CBSE BoardEnglish MediumSTD 10MathsReal Numbers3 Marks
Question
Prove that $\sqrt{3}$ is an irrational number.
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Answer
Let $\sqrt{3}$ be arational number, then its simplest form is $\frac{a}{b} ($where $a$ and $b$ are co$-$primes$)(\sqrt{3})^2=\frac{a^2}{b^2}$
$\Rightarrow a^2=3 b^2$
$\Rightarrow 3$ divides $a^2$
$\Rightarrow 3$ divides $a$
Let $a =3 c$, for some integer $c$
Putting $a =3 c$ in $(1),3 b^2=9 c^2$
$\Rightarrow b^2=3 c^2$
$\Rightarrow 3$ divides $b^2$
$\Rightarrow 3$ divides $b$
Thus, $3$ is a common factor of $a$ and $b$
But this is not possible as $a$ and $b$ are co$-$primes
$\Rightarrow$ Our assumption that $\sqrt{3}$ is rational is wrong
$\Rightarrow \sqrt{3}$ is irrational
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