Math homework help

[aporter1]

I am trying to solve this problem:

Prove the impossibility of constructing with ruler and compass the number ∛3

[AceInfinity]

An irrational number cannot be written as a fraction... Square root 3 is irrational, that is your hint.

Imagine if (square root 3) √3 were a rational number however.

Represented this as a fraction: √3=x/b, with x and b in simplest form (no common factors). Then we can assume, b√3=x and thus 3b2=x2. This is definitely a form of "3n", therefore a multiple of 3, or 3n+1, or 3n+2.

If x=3n+1: x2=(3n+1)2=9n2+6n+1=3(3n2+2n)+1
If x=3n+2: x2=(3n+2)3=9n2+12n+1=3(3n2+4n)+1

Both of which do not represent a multiple of 3...

And because x2=3b2, x itself has to be a multiple of 3, because x2=3b2 simplified (divide both sides of the equation operator by 2) is x=3n.

We now know x2=(3n)2=9n2=3b2 and b2=3n2 are also multiples of 3. Although previously, taking into account if applied to b instead of x, you should reach the conclusion that b is also a multiple of 3, contradictiong the fact that x and b had no common factors, because both have a factor of 3.