How do you find the set in which the real number `sqrt93` belongs?

The prime factorisation of `93` is:

`93 = 3*31`

Since not all of the prime factors occur an even number of times (in fact neither of them do), `sqrt(93)` is irrational.

One way of proving that might go as follows:

Suppose `sqrt(93) = p/q` for a pair of positive integers `p, q` with `p > q`.

Without loss of generality we can assume that this is the smallest such pair of integers.

Then:

`(p/q)^2 = 93`

So:

`p^2 = 93 q^2`

Hence `p^2` is divisible by both `3` and `31`.

Since `3` and `31` are prime, `p` must also be divisible by `3` and `31` (since factorisation into a product of primes is unique up to a unit).

Hence:

`p = 93 k` for some integer `k` and we find:

`93q^2 = p^2 = (93 k)^2 = 93^2 k^2`

Then dividing both ends by `93` we find:

`q^2 = 93 k^2`

Hence:

`(q/k)^2 = 93`

So `sqrt(93) = q/k`

Now `p > q > k`, so `q, k` are a smaller pair of integers whose quotient is `sqrt(93)`, contradicting our assertion that `p, q` was the smallest such pair.

Hence there is no such pair of integers and `sqrt(93)` is therefore irrational.

`sqrt(93)` is also an algebraic number. That is, it is the root of a polynomial equation with rational coefficients, namely:

`x^2-93 = 0`

Some irrational numbers are algebraic, but most are transcendental. Numbers like `e` and `pi` are transcendental: They satisfy no polynomial equation with rational coefficients.