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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 5 · Set 1 · Q26
Prove that $\displaystyle \sqrt{3}$ is an irrational number.Prove that $\displaystyle (\sqrt{2}+\sqrt{3})^{2}$ is an irrational number, given that $\displaystyle \sqrt{6}$ is an irrational number.
Prove that $\displaystyle \sqrt{3}$ is an irrational number.
Prove that $\displaystyle (\sqrt{2}+\sqrt{3})^{2}$ is an irrational number, given that $\displaystyle \sqrt{6}$ is an irrational number.
Marking-scheme solution
Let \(\displaystyle \sqrt{3}\) be a rational number.
\(\displaystyle \therefore \sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and p & q are coprime.
\(\displaystyle 3 \mathrm{q}^{2}=\mathrm{p}^{2} \Rightarrow \mathrm{p}^{2}\) is divisible by $\displaystyle 3$
⇒ p is divisible by $\displaystyle 3$----- (i)
\(\displaystyle \Rightarrow \mathrm{p}=3 \mathrm{a}\), where 'a' is a postive integer
\(\displaystyle 9 \mathrm{a}^{2}=3 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=3 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2}\) is divisible by $\displaystyle 3$
⇒ q is divisible by $\displaystyle 3$ ----- (ii)
(i)
and (ii) leads to contradiction as 'p' and 'q' are coprime.
\(\displaystyle \therefore \sqrt{3}\) is an irrational number.
\(\displaystyle (\sqrt{2}+\sqrt{3})^{2}=2+3+2 \sqrt{6}=5+2 \sqrt{6}\)
Let us assume, to the contrary, that \(\displaystyle 5+2 \sqrt{6}\) is rational
\(\displaystyle \therefore 5+2 \sqrt{6}=\frac{\mathrm{a}}{\mathrm{b}} ; \mathrm{a}, \mathrm{b}\) are integers, \(\displaystyle \mathrm{b} \neq 0\)
\(\displaystyle \therefore \sqrt{6}=\frac{\mathrm{a}-5 \mathrm{~b}}{2 \mathrm{~b}}\)
RHS is a rational number, whereas LHS is an irrational number.
∴ Our assumption is wrong.
\(\displaystyle \Rightarrow 5+2 \sqrt{6}=(\sqrt{2}+\sqrt{3})^{2}\) is an irrational number
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.