Hint Let us suppose that \( \sqrt{p} \)+ \( \sqrt{q} \)is a rational
number, Again, let  \( \sqrt{p} \)+ \( \sqrt{q} \) = a, where a is rational.
Therefore, \( \sqrt{q} \)  = a - \( \sqrt{p} \)
On squaring both sides, we get
\({l} q=a^{2}+p-2 a \sqrt{p}\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\\ \sqrt {p}=\frac{a^{2}+p-q}{2 a} \)
Since, p and q are primes and a is a rational number, so
\(\frac{a^{2}+p-q}{2 a}\) is rational, therefore \( \sqrt{p} \) is a rational number.
But this contradicts the fact that \( \sqrt{p} \) is irrational
number as p is prime. So, our assumption was incorrect.
Hence, \( \sqrt{p} \) + \( \sqrt{q} \)  is irrational

Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer. On dividing n by 5, let q be the quotient and r be the remainder.
Then, n = 5q + r, where \(0\le r<5\)
[ by Euclid's division lemma]
\(\Rightarrow \)  n = 5q + r, where r = 0, 1, 2, 3, 4
\(\Rightarrow \)  n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4
If n = 5q, then n is only divisible by 5.
If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) which is divisible by 5. So, (n + 4) is only divisible by 5.
If n = 5q + 2, then n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) which is divisible by 5. So, (n + 8) is only divisible by 5.
If n = 5q + 3, then n + 12 = 5q + 3 + 12  = 5q + 15 = 5(q + 3) which is divisible by 5. So, (n + 12) is only divisible by 5.
If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) which is divisible by 5. So, (n + 16) is only divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

No, because HCF does not divide LCM.

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

Let a be any odd positive integer, then on dividing a by b, we have a = bq + r,\( \ 0\le r ..\). (i) [by Euclid's division lemma]
On putting b = 2 in Eq. (i), we get
a = bq + r,          \(0\le\) r   \(\Rightarrow \)  r = 0 or 1
If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.
If r = 1, then a = 2q + 1, which is not divisible by 2.
\(\therefore \quad \left( 2q+1 \right) \) is odd.
Now, as a is odd, so it cannot be of the form 2q. Thus, any odd positive integer a is of the form (2q + 1).
Now, consider \({ a }^{ 2 }=\left( 2q+1 \right) ^{ 2 }={ 4q }^{ 2 }+1+4q\quad \left[ \because \quad \left( x+y \right) ^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy \right] \) 
= 4(q² + q) + 1 = 4 m + 1,   where m = q² + q
Hence, for some integer m, the square of any odd integer is of the form 4m + 1.