Why is the square root of 5 an irrational number?

Question

Luna Angel · Accepted Answer

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This is a sketch of a contradiction-based proof: Suppose #sqrt(5) = p/q# for some positive integers #p# and #q#. Without loss of generality, we may suppose that #p, q# are the smallest such numbers. Therefore, by definition: #5 = (p/q)^2 = p^2/q^2# Multiply both ends by #q^2# to get: #5 q^2 = p^2# So #p^2# is divisible by #5#. Then since #5# is prime, #p# must be divisible by #5# too. So #p = 5m# for some positive integer #m#. Thus, we have: #5 q^2 = p^2 = (5m)^2 = 5*5*m^2# Divide both ends by #5# to get: #q^2 = 5 m^2# Divide both ends by #m^2# to get: #5 = q^2/m^2 = (q/m)^2# So #sqrt(5) = q/m# Now #p > q > m#, so #q, m# is a smaller pair of integers whose quotient is #sqrt(5)#, contradicting our hypothesis. So our hypothesis that #sqrt(5)# can be represented by #p/q# for some integers #p# and #q# is false. That is, #sqrt(5)# is not rational. That is, #sqrt(5)# is irrational.

Skylar Adams · Answer

undefined The square root of 5 is an irrational number because it cannot be expressed as a fraction of two integers where the denominator is not zero, and the numerator and denominator have no common factors other than 1. In other words, the decimal expansion of the square root of 5 neither terminates nor repeats in a pattern. This property is characteristic of irrational numbers.