Question

Prove that √5-√3is an irrational number

Answer 1

check the attachment

at last Irational -Irational number = Irational number

hence as root 5 and root 3 are proven irrational so when they substracted there answer is Irrational.

Answer 2

Let us assume that √5-√3 is a rational number

Rational numbers are in the form, p/q where p and q are co-factors and q ≠ 0

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

Squaring both sides, we get,

$(\sqrt{5} - \sqrt{3})^{2} = \dfrac{a^{2} }{b^{2} }$

$5 - 2\sqrt{15} + 3 = \dfrac{a^{2} }{b^{2} }$

$8-2\sqrt{15} = \dfrac{a^{2} }{b^{2} }$

$-2\sqrt{15} = \dfrac{a^{2} }{b^{2} } - 8$

$-2\sqrt{15} = \dfrac{a^{2} -8b^{2} }{b^{2} }$

$\sqrt{15} = \dfrac{a^{2} -8b^{2} }{-2b^{2} }$

The R.H.S is a rational number

=> √15 is a rational number

But  this contradicts to the fact that it is an irrational number.

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ \sqrt{5} - \sqrt{3} \ is \ an \ irrational \ number.}}}}}$