Show that the relation $S$ in the set $R$ of real numbers, defined as $S=\{(a,b):a,b \in R \text{ and } a \leq b^{3}\}$, is neither reflexive, nor symmetric, nor transitive.

CBSE Class 12 · 2010 · Relations And Functions XIILong Answer

Show that the relation

S

in the set

R

of real numbers, defined as

S = {(a, b) : a, b \in R and a \leq b^{3}}

, is neither reflexive, nor symmetric, nor transitive.

Answer: $S$ is neither reflexive, nor symmetric, nor transitive.

We exhibit a counterexample for each property.
Not reflexive: take

a = \frac{1}{2}

. Then

a \leq a^{3}

means

\frac{1}{2} \leq \frac{1}{8}

, which is false, so

(\frac{1}{2}, \frac{1}{2}) \in / S

.
Not symmetric: take

a = 1, b = 2

. Then

1 \leq 2^{3} = 8

is true, so

(1, 2) \in S

; but

2 \leq 1^{3} = 1

is false, so

(2, 1) \in / S

.
Not transitive: take

a = 10, b = \frac{5}{2}, c = \frac{3}{2}

10 \leq (\frac{5}{2})^{3} = \frac{125}{8} = 15.625 \Rightarrow (a, b) \in S

\frac{5}{2} \leq (\frac{3}{2})^{3} = \frac{27}{8} = 3.375 \Rightarrow (b, c) \in S

but 10 \leq (\frac{3}{2})^{3} = 3.375 is false \Rightarrow (a, c) \in / S

S is neither reflexive, nor symmetric, nor transitive

CBSE Class 12 Board · 2010 · Subjective