mathbb{Q}$. So now we have the required result.

So, how about it? Why is $\sqrt2^{\sqrt2}$ irrational? And would it be just as easy to prove that $2^{\sqrt2}$ is irrational, thus establishing the result much more simply, with $a = 2,\, b =\sqrt2$?

Or am I still missing something obvious? If so, please spell it out for me.

8. I think Grandad is absolutely right. What some posters were showing is that there exist two irrational numbers $a, b$ so that $a^b$ is rational. The original question was about a rational $a$.

In fact, $\sqrt{2}^{\sqrt{2}}$ is irrational (see Gelfond–Schneider theorem in Wikipedia), but the proof is in no way easy. Anyway, using Grandad's addition to the argument is one way to give a complete solution.

As for the second problem, Turiski has a nice idea, except that the sum of two squares is not always positive. It is easy to show, though, that at least one of the summands is non-zero.

Evgeny

9. ## First problem

We've already seen the proof of $\exists a, b\notin\mathbb{Q}.\,a^b\in\mathbb{Q}$. In it we considered three numbers, all of which are equal to $\sqrt{2}$.

What if we apply the same reasoning to the following three numbers: 2, $\sqrt{2}$, and $1/(2\sqrt{2})$?

Evgeny

OK, but this is incomplete. If $\sqrt2^{\sqrt2}$ is irrational (and, although I assume it is, how do we prove it?), all it proves is that there exist $a\, (=\sqrt2^{\sqrt2})$ and $b\,(=\sqrt2)$, both irrational, for which $a^b$ is rational. Which is not what the question asks for. We need to establish the existence of a rational, $a$, and an irrational, $b$, for which $a^b$ is irrational.

However, even if we can establish that $\sqrt2^{\sqrt2}$ is irrational, we would still need to go on to say:
We have found $a,\, b$ such that $a^b = c$, where $a,b \notin \mathbb{Q}$ and $c \in \mathbb{Q}$

Now $a^b=c\Rightarrow a = c^{\frac1b}$
But if $b \notin \mathbb{Q}$, then $\frac1b \notin \mathbb{Q}$. So now we have the required result.

So, how about it? Why is $\sqrt2^{\sqrt2}$ irrational? And would it be just as easy to prove that $2^{\sqrt2}$ is irrational, thus establishing the result much more simply, with $a = 2,\, b =\sqrt2$?

Or am I still missing something obvious? If so, please spell it out for me.

Hi Grandad - Let try to explain.

Let $\sqrt2^{\sqrt2}$ = x

Case 1: x is irrational

Consider $\sqrt2^{{\sqrt2}^{\sqrt2}}=2$.
We can write it as
$2^{x/2}$

Case 2: x is rational

Consider $\sqrt2^{{\sqrt2}^{\sqrt2}}=2$.
We can write it as
$x^{\sqrt2}$

Hence we are done in both cases.

11. Hello aman_cc
Originally Posted by aman_cc
Hi Grandad - Let try to explain.

Let $\sqrt2^{\sqrt2}$ = x

Case 1: x is irrational

Consider $\sqrt2^{{\sqrt2}^{\sqrt2}}=2$.
We can write it as
$2^{x/2}$

Case 2: x is rational

Consider $\sqrt2^{{\sqrt2}^{\sqrt2}}=2$.
We can write it as
$x^{\sqrt2}$

Hence we are done in both cases.
OK, but I think you're still missing the point, which is that in each case you end up with a rational number, namely 2. Which, I repeat, is not what the question asks for!

Case 1 is then OK, because you can then go on to re-arrange in the way I indicated in my last reply.

But Case 2...?

Hello aman_ccOK, but I think you're still missing the point, which is that in each case you end up with a rational number, namely 2. Which, I repeat, is not what the question asks for!

Case 1 is then OK, because you can then go on to re-arrange in the way I indicated in my last reply.

But Case 2...?

Hi - Sorry I missed the original question. But we can so it the same way.

Let $\sqrt2^{\sqrt2}=x$

case 1: x is irrational
x = $\sqrt2^{\sqrt2}$= $2^{\sqrt2/2}$

case 2: x is rational

Consider $(\sqrt2^{{\sqrt2}^{\sqrt2}})^{1/2}=(\sqrt2^{\sqrt2})^{\sqrt2/2}=\sqrt2
$

Note: I have used that $\sqrt2^{{\sqrt2}^{\sqrt2}}=2$

13. I think this works. Just a remark: one should be careful about notation. As far as I know, power operation associates to the right, so $\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}=\sqrt{2}^{\left(\sq rt{2}^{\sqrt{2}}\right)}
\ne\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=2$
.

For example, you wrote earlier:
Let = x... Consider . We can write it as
Here, for $\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$ to be equal to 2, the power operation has to associate to the left, but for it to be equal to $2^{x/2}$, power has to associate to the right.

14. Originally Posted by emakarov
I think this works. Just a remark: one should be careful about notation. As far as I know, power operation associates to the right, so $\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}=\sqrt{2}^{\left(\sq rt{2}^{\sqrt{2}}\right)}
\ne\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=2$
.

For example, you wrote earlier:
Here, for $\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$ to be equal to 2, the power operation has to associate to the left, but for it to be equal to $2^{x/2}$, power has to associate to the right.
Yes you are correct. Sorry not too conversant with Latex. When I wrote
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$
I meant $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}$

Thanks for pointing out

15. Hello everyone

Originally Posted by j5sawicki
...Prove there exists a rational number a and an irrational number b such that a^b is irrational...
The problem as I see it with all the solutions so far is that none has spelt out the values of $a$ and $b$ with any clarity. This may have been deliberate on the part of those offering the solutions, since they preferred to leave the rest of us to sort it out for ourselves. But I suspect in some cases, it has been due to an inadequate reading of the question.

So, for my own benefit, and for the benefit of anyone else who has been bewildered by all this, may I amplify (and hopefully clarify) aman_cc's final solution (which is correct). Here it is again, with my additions
in red.

Let $x = \sqrt2^{\sqrt2}$.

We do not know whether $\color{red}x$ is rational or irrational, but in either case we can formulate a solution.

Case 1:
$x$ is irrational

$x = \sqrt2^{\sqrt2}\color{red}=\Big(2^{\tfrac12}\Big)^ {\sqrt2}\color{black}=(2)^{({\frac{\sqrt2}{2})}}$

So this gives the solution $\color{red}a = 2$ (rational), $\color{red}b=\frac{\sqrt2}{2}$ (irrational), $\color{red}a^b = x$ (irrational, by assumption)

Case 2:
$x$ is rational

Note first that
$\color{red}x^{\sqrt2}=(\sqrt2^{\sqrt2})^{\sqrt2}=( \sqrt2)^{\sqrt2\times\sqrt2}=(\sqrt2)^2 = 2$

Consider
$\Big((\sqrt2^{\sqrt2})^{\sqrt2}\Big)^{\frac12}\col or{red}=\Big((x)^{\sqrt2}\Big)^{\frac12}=2^{\frac1 2}\color{black}=\Big((\sqrt2^{\sqrt2})\Big)^{\frac {\sqrt2}{2}}=\sqrt2$

So we now have the solution
$\color{red}a=x$ (rational, by assumption), $\color{red}b= \frac{\sqrt2}{2}$ (irrational), $\color{red}a^b = \sqrt2$ (irrational).

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