Question

Prove that 6-2√7 is irrational

Answer 1

ANSWER:

• 6-2√7 is an Irrational number.

GIVEN:

• Number = 6-2√7.

TO PROVE:

• (6-2√7) is an irrational number.

SOLUTION:

Let (6-2√7 ) 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.

$\implies \: 6 - 2 \sqrt{7} = \dfrac{p}{q} \\ \\ \implies \: 6 - \dfrac{p}{q} = 2 \sqrt{7} \\ \\ \implies \: \dfrac{6q - p}{q} = 2 \sqrt{7} \\ \\ \implies \: \dfrac{6q - p}{2q} = \sqrt{7}$

Here:

• (6q-p)/2q is rational while √7 is an Irrational number.
• Thus our contradiction is Wrong.
• 6-2√7 is an Irrational number.

Answer 2

$\huge\mathfrak{Answer:}$

Given:

• We have been given a number 6 - 2√7.

To Prove:

• We need to prove that 6 - 2√7 is irrational.

Solution:

Let us assume that 6 - 2√7 is a rational number.

Therefore, it can be written in the form of p/q where p and q are coprime.

$\longrightarrow \sf{6 - 2 \sqrt{7} = \dfrac{p}{q}}$

$\implies\sf{ 6 - \dfrac{p}{q} = 2\sqrt{7}}$

$\implies\sf{ \dfrac{6q - p}{q} = 2\sqrt{7}}$

$\implies\sf{\dfrac{6q - p}{2q} = \sqrt{7}}$

Here, (6q - p)/(2) is rational but √7 is irrational.

Rational number can never be equal to an irrational number.

Therefore, our assumption was wrong.

Hence, 6 - 2√7 is irrational.