CBSE 2025 · Region 1 · Set 3 · Q26 · 3 marks
Prove that $\displaystyle \sqrt{5}$ is an irrational number.
Marking-scheme solution
Let \(\displaystyle \sqrt{5}\) be a rational number.
\[\therefore \sqrt{5}=\frac{p}{q}, \text { where } \mathrm{q} \neq 0 \text { and let } \mathrm{p} \ \& \ \mathrm{q} \text { are co-primes. }
\]
\[5 \mathrm{q}^{2}=\mathrm{p}^{2} \Rightarrow \mathrm{p}^{2} \text { is divisible by } 5
\]
⇒ p is divisible by $\displaystyle 5$ ----- (i)
⇒ let \(\displaystyle \mathrm{p}=5 \mathrm{a}\), where 'a' is some integer
\[25 \mathrm{a}^{2}=5 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=5 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2} \text { is divisible by } 5 .
\]
⇒ q is divisible by 5. ----- (ii)
(i)
and (ii) leads to contradiction as p and q are coprimes.
\(\displaystyle \therefore \sqrt{5}\) is an irrational number
Real NumbersProving a Number IrrationalUnderstandshort_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.