Prove that (2 + ROOT under …

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Sia ? 6 years, 7 months ago
We will prove this result by contradiction method.
Let assume that {tex}( 2 + \sqrt { 3 } ){/tex} be rational.
Then 2 and {tex}\sqrt { 3 }{/tex} are rational.
{tex}\Rightarrow 2 + \sqrt { 3 } - 2{/tex} is rational ..... ({tex}\because{/tex} difference of two rational numbers is rational)
{tex}\Rightarrow \sqrt { 3 }{/tex} is rational
This contradicts the fact that {tex}\sqrt { 3 }{/tex} is irrational.
The Contradiction arises because we assume that {tex}( 2 + \sqrt { 3 } ){/tex} is rational.
Hence, {tex}2 + \sqrt { 3 }{/tex} is not a rational but an irrational number.
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