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Old 2018-09-06, 13:02   #1
jnml
 
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Default Rationality of an expression

Given $$x \in \mathbb{Q}, x \neq 0,$$ can or cannot $$\sqrt[3]{x^3+1} \in \mathbb{Q}$$ be true?
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Old 2018-09-06, 13:12   #2
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Let x=1, then 2^{1/3} is irrational.

FLT might have something to do with the stated problem.

Last fiddled with by paulunderwood on 2018-09-06 at 13:18
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Old 2018-09-06, 13:48   #3
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fermat's last theorem

EDIT:- Loophole abuse - x=-1

Last fiddled with by axn on 2018-09-06 at 14:00
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Old 2018-09-06, 14:51   #4
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Say \(x\in\mathbb{Q}\) not 0, then \(x\) can be written \(\frac ab\), so you have \(\ ^3\sqrt{(\frac ab)^3+1}=\ ^3\sqrt{\frac{a^3}{b^3}+1}=\ ^3\sqrt{\frac{a^3+b^3}{b^3}} = \frac{\ ^3\sqrt{a^3+b^3}}{b}\) which is rational, so it is \(=\frac mn\), therefore \(\ ^3\sqrt{a^3+b^3}=\frac{bm}{n}\), and renaming the last with \(c\), we have \(\ ^3\sqrt{a^3+b^3}=c\) or \(a^3+b^3=c^3\) and as Paul said, we just found a counterexample for FLT.



Therefore, that can never be true.

Last fiddled with by LaurV on 2018-09-06 at 14:53
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Old 2018-09-06, 15:13   #5
R. Gerbicz
 
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Quote:
Originally Posted by axn View Post
EDIT:- Loophole abuse - x=-1
and there is no more solution.
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Old 2018-09-06, 15:27   #6
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Quote:
Originally Posted by LaurV View Post
... =\ ^3\sqrt{\frac{a^3+b^3}{b^3}} = \frac{\ ^3\sqrt{a^3+b^3}}{b} which is rational ...
Did you mean irrational?
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Old 2018-09-06, 15:47   #7
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Quote:
Originally Posted by jnml View Post
Did you mean irrational?
It is presumed rational until the contradiction by FLT is established.
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Old 2018-09-06, 15:57   #8
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Quote:
Originally Posted by paulunderwood View Post
It is presumed rational until the contradiction by FLT is established.
I see, thanks. I did not assume that particular implicit assumption ;-)
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