Prove the  is irrational.

Proof Number 1:

By contradiction, assume  is rational.   s.t.   gcd(p,q)=1.  Then

q=p.  Squaring both sides: 3q²=p².  Since 3|p² then 3|p (which was proven in class).

Since 3|q² then 3|q.  Therefore, gcd(p,q)=1 is contradicted so  must be irrational.

Proof Number 2:

The easiest way to prove  is irrational is through contradiction.  Assume  is rational.  In mathematics notation this means  where  and q≠0.  Because the fraction is in its most reduced form, gcd(p,q)=1.  Squaring the equation results in  or p²=3q².  If 3|p², then 3|p (given).  If 3|p there is an integer k, s.t. p=3k.  Substituting back into the equation, (3k)²=3q² or 9k²=3q².  Simplified once more, q²=3k².  This means 3 also divides q².  Because the gcd(p,q) ≠1,  must be irrational (by contradiction.)

Proof Number 3:

By contradiction.  Let p be irrational and p²=3.  If p is rational #  a “r” and “q” s.t.   So if p²=3 and  then   If r²=3q² then 3|r².  By the fact above (if 3|n² then 3|n) then 3|r which means  some integer k s.t. r=3k.  If r=3k and r²=3q² then (3k)²=3q².  So 3k²=q² which means that 3|q² then by fact from above again 3|q.  If 3|q then  some integer j s.t. 3j=q.  Since 3 divides both q & r the q & r are integers so  which means p or  is irrational.