Given that root 3 is an …
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Sia ? 6 years, 5 months ago
To Prove: 2+{tex}\sqrt3{/tex} is an irratinal number.
Given: {tex}\sqrt3{/tex} is irrational number.
Proof: Let 2 + {tex}\sqrt{3}{/tex} be a rational number.
{tex}\Rightarrow{/tex} 2 + {tex}\sqrt{3}{/tex} = {tex}\frac{p}{q}{/tex}, p, q {tex}\in{/tex} I, q {tex}\ne{/tex} 0
{tex}\Rightarrow{/tex} {tex}\sqrt{3}{/tex} = {tex}\frac{p}{q}{/tex} - 2
= {tex}\frac{p - 2q}{q}{/tex}
= {tex}\frac{integer}{integer}{/tex}
{tex}\implies{/tex}{tex}\sqrt{3}{/tex} is rational number
{tex}\Rightarrow{/tex} which is a contradiction to the fact that {tex}\sqrt{3}{/tex} is a rational
hence 2 + {tex}\sqrt{3}{/tex} is irrational number.
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