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In my book there is a section about rational numbers.

The first example show that:6^(1/3) cannot represent a rational number. In the proof it says that the possible solutions are ±1,±2,±3,±6. Since that none of these numbers satisfies the equation x^(3)-6=0

The second example show that :b=((4-2*sqrt(3))/7)^(1/2) does not represent a rational number.

In the proof it says that the possible solutions of this equation are ±2,±2/7,±2/49,±4,±4/7,±4/49

My question is: how i arrive to the solutions like ±2,±2/7,±2/49,±4,±4/7,±4/49 given the equation?

BioShock
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  • And do you understand the first statement? To be honest, I don't understand why $\frac{1 817 120 593}{1 000 000 000}$ is not a suspected solution. If I did not know about irrational numbers or I could not count beyond $1 000 000 000$ then I would be convinced that we are talking about $6^{\frac{1}{3}}$. – zoli Apr 06 '15 at 11:04

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Set $\,x=\sqrt{\dfrac{4-2\sqrt 3\strut}7}$, and eliminate progressively the radicals. You'll obtain the equation: $$49x^4-56x^2+4=0$$ If $x$  is rational, say $x=\dfrac pq$, $p$ and $q$ coprime, we know $p$ is a divisor of $4$, and $q$ a divisor of $49$, whence the statement.

Bernard
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