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

##### 1 Answer

#### Explanation:

The prime factorisation of

#93 = 3*31#

Since not all of the prime factors occur an even number of times (in fact neither of them do),

One way of proving that might go as follows:

Suppose

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

Since

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

#q^2 = 93 k^2#

Hence:

#(q/k)^2 = 93#

So

Now

Hence there is no such pair of integers and

#x^2-93 = 0#

Some irrational numbers are algebraic, but most are transcendental. Numbers like