Let’s assume on the contrary that √5 + √3 is a rational number. Then, there exist co prime positive integers a and b such that 

√5 + √3 = \(\frac{a}{b}\)

⇒ √5 = (\(\frac{a}{b}\)) – √3

⇒ (√5)2 = ((\(\frac{a}{b}\)) – √3)² [Squaring on both sides] 

⇒ 5 = (\(\frac{a^2}{b^2}\)) + 3 – (2√3\(\frac{a}{b}\)) 

⇒ (\(\frac{a^2}{b^2}\)) – 2 = (2√3\(\frac{a}{b}\)) 

⇒ (\(\frac{a}{b}\)) – (2\(\frac{b}{a}\)) = 2√3 

⇒ \(\frac{(a^2 – 2b^2)}{2ab}\) = √3 

⇒ √3 is rational [∵ a and b are integers ∴ \(\frac{(a^2 – 2b^2)}{2ab}\) is rational] 

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, √5 + √3 is an irrational number.

Let 5 – √3 is a rational number.

we have to find out two integers a and b such as

5 – √3 = a/b

– √3 = a/b – 5

√3 = 5 – a/b

√3 = (5b – a)/b

a, b and 5 all are integers

(5b – a)/b is a rational number.

√3 will be also rational number.

But this contradicts the fact that √3 is an irrational number.

So our hypothesis is wrong.

Hence, 5 – √3 is an irrational number.