Class 12 Maths NCERT Solutions Chapter 1 Ex 1.1 – Types of Relations

For reflexivity we'd need 3x-x=0\Rightarrow x=0, which fails for every element (e.g. (1,1): 3(1)-1=2\neq0).

(1,3)\in R since 3(1)-3=0, but (3,1)\notin R since 3(3)-1=8\neq0 — not symmetric.

(1,3)\in R and (3,9)\in R, but (1,9)\notin R since 3(1)-9=-6\neq0 — not transitive.

The only pairs satisfying this are (1,6),(2,7),(3,8). None of these have equal coordinates, so R isn't reflexive.

(1,6)\in R but (6,1) would need 1=11, false, so not symmetric.

Since none of 6,7,8 appear as a first coordinate of any pair, there's no chain to test, so transitivity holds vacuously.

Every x is divisible by itself, so R is reflexive.

(1,2)\in R but (2,1)\notin R (1 isn't divisible by 2) — not symmetric.

If y is divisible by x and z by y, then z is divisible by x — transitive.

Since x,y\in\mathbb{Z}, x-y is always an integer, so R is the universal relation \mathbb{Z}\times\mathbb{Z}.

(a) Same workplace and (b) same locality: both are reflexive, symmetric and transitive by the same "shared category" reasoning.

(c) x is exactly 7 cm taller than y: x can't be taller than itself (not reflexive); if x is 7 cm taller than y, y is 7 cm shorter, not taller (not symmetric); if x is 7 cm taller than y and y 7 cm taller than z, x is 14 cm taller than z (not transitive).

(d) x is wife of y: not reflexive, not symmetric (y is husband, not wife), but transitive vacuously since no chain is possible.

(e) x is father of y: not reflexive, not symmetric, and not transitive (x would be z's grandfather, not father).

Taking a=\frac12: is \frac12\le\frac14? No — not reflexive.

Taking a=1,b=2: 1\le4 is true, so (1,2)\in R, but (2,1) fails since 2\le1 is false — not symmetric.

Taking a=3,b=-2,c=1: (3,-2)\in R since 3\le4, and (-2,1)\in R since -2\le1, but (3,1) fails since 3\le1 is false — not transitive.

(1,1) would need 1=2, false — not reflexive.

(1,2)\in R since 2=1+1, but (2,1) would need 1=3, false — not symmetric.

(1,2)\in R and (2,3)\in R, but (1,3) would need 3=2, false — not transitive.

a\le a is always true — reflexive.

a\le b and b\le c together give a\le c — transitive.

But 1\le2 is true while 2\le1 is false, so a\le b doesn't imply b\le a — not symmetric.

Taking a=\frac12: is \frac12\le\frac18? No — not reflexive.

Taking a=-2,b=1: -2\le1 is true, but (1,-2) fails since 1\le-8 is false — not symmetric.

Taking a=27,b=3,c=2: 27\le27 and 3\le8 both hold, but 27\le8 is false — not transitive.

None of (1,1),(2,2),(3,3) appear in R — not reflexive.

The only two pairs present, (1,2) and (2,1), are each other's reverse — symmetric.

But (1,2)\in R and (2,1)\in R would require (1,1)\in R, which it isn't — not transitive.

Every book has the same page count as itself — reflexive.

If x matches y's page count, y matches x's — symmetric.

If x matches y and y matches z, then x matches z — transitive.

|a-a|=0, even — reflexive.

|a-b|=|b-a| always — symmetric.

|a-b| even means a,b share parity; likewise |b-c| even means b,c share parity, so a,b,c all share parity, making |a-c| even too — transitive.

Since 1,3,5 are all odd, every pairwise difference is even — they're all related. 2,4 are both even, so related.

Any odd minus any even is odd, not even, so no element of \{1,3,5\} relates to any element of \{2,4\}.

|a-a|=0, a multiple of 4 — reflexive. |a-b|=|b-a| — symmetric. If a-b and b-c are both multiples of 4, so is their sum a-c — transitive.

Elements related to 1: a\in\{0,\ldots,12\} with |a-1| a multiple of 4 are a=1,5,9 (differences 0,4,8); a=13 is out of range.

The equality relation is trivially reflexive, symmetric and transitive.

On A=\{1,2,3\}, R=\{(1,2),(2,1)\} — this is the relation from Q6, already shown to be symmetric only.

On A=\{1,2,3\}, R=\{(1,2)\}. There's no pair (2,x)\in R, so transitivity holds vacuously, while (1,1) and (2,1) are both missing.

On A=\{1,2,3\}, R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}.

All three self-pairs are present (reflexive) and every pair's reverse is present (symmetric), but (1,2),(2,3)\in R while (1,3)\notin R — not transitive.

The relation R=\{(a,b): a\le b\} on \mathbb{R} from Q4 already does this.

On A=\{1,2,3\}, R=\{(1,1),(2,2),(1,2),(2,1)\}, deliberately leaving out (3,3).

Every pair's reverse is present (symmetric), and every chain checks out (transitive), but (3,3)\notin R — not reflexive.

Every point is trivially equidistant from the origin as itself — reflexive.

If P,Q are equidistant, so are Q,P — symmetric.

If P,Q equidistant and Q,R' equidistant, then P,R' share that distance — transitive.

Every point related to P has distance |OP| from the origin — by definition, that's the circle centred at the origin with radius |OP|, which passes through P itself.

Every triangle is similar to itself (reflexive); similarity runs both ways between two triangles (symmetric); and similarity chains through a third triangle (transitive) — so R is an equivalence relation.

T_1's sides 3:4:5 and T_3's sides 6:8:10=2\times(3:4:5) are in the same ratio — similar.

T_2's sides 5:12:13 aren't proportional to 3:4:5 (e.g. 5/3\neq12/4), so T_2 is unrelated to either.

Matching side counts is reflexive, symmetric and transitive by the same "shared category" reasoning as Q7. Since T is a triangle with 3 sides, its equivalence class is every polygon in A with exactly 3 sides.

Every line is (by convention) parallel to itself (reflexive); parallelism runs both ways (symmetric); and chains through a third line (transitive) — R is an equivalence relation.

The line y=2x+4 has slope 2, so any parallel line must share that slope.

All four self-pairs (1,1),(2,2),(3,3),(4,4) are present — reflexive. (1,2)\in R but (2,1)\notin R — not symmetric.

Checking every possible chain — (1,3),(3,2)\to(1,2) ✓, (1,3),(3,3)\to(1,3) ✓, (3,3),(3,2)\to(3,2) ✓, (1,2),(2,2)\to(1,2) ✓, (3,2),(2,2)\to(3,2) ✓ — every chain checks out, so R is transitive.

(2,4): 2=4-2 ✓, but 4\gt6 is false — rejected.

(3,8): 3=8-2=6? False — rejected.

(6,8): 6=8-2=6 ✓ and 8\gt6 ✓ — both hold.

(8,7): 8=7-2=5? False — rejected.