Question

prove that under root 2 is not a rational number​

Answer 1

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 other common factor than 1.

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

From eq (ii) and (iii)

• 2 is the common factor of p and q.
• Thus our contradiction is wrong.
• So √2 is an Irrational number.

Answer 2

Answer:

To Prove:

√2 is irrational.

Solution:

Let us assume that √2 is a rational number.

We know that any rational number is of the form p/q where q is not equal to zero.

Let a and b have no common factor other than 1

√2 = a/b

Squaring both sides, we get

2 = a^2/b^2

Transposing b^2 to LHS we get,

2b^2 = a^2______(1)

=> 2 divides a^2 => 2 divides a.

Therefore a can be written as 2m where m is an integer.

Substituting the value of m in equation 1 we have

2b^2 = (2m)^2 = 4m^2

=> b^2 = 2m^2

Where 2 divides b^2 => 2 dives b.

Now, as 2 is the common factor other than a and b so our supposition is wrong.

Hence, √2 is irrational.

Hence, √2 is irrational. Hence proved!