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

1 Answer
Nov 29, 2016

Answer:

#sqrt(93)# is an algebraic irrational number.

Explanation:

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.