Question

Prove that (3 - 4 root 2) by 7 is an irrational number given that root 2 is an irrational number.

Correct explanation will be marked as brainliest.​

Answer 1

Answer:

Step-by-step explanation;

let us assume the given no. is a rational no.

therefore 3-4 root 2/7 =a/b.

3-4 root 2 =7a/b

4 root 2 =3 - 7a/b

root 2 = (3 - 7a/b)/4

but root 2 is an irrational no.

this contradicts our assumption.

therefore the given no. is an irrational no.

Answer 2

Concept:

All real numbers that are not rational numbers are referred to be irrational numbers in mathematics. Irrational numbers, on the other hand, can't be stated as a ratio of two integers.

Given:

$\sqrt{2}$ is an irrational number.

Prove:

$\frac{3-4\sqrt{2} }{7}$ is an irrational number.

Solution:

Let us assume that the given number $\frac{3-4\sqrt{2} }{7}$ is a rational number.

$\frac{3-4\sqrt{2} }{7}=\frac{p}{q}$

where p and q are any integers.

(∵ Rational numbers can be represented as the ratio of two integers)

$3-4\sqrt{2} =\frac{7p}{q}$

$4\sqrt{2}=3-\frac{7p}{q}$

$\sqrt{2} =\frac{1}{4}(3-\frac{7p}{q} )$

But it is given that $\sqrt{2}$ is an irrational number.

This contradicts our assumption.

∴ $\frac{3-4\sqrt{2} }{7}$ is an irrational number.