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?
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.