Question

Prove that root 3 is irrational

Answer 1

Say 3–√3 is rational. Then 3–√3 can be represented as abab, where a and b have no common factors.

So 3=a2b23=a2b2 and 3b2=a23b2=a2. Now a2a2 must be divisible by 33, but then so must aa(fundamental theorem of arithmetic). So we have 3b2=(3k)23b2=(3k)2 and 3b2=9k23b2=9k2 or even b2=3k2b2=3k2 and now we have a contradiction.

Answer 2

First notice that 3 is a prime number. Thus if  a2a2is divible by 3,  aa must also be divisible by 3.

Proof by contradiction:
Let's assume that 3–√3 is rational. i.e
 3–√=m/n3=m/n, where m and n have no common factors.

Multiplying both side by nn and squaring we get.
 3n2=m23n2=m2

Since lhs is divisible by 3, rhs should also be divisible by 3. Thus, we can write mm as 3p3p.

 3n2=(3p)23n2=(3p)2
=> 3n2=9p23n2=9p2
=> n2=3p2n2=3p2

Since rhs is divisible by 3, lhs should also be divisible by 3. Thus  nn is divisible by 3.

Both  n,mn,m are divisible by 3, contradicting our original assumption that m and n are co-prime.

Thus, 3–√3 is irrational. Hence Proved.