Question

1. show that √n is irrational if n is prime or not a perfect square.
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Answer 1

Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.

We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So n−−√=ab, where a,b∈Z+,b≠0. We also suppose that a≠0, otherwise ab=0 , and n will be a square number, which is rational.

Hence n=a2b2, so nb2=a2.

Suppose b=1. Then n−−√=a , which shows that n is a square number. So b≠1. Since n−−√>1, then a>b>1.

By the unique factorization of integers theorem, every positive integer greater than 1 can be expressed as the product of its primes. Therefore, we can write a as a product of primes and for every prime number that exists in a, there will be an even number of primes in a2. Similarly, we can express b as a product of primes and for every prime number that exists in b, there will be an even number of primes in b2.

However, we can also express n as a product of primes. Since n is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of nb2 that has an odd number of primes. Since nb2=a2 , then this contradicts the fact that there is an even number of primes in a2 since a number can neither be even and odd.

Therefore, this contradicts the fact that n−−√ is rational. Therefore, n−−√ must be irrational.