Question

show that 3+√5 is an irrational​

Answer 1

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

Answer 2

Step-by-step explanation:

• I think it is helpful to you