Question

prove that √5 is irrational​

Answer 1

$\sf Let\ us\ assume\ that\ \sqrt{5}\ is\ rational.\\ \\\sqrt{5}\ =\ \frac{a}{b},\ where\ a\ and\ b\ are\ co - prime\ integers\ and\ b\ \neq\ 0.\\\\Squaring\ both\ the\ sides,\\\\\ (\sqrt{5})^{2}\ =\ (\frac{a}{y})^{2}\\\\\sf 5\ =\ \frac{a^{2}}{b^{2}}\ \\\\a^{2}\ =\ 5b^{2}\ (Taking\ this\ as\ equation\ 1.)\\\\This\ implies\ that\ 5\ divides\ a^{2}\ and\ that\ 5\ divides\ a\ as\ well.\\\\ \sf Let\ a\ =\ 5c,\ for\ some\ integer\ c.\\\\Substituting\ a\ =\ 5c\ in\ equation\ 1,\\\\(5c^{2})\ =\ 5b^{2}\\\\25b^{2}\ =\ 5b^{2} \\\\b^{2}\ =\ 5c^{2}\\\\This\ implies\ that\ 5\ divides\ b^{2}\ and\ that\ 5\ divides\ b.\\\\Thus,\ 5\ is\ a\ factor\ of\ both\ a\ and\ b.\\\\This\ contradicts\ the\ fact\ that\ a\ and\ b\ are\ co - prime\ integers.\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ asummption.\\\\Hence, \sqrt{5}\ is\ irrational.$

Answer 2

║⊕ANSWER⊕║

To prove that √5 is irrational number 

Let us assume that √5 is rational 

Then √5 =  

(a and b are co primes, with only 1 common factor and b≠0) 

⇒ √5 =  

(cross multiply) 

⇒ a = √5b 

⇒ a² = 5b² -------> α

⇒ 5/a² 

(by theorem if p divides q then p can also divide q²) 

⇒ 5/a ----> 1 

⇒ a = 5c 

(squaring on both sides) 

⇒ a² = 25c² ----> β 

From equations α and β 

⇒ 5b² = 25c²

⇒ b² = 5c² 

⇒ 5/b² 

(again by theorem) 

⇒ 5/b-------> 2 

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong. 

This contradiction arises because we assumed that √5 is a rational number 

∴ our assumption is wrong 

∴ √5 is irrational number