Question

prove that √2 is an irrational number ?
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Answer 1

Answer:

Let √2 be a rational number 

Therefore, √2= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 

                   p²= 2q²                                                                                    ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p²                                                                    [since, 2q²=p²]

⇒ 2 is a factor of p

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

Squaring both sides, we get 

            p²= 4 m²                                                                                          ...(2)

From (1) and (2), we get 

           2q² = 4m²      ⇒      q²= 2m²

Clearly, 2 is a factor of 2m²

⇒       2 is a factor of q²                                                      [since, q² = 2m²]

⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

Answer 2

Question should be:

• Prove that √2 is an Irrational number.

ANSWER:

• √2 is an Irrational number.

GIVEN:

• Number = √2

TO PROVE:

• √2 is an Irrational number.

SOLUTION:

Let √2 be a rational number which can be expressed in the form of p/q where p and q have no common factor other than 1.

$\implies \: \sqrt{2} = \dfrac{p}{q} \\ \\ \implies \: \sqrt{2} q = p \\ \: \: \: \: squaring \: both \: sides \: we \: get. \\ \implies \: ( \sqrt{2} q) {}^{2} = (p) {}^{2} \\ \implies \: 2q {}^{2} = p {}^{2} ...(i)\\ \\ \: \: \: 2 \: divides \: p {}^{2} \\ \: \: 2 \: divides \: p \\ \\ \: \: \: let \: p \: = 2m \: in \: eq(i) \\ \implies \: 2q {}^{2} = (2m) {}^{2} \\ \implies \: 2q {}^{2} = 4m {}^{2} \\ \implies \: q {}^{2} = 2m {}^{2} \\ \\ \: \: 2 \: divides \: q {}^{2} \\ \: \: \: 2 \: divides \: q$

• 2 is the common factor of both p and q.
• Our contradiction is wrong.
• √2 is an Irrational number.