How to prove that √3 is an irrational number ?

Assume #sqrt3# is rational

#:. p/q =sqrt3 :{p,q} in ZZ, q!=0#

Also assume that #p and q# are in the simplest form (coprime) such that the Highest Common Factor #{p, q} =1#

Cross multiply:

#p=sqrt3 q#

Square both sides:

#p^2= 3q^2# [A]

#:. q^2 = p^2/3#

This implies that #3|p^2# (i.e. 3 divides #p^2#)

Since #3|p^2# then #3|p#

I.e #3# is a factor of #p#

So, we can write #p=3k# where #k# is some constant.

Replace #p=3k# in [A] above:

#(3k)^2 = 3q^2#

#9k^2 = 3q^2#

#3k^2 = q^2#

#:. k^2 = q^2/3 -> 3|q^2 -> 3|q#

Hence, #3# is also a factor of #q#

Now, since #3# is a factor of both #p and q# they cannot be coprime (as their HCF is 3 rather than 1)

Hence our assumption that #sqrt3# is rational has led to a contradiction.

Therefore we must conclude that #sqrt3# is irrational.

Here is a hint from " Mathematical Proofs " by Gary Chartrand, Albert Polimeni and Ping Zhang: page 90, Exercise 5.10 in what must be the 1st edition (green cover):

"Prove that #sqrt(3)# is irrational. [Hint: First prove that for an integer #a#, #3|a^2 iff 3|a#. Recall that every integer can be written as #3q, 3q+1# or #3q + 2# for some integer #q#.]"

Plainer English, "First prove that for an integer #a#, #3# divides #a^2# if and only if #3# divides #a#".

Here's another proof, based on continued fractions.

First note that the continued fraction of any rational number will terminate.

Next, suppose #x > 0# satisfies:

#x = 1+1/(1+1/(1+x))#

#color(white)(x) = 1+(x+1)/(x+2)#

#color(white)(x) = (2x+3)/(x+2)#

Then multiplying both ends by #(x+2)# we see:

#x^2+2x = 2x+3#

Hence:

#x^2=3#

So:

#x = sqrt(3)#

So we have found:

#sqrt(3) = 1+1/(1+1/(1+sqrt(3)))#

#color(white)(sqrt(3)) = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+...))))))#

Then since the continued fraction does not terminate, it cannot represent a rational number.