Question

Prove that 5 - √3 is irrational, given that √3 is irrational​

Answer 1

$\bold {\huge{ \underline{\underline{ \sf{Detailed\:Answer}}}}}$

• To Prove : 5 - $\sqrt{3}$is irrational.

• Given : $\sqrt{3} \:$is already irrational.

Solution :

5 - $\sqrt{3}$ = $\frac{a}{b}$ ( Where a & b are rational numbers )

Just an information - The numbers which can be written in the form of $\frac{p}{q}$ ; where q ≠ 0 are called Rational Numbers.

Moving 5 to the other side,

$\sqrt{3}$ = $\frac{a}{b} - 5$

Taking the LCM,

$\sqrt{3} \: = \frac{a - 5b}{b}$

We know that,

$\frac{a - 5b}{b}$ is a rational number, but it's already given that $\sqrt{3}$ is irrational.

So, we can now say that $5 - \sqrt{3}$ is irrational.

$\sf{Hence \: Proved!}$

Answer 2

$\Huge{\underline{\boxed{\bf{\blue{Answer\::}}}}}$

• $\large{\underline{\textbf{Given\::}}}$

$\Longrightarrow$ √3 is already irrational.

$\large{\underline{\textbf{To\:prove\::}}}$

$\Longrightarrow$ 5 - √3 is a irrational.

Solution :-

$\Longrightarrow$ 5 - √3 = a/b ( Where a & b are rational number).

● We know that the number which can be written in form of a/b where b is not equal to zero are called rational number.

● Moving 5 to the other side.

$\Longrightarrow$ √3 = a/b - 5

● By taking LCM.

$\Longrightarrow$ √3 = a - 5b/b.

We know that a - 5b/b is a irrational number but it is already given that √3 is irrational.

● We can now say that 5 - √3 is irrational.

$\red{Hence\:proved\:!}$