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Prove that \sqrt{2} is an irrational number.

 

 
 
 
 
 

Answers (1)

Lets assume \sqrt{2} is a rational number.

For rational number,

it should be in a form of p/q where q\neq 0 and HCF of P and q should be 1 in their least terms.

So,

\sqrt{2} = p/q.

After squaring both sides

2=\frac{p^2}{q^2}

p^2=2q^2 ---(1)

From the above eqn, 2 is the divisor of 2q^2

Here 2 is a factor of p^2   \left (\because p^2=2q^2 \right )

and atlast we can say 2 is a factor of p.

Let p=2a for all a (where a is a positive integer)

Squaring both sides,

p^2=4a^2 ---(2)

From eqn(1) and eqn(2)

2q^2=4a^2

q^2=2a^2

From this, 2 is a divisor of 2a^2

So 2 is a divisor of q^2\: \: \left ( \because q^2=2a^2 \right )

We can conclude 2 is a factor of q,

We have seen that 2 is a factor of p as well as q which is against the condition HCF(p,q) =1.

Therefore there is a contradiction because our assumption is wrong.

Hence \sqrt{2} is an irrational number.

Posted by

Safeer PP

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