Considering that the root is positive, it can be expressed as P/Q(P, q is a positive integer of coprime).

So p 3+AP 2q+BQ 3 = 0.

P^3 = Q^2

P if the factor aP-bQ is not included, then the equation cannot be established, so let p = (AP-bq) r.

Then (AP-bq) 2 * r 3 = q 2.

Similarly, q must also contain a factor (aP-bQ).

It is contradictory to p and q, so the root can't be misspelled in the form of p/q.

Negative roots are similar. Just change it to-p/q.

(This practice is similar to proving that √2 is irrational. )

(b) Prove that the {log2 n: n∈N} set is either an integer or an irrational number.

Certificate:

When n = 2^k, log2n is an integer,

In other cases, using the reduction to absurdity, it is assumed that log2 n can be expressed as P/Q(P, q is a coprime positive integer).

Then n q = 2 p.

N can be expressed as 2 r * t (r, t is an integer, making 2, t coprime).

So 2 rq * t q = 2 p.

But because of 2, t coprime, the equation cannot be established. Get a license.

(The writing may not be very clear, but they are all very simple mathematics, and these two questions are both number theory. )