Proof that Sqrt(3) is Irrational

Date: 08/14/97 at 14:03:47
From: Stephen Gardner
Subject: Proof: no Q expr. for sqrt(3)

I am a third year pure mathematics major, and I have a question 
pertaining to number theory. 

I know a proof why sqrt(2) cannot be expressed as a rational number; 
however, it involves the use of "evenness" of the denominator and 
numerator; i.e., it suggests that if sqrt(2) = a/b, and we let 
gcd(a,b) = 1; i.e., the fraction is in its reduced (simplest) form, 
then 2 = a^2 / b^2, a^2 = 2b^2, therefore 2 | a^2, therefore 2 | a.  
Then we can express a as 2c => a^2 = 4c^2 = 2b^2 => b^2 is ever => 
2 | b => contradiction.  

I was excited when I learned this proof, and naturally it led to the 
question, How does one prove that sqrt(3) is irrational? or others?  
Is there a general algorithm for proving such things?  How about just 
for primes?

Date: 08/14/97 at 19:40:01
From: Doctor Anthony
Subject: Re: Proof: no Q expr. for sqrt(3)

A proof can be conducted along much the same lines as when proving 
that sqrt(2) is irrational.

Let sqrt(3) = a/b where a and b are integers and co-prime. (fraction 
in its lowest terms)

squaring   3 = a^2/b^2    and so  a^2 = 3b^2

Now a^2 must have its factors as even powers of primes. So it must 
have 3^2 as one of its factors, and so a must have 3 as a factor.

Now 3b^2, as just seen, has a factor 3^2, this means that b^2 has a 
factor 3.  But if b^2 has a factor 3, then since powers must be even, 
it has 3^2 as a factor.  This means that b must have 3 as a factor.

So we have shown that both a and b have 3 as a factor. But we started 
by saying that a/b was a fraction in its lowest terms, (i.e. no common 
factors of a and b). Thus we have a contradiction, and it follows that 
we cannot express sqrt(3) as a rational fraction.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/