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

Question

Hazel Anglin · Accepted Answer

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

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. An approach to demonstrating that could be as follows: Suppose #sqrt(93) = p/q# for a pair of positive integers #p, q# with #p > q#. We can presume that this is the smallest such pair of integers without losing generality. Next: #(p/q)^2 = 93# Thus: #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). Thus: #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# Thus: #(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.

Eva Abramson · Answer

undefined The real number $\sqrt{93}$ belongs to the set of irrational numbers.