Question

Prove that "root 5 – 2 × root 3" is a irrational no

Answer 1

$\huge\blue{\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}}$

TO PROVE:

$\sqrt{5}-2\sqrt{3} \:is$ a irrational number.

PROOF:

On the contrary let us assume that $\sqrt{5}-2\sqrt{3} \:is$ is a rational number and p and q are coprimes.

So, it can be expressed

in the form of $\dfrac{p}{q}$where p and q are integers and q is not equal to zero .

______________________________________

=>$\sqrt{5}-2\sqrt{3} =\dfrac{p}{q}$

On squaring both sides,

=>$(\sqrt{5}-2\sqrt{3} )^{2}=(\dfrac{p}{q}) ^{2}$

=>$(\sqrt{5}^{2})+(2\sqrt{3})^{2}$- 2×2√3×√5 = $(\dfrac{p}{q})^2$

$\large\green{\boxed{(a-b)^{2}=a^{2}+b^{2}-2ab}}$

=>$5+12-4\sqrt{15}= (\dfrac{p}{q})^{2}$

=>$17-4\sqrt{15}= (\dfrac{p}{q})^{2}$

=>$4\sqrt{15}=17-\dfrac{p^{2}}{q^{2}}$

=>$4\sqrt{15}=\dfrac{17q^{2}-p^{2}}{q^{2}}$

=>$\sqrt{15}=\dfrac{17q^{2}-p^{2}}{4q^{2}}$

But a contradiction has a right under $\sqrt{15}$is a irrational number and$\dfrac{17q^{2}-p^{2}}{4q^{2}}$ is a rational number .

________________________________

Therefore assumption was wrong the given number is a irrational number .