Is the following a rational number?

• Dec 3rd 2009, 03:00 PM
Pinkk
Is the following a rational number?
Is $x = (\sqrt{2 + 3^{1/5}} + 7^{1/3})^{1/6}$ a rational number? Prove your answer.

I know it isn't and that I need to rewrite $x$ in the form $a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} + a_{n}x^{n} = 0$, but I'm having trouble doing so.
• Dec 3rd 2009, 05:09 PM
Drexel28
Quote:

Originally Posted by Pinkk
Is $x = (\sqrt{2 + 3^{1/5}} + 7^{1/3})^{1/6}$ a rational number? Prove your answer.

I know it isn't and that I need to rewrite $x$ in the form $a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} + a_{n}x^{n} = 0$, but I'm having trouble doing so.

Suppose that $x\in\mathbb{Q}$, since the rationals are closed under multiplcation so is $x^6=\sqrt{2+\sqrt[5]{3}}+\sqrt[3]{7}$. Clearly though $\sqrt[3]{7}\notin\mathbb{Q}$ once again if $\sqrt{2+\sqrt[5]{3}}\in\mathbb{Q}\implies 2+\sqrt[5]{3}\in\mathbb{Q}$ which is a contradiction. Lastly seeing that the irrational part of $\sqrt{2+\sqrt[5]{3}}$ does not cancel with $\sqrt[3]{7}$ we see that $x^6\notin \mathbb{Q}$ which contradicts $x\in\mathbb{Q}$
• Dec 3rd 2009, 05:40 PM
Drexel28
• Dec 3rd 2009, 06:04 PM
Pinkk
Quote:

Originally Posted by Drexel28
Suppose that $x\in\mathbb{Q}$, since the rationals are closed under multiplcation so is $x^6=\sqrt{2+\sqrt[5]{3}}+\sqrt[3]{7}$. Clearly though $\sqrt[3]{7}\notin\mathbb{Q}$ once again if $\sqrt{2+\sqrt[5]{3}}\in\mathbb{Q}\implies 2+\sqrt[5]{3}\in\mathbb{Q}$ which is a contradiction. Lastly seeing that the irrational part of $\sqrt{2+\sqrt[5]{3}}$ does not cancel with $\sqrt[3]{7}$ we see that $x^6\notin \mathbb{Q}$ which contradicts $x\in\mathbb{Q}$

I follow this up to your last statement that the two terms don't cancel out. It's obvious, but how can I support that claim in a proof?
• Dec 3rd 2009, 06:06 PM
Drexel28
Quote:

Originally Posted by Pinkk
I follow this up to your last statement that the two terms don't cancel out. It's obvious, but how can I support that claim in a proof?

Tell me what you think.
• Dec 3rd 2009, 06:28 PM
Pinkk
Hmm, well, assuming that $\sqrt{2+3^{1/5}} + 7^{1/3} = r,\,r \in \mathbb{Q}$, then
$2 + 3^{1/5} = r^{2} - 2r(7^{1/3}) + 7^{2/3}$
$343(2 + 3^{1/5}) = 343r^{2} - 14r + 49$, which is a contradiction since the product of a rational and an irrational cannot be rational.
• Dec 4th 2009, 08:56 PM
Pinkk
Argh, I made a careless mistake, so what I posted above is clearly wrong. I am stuck on how to show the sum of those two irrationals must be irrational. (Headbang)(Headbang)(Headbang)
• Dec 4th 2009, 09:07 PM
Sampras
Quote:

Originally Posted by Pinkk
Argh, I made a careless mistake, so what I posted above is clearly wrong. I am stuck on how to show the sum of those two irrationals must be irrational. (Headbang)(Headbang)(Headbang)

Counterexample: $(6-\pi)+ \pi = 6$.
• Dec 4th 2009, 09:10 PM
Pinkk
Well yes, which is exactly why I don't know how to make the conclusion that $x^6=\sqrt{2+\sqrt[5]{3}}+\sqrt[3]{7}$ is irrational.
• Dec 4th 2009, 09:17 PM
Drexel28
Quote:

Originally Posted by Pinkk
Hmm, well, assuming that $\sqrt{2+3^{1/5}} + 7^{1/3} = r,\,r \in \mathbb{Q}$, then
$2 + 3^{1/5} = r^{2} - 2r(7^{1/3}) + 7^{2/3}$
$343(2 + 3^{1/5}) = 343r^{2} - 14r + 49$, which is a contradiction since the product of a rational and an irrational cannot be rational.

Quote:

Originally Posted by Pinkk
Argh, I made a careless mistake, so what I posted above is clearly wrong. I am stuck on how to show the sum of those two irrationals must be irrational. (Headbang)(Headbang)(Headbang)

Maybe a hint? I half-way worked through it and it seems as though you should be able to prove this by noting that $\sqrt{2+\sqrt[5]{3}}+\sqrt[3]{7}=r\implies \left(r-\sqrt{2+\sqrt[5]{3}}\right)^3\in\mathbb{Z}$
• Dec 5th 2009, 07:03 AM
Pinkk
Bleh, I can't see how that helps. (Headbang)

Any suggestions please? I'm sorry for the excessive posting but I really need to know how to show a formal proof for this.
• Dec 6th 2009, 01:13 PM
evry1cndie
Hey Pinkk, I've gone as far as expanding the polynomial to x^72, but still have some ways to go in order to use the rational roots theorem. I'm just as clueless as to how the hell to work this out any other way, seeing as we can't assume that the sum of two irrationals is going to be irrational. I think there was some sort of underestimation as to how difficult this problem really is using the knowlege we have. I think it's just a combination of a few problems from the yellow book and was figured that it couldn't be too much harder than the others... i dunno, let me know if you make any progress, i've just about given up.
• Dec 6th 2009, 01:38 PM
Pinkk
Nope, I've given up on it, it's just way too tedious and there's no guarantee that approach will work.

Edit: A professor at our college provided hints/solutions for these problems:
http://mathbin.net/38158
http://mathbin.net/38159
http://mathbin.net/38160