Prove that √3+√5 is irratinal no

1.	Prove that √3+√5 is irratinal no
Answer» Let us consider\xa0{tex} \\sqrt { 3 } + \\sqrt { 5 }{/tex} is a rational number that can be written as{tex} \\sqrt { 3 } + \\sqrt { 5 }{/tex}\xa0= a{tex} \\Rightarrow \\sqrt { 5 } = a - \\sqrt { 3 }{/tex}Squaring both sides, we get{tex} ( \\sqrt { 5 } ) ^ { 2 } = ( a - \\sqrt { 3 } ) ^ { 2 }{/tex}⇒ 5 = (a)2{tex} + ( \\sqrt { 3 } ) ^ { 2 } - 2 ( a ) ( \\sqrt { 3 } ){/tex}⇒ {tex} 2 a \\sqrt { 3 } {/tex} = a2 + 3 - 5⇒ {tex} 2 a \\sqrt { 3 } {/tex} = a2 - 2{tex} \\Rightarrow \\sqrt { 3 } = \\frac { a ^ { 2 } - 2 } { 2 a }{/tex}As a2\xa0– 2, 2a are integers .So\xa0{tex} \\frac { a ^ { 2 } - 2 } { 2 a }{/tex}\xa0is also rational but\xa0{tex} \\sqrt { 3 }{/tex}\xa0is not rational which contradicts our consideration.Since a rational number cannot be equal to an irrational number . Our assumption that\xa0√3 +\xa0√5 is rational wrong .So,\xa0{tex} \\sqrt { 3 } + \\sqrt { 5 }{/tex}\xa0is irrational.

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