Any positive integer is of the form 2q or 2q + 1, for some integer q.
∴  When   n=2q
n²-n = 2q(2q-1)
= 2m, where m=q(2q-1)
which is divisible by 2.
When     n = 2q+1
n²-n = (2q+1)(2q+1-1)
= 2q (2q+1)
= 2m,  when m = q(2q+1)
which is divisible by 2.
Hence, n²-n is divisible by 2 for every positive integer  n.

(i) 36  
(ii) 42

Consider an arbitrary positive integer y. Then, by taking y as dividend and 3 as divisor, we can write y as
y = 3q + r, where \(0\le r<3\)
[by Euclid's division lemma]
i.e. r = 0, 1, 2
Now, if r = 0, then y = 3q
if r = 1, then y = 3q + 1
and if r = 2, then y = 3q + 2
Thus, the positive integer y is of the form 3q, 3q + 1 or 3q + 2.
Consider, y = 3q and squaring on both sides, we get
y² = (3q)²  \(\Rightarrow \)  y² + 9q² = 3(3q)² = 3m .... (i)
[taking 3q² = m, where m is some integer]
Consider, y = 3q + 1 and squaring on both sides, we get
y² = (3q + 1)²  
\(\Rightarrow \)  y² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1   .... (ii)
[taking 3q² + 2q = m, where m is some integer]
Consider, y = 3q + 2 and squaring on both sides, we get
y² = (3q + 2)²  \(\Rightarrow \)  y² = 9q² + 12q + 4
\(\Rightarrow \)  y² = (9q² + 12q + 3) + 1 = 3(3q² + 4q + 1) + 1
\(\Rightarrow \)  y² = 3m + 1  .... (iii)
From Eqs. (i), (ii) and (iii), we get y² = 3m or 3m + 1
Thus, the square of any positive integer can be either of the form 3m or 3m + 1, for some integer m.

Irrational

HCF (60, 84, 108) = 12 and LCM = 3780

 P = ab² = a x b x b and q = a³b  = a x a x a x b
LCM (p, q) = a³ b²

Consider the greater number as a and smaller as b and then apply Euclid's division algorithm to get the required HCF.
(i) 130   (ii) 15

True, because the product of any two consecutive numbers, say n(n + 1) will always be even as one out of n or (n+1) must be even.

Use, dividend = divisor x quotient + remainder
= 53 x 34 + 21
= 1823

Let us assume to the contrary that \(3+2\sqrt { 5 } \)  is a rational number. Then, it can be expressed in the form \(\frac{a}{b}\), where a, b are coprime integers and \(b\neq 0\)
Now,  \(3+2\sqrt { 5 } \) = a/b, where a,b are integers and \(b\neq 0\)
On rearranging, we get
\(2\sqrt { 5 } =\frac { a }{ b } -3\quad or\quad \sqrt { 5 } =\frac { a }{ 2b } -\frac { 3 }{ 2 } \)
Since, a, b are integers and \(b\neq 0\) , therefore \(\frac{a}{2b}\) is rational number and so \(\frac{a}{2b}\) - \(\frac{3}{2}\) is a rational number.
[since, difference of two rational numbers is also a rational number]
\(\Rightarrow \sqrt { 5 } \) is a rational number. But \(\sqrt { 5 } \) is an irrational number.
This shows that our assumption is incorrect.
So, \(3+2\sqrt { 5 } \) is irrational.