Question

Prove that (2√3+√5) is an irrational number .also check whether (2√3+√5)(2√3-√5) is rational or irrational number

Answer 1

Answer:

prime and derived that they have common factors 2 and 5. Our assumption that the given number is a rational number , must be wrong. Hence, 3 √2 + √5 is irrational. So the product of the two irrational numbers is rational and in fact, is a positive integer.

Answer 2

Assume that (2√3+√5) is rational number equal to p/q where p and q is integer.

Let,

$\implies\rm{2\sqrt{3} + \sqrt{5} = \dfrac{p}{q}}$

• Squaring the both sides.

$\implies\rm{ (2\sqrt{3} + \sqrt{5})^2= (\dfrac{p}{q})^2}$

$\implies\rm{(2\sqrt{3})^2 + 2\times{2\sqrt{3}}\times{\sqrt{5}} + \sqrt{(5)^2} = \dfrac{p^2}{q^2}}$

$\implies\rm{ 12 + 4\sqrt{15} + 5 = \dfrac{p^2}{q^2}}$

$\implies\rm{ 17 + 4\sqrt{15} = \dfrac{p^2}{q^2}}$

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

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

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

So, $\rm{\dfrac{p^2 - 17q^2}{4q^2}}$ is rational , then $\rm{\sqrt{15}}$ must be rational.

But it contradict the fact that √15 is irrational number.

Now,

$\implies\rm{(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5})}$

$\implies\rm{ 12- 2\sqrt{15} + 2\sqrt{15} - 5 }$

$\implies\rm{ 7 }$.

Hence, 7 is rational Number.