Question

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

Answer 1

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

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

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

Let us assume that $5 - \sqrt{3}$ is rational.

So now,

$5 - \sqrt{3}$ = $\frac{a}{b}$ ( Where $\frac{a}{b}$ is a rational number )

Taking 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 $\sqrt{3}\:$ is irrational, which is given.

So our assumption is wrong, and $5 - \sqrt{3}$ is an irrational number.

$\sf{Hence \: Proved! \: }$

Answer 2

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

we have to prove that 5-√3 is irrational.

let us assume that 5-√3 is irrational.

we know that if any no is rational then it can be written in the form $p/q$

Hence,

$5 - \sqrt{3} \: can \: be \: written \: in \: the \: form \: \frac{a}{b}$

where a,b(b≠0 ) are co- prime .

thus ,

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

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

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

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

Here ,

$\frac{ - a + 5b}{b} \: is \: rational$

But √3 is irrational.

since, Rational≠iraational

this is a contradiction

∴ our assumption is wrong

Hence , 5-√3 is irrational.

$\huge{\bold{ Hence\: proved }}$