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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 6 · Set 1 · Q26
(a) Prove that √ 5 is an irrational number.
Prove that $\displaystyle \sqrt{5}$ is an irrational number.
Let p, q and r be three distinct prime numbers. Check whether $\displaystyle \mathrm{p} \cdot \mathrm{q} \cdot \mathrm{r}+\mathrm{q}$ is a composite number or not. Further, give an example for $\displaystyle 3$ distinct primes $\displaystyle \mathrm{p}, \mathrm{q}, \mathrm{r}$ such that
(i)
$\displaystyle \mathrm{p} \cdot \mathrm{q} \cdot \mathrm{r}+1$ is a composite number.
(ii)
$\displaystyle \mathrm{p} \cdot \mathrm{q} \cdot \mathrm{r}+1$ is a prime number.
Marking-scheme solution
Let \(\displaystyle \sqrt{5}\) be a rational number.
\(\displaystyle \therefore \sqrt{5}=\frac{\mathrm{p}}{\mathrm{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and let p & q be the coprimes.
\(\displaystyle \Rightarrow 5 \mathrm{q}^{2}=\mathrm{p}^{2}\)
\(\displaystyle \Rightarrow \mathrm{p}^{2}\) is divisible by 5.
\(\displaystyle \Rightarrow\) p is divisible by 5. ----- ①
Let \(\displaystyle \mathrm{p}=5 \mathrm{a}\), where 'a' is some integer
\(\displaystyle \therefore 25 \mathrm{a}^{2}=5 \mathrm{q}^{2}\)
\(\displaystyle \Rightarrow \mathrm{q}^{2}=5 \mathrm{a}^{2}\)
\(\displaystyle \Rightarrow \mathrm{q}^{2}\) is divisible by 5.
\(\displaystyle \Rightarrow\) q is divisible by 5. ----- ②
\(\displaystyle \therefore\) $\displaystyle 5$ divides both p & q.
① and ② leads to contradiction as p and q are coprimes.
Hence, \(\displaystyle \sqrt{5}\) is an irrational number.
OR
\(\displaystyle \mathrm{p} . \mathrm{q} . \mathrm{r}+\mathrm{q}=\mathrm{q}(\mathrm{pr}+1)\)
Thus, the given number has more than $\displaystyle 2$ factors.
Hence it is composite.
(i)
Taking \(\displaystyle \mathrm{p}=3, \mathrm{q}=5\) and \(\displaystyle \mathrm{r}=7\)
\(\displaystyle \mathrm{pqr}+1=3.5 .7+1=106\) is a composite number
(ii)
Taking \(\displaystyle \mathrm{p}=2, \mathrm{q}=3\) and \(\displaystyle \mathrm{r}=5\)
\(\displaystyle \mathrm{pqr}+1=2.3 .5+1=31\) is a prime number
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