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Exercise 1.1 · Q5

Q.Check whether the relation R in R\mathbf{R} defined by R={(a,b):a≤b3}R = \{(a, b) : a \le b^3\} is reflexive, symmetric or transitive.

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The relation R={(a,b):a≤b3}R = \{(a,b) : a \le b^3\} on R\mathbf{R} is neither reflexive, nor symmetric, nor transitive.

We test each property directly against the defining rule a≤b3a \le b^3. A single counterexample is enough to disprove a property.

1. Reflexivity

Reflexivity would require a≤a3a \le a^3 for every real number aa.

Take a=12a = \tfrac{1}{2}:

12≤(12)3=18 ?\tfrac{1}{2} \le \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}\,?

This is false, since 12>18\tfrac{1}{2} > \tfrac{1}{8}. So (12,12)∉R\left(\tfrac{1}{2}, \tfrac{1}{2}\right) \notin R.

RR is not reflexive.

Tip

For 0<a<10 < a < 1 the cube is smaller than the number itself, so a≤a3a \le a^3 fails. Cubes only "grow" for a>1a > 1.

2. Symmetry

Symmetry would require: whenever a≤b3a \le b^3, also b≤a3b \le a^3.

Take a=1, b=2a = 1,\ b = 2:

a≤b3:1≤23=8✓so (1,2)∈R.a \le b^3 : \quad 1 \le 2^3 = 8 \quad \checkmark \qquad\text{so } (1,2) \in R.

But

b≤a3:2≤13=1×so (2,1)∉R.b \le a^3 : \quad 2 \le 1^3 = 1 \quad \times \qquad\text{so } (2,1) \notin R.

RR is not symmetric.

3. Transitivity …

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