Question

prove that 3+√5 is an irrational number​

Answer 1

$\sf We\ know\ that\ \sqrt{5}\ is\ irrational.\\\\ \sf Let\ us\ assume\ that\ \sqrt{5}\ \sf is\ a\ rational\ number.\\ \\\sqrt{5}\ =\ \frac{a}{b},\ where\ a\ and\ b\ are\ co-prime\ integers\ and\ b\ \neq\ 0.\\\\ \sf 3\ +\ \sqrt{5}\ =\ \sf \frac{a}{b}\\ \\\sqrt{5}\ =\ \frac{a}{b}\ -\ 3 \\\\\sqrt{5}\ =\ \sf \frac{a\ -\ 3b}{b}\\ \\\sf Since\ a\ and\ be\ are\ integers,\ \frac{a\ -\ 3b}{b}\ is\ rational.\\ \\This\ implies\ that\ \sqrt{5}\ is\ rational\ too!\\\\ \sf This\ contradicts\ the\ fact\ that\ \sqrt{5}\ is\ irrational.\\ \\Therefore,\ our\ assumption\ is\ wrong.\\\\3\ +\ \sqrt{5}\ is\ irrational.$

Answer 2

Solution :-

Let us assume that 3 + √5 is rational

i.e, 3 + √5 = a/b (where a and b are coprimes and b ≠ 0)

⇒ √5 = a/b - 3

⇒ √5 = (a - 3b)/b

Since a and b are integers, the RHS of the equation i.e (a - 3b)/b is a rational number.

So, LHS of the equation i.e √5 must be a rational number.

But this contradicts the fact that √3 is irrational.

This contradiction has arised because of our wrong assumption that 3 + √5 is rational.

So we can conclude that 3 + √5 is an irrational number.