If p is a prime then √p is irrational

To prove √p is irrational, where p is a prime, we will need the following theorem:

Theorem: If p is prime then p|ab, then p|a or p|b.

Proof: If p|a, we are done. So let us assume that pmid a. Therefore, gcd(p,a)=1. Hence, by Euclid’s lemma, p|b. [Euclid’s lemma: If a|bc, with gcd(a,b)=1, then a|c.]

 

Now to prove √p is irrational (where p is a prime):

Assume that √p is rational.

Then there exists two integers a, b such that frac{a}{b}=sqrt{p} where gcd(a,b)=1 and bne 0.

Then frac{a^2}{b^2}=p. Which implies

a^2=pb^2. __________(1)

Rightarrow p divides a^2.

Rightarrow p divides a. (By the above theorem.)

So there exists an integer a_1 such that a=pa_1.

So, from (1), we get (pa_1)^2=pb^2.

Rightarrow p^2a_1^2=pb^2.

Rightarrow pa_1^2=b^2. (Dividing both sides by p.)

This implies p divides b. This is a contradiction since gcd(a,b)=1.

Thus, when p is a prime, p is irrational.

 

Particular Case:- √2 is irrational.

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Pankaj Jyoti Mahanta

READ:   Amalendu Krishna and Naveen Garg awarded the 2016 Shanti Swarup Bhatnagar Prize for Mathematical Sciences
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