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The relation R in R\mathbb{R} (set of real numbers) is defined by R={(a,b):a≤b3}R = \{(a,b): a \leq b^3\}, then R is

Solution

✅ Correct Option: 4

Given: R={(a,b):a≤b3}R = \{(a,b): a \leq b^3\} where a,b∈Ra, b \in \mathbb{R}

To determine the properties of relation RR, we check if it is reflexive, symmetric, and transitive.


For RR to be reflexive, (a,a)(a,a) must be in RR for all values of aa.

This requires a≤a3a \leq a^3 for all a∈Ra \in \mathbb{R}

Take a=0.5a = 0.5:

0.5≤(0.5)30.5 \leq (0.5)^3

0.5≤0.1250.5 \leq 0.125 is false.

RR is not reflexive.


For RR to be symmetric, if (a,b)∈R(a,b) \in R, then (b,a)∈R(b,a) \in R.

This requires: if a≤b3a \leq b^3, then b≤a3b \leq a^3.

Take a=0,b=1a = 0, b = 1:

Check (a,b)(a,b): 0≤130 \leq 1^3

0≤10 \leq 1 is true, so (0,1)∈R(0,1) \in R

Check (b,a)(b,a): 1≤031 \leq 0^3

1≤01 \leq 0 is false, so (1,0)∉R(1,0) \notin R

RR is not symmetric.


For RR to be transitive, if (a,b)∈R(a,b) \in R and (b,c)∈R(b,c) \in R, then (a,c)∈R(a,c) \in R.

This requires: if a≤b3a \leq b^3 and b≤c3b \leq c^3, then a≤c3a \leq c^3.

Take a=100,b=5,c=2a = 100, b = 5, c = 2:

Check (a,b)(a,b): 100≤53100 \leq 5^3

100≤125100 \leq 125 is true, so (100,5)∈R(100, 5) \in R

Check (b,c)(b,c): 5≤235 \leq 2^3

5≤85 \leq 8 is true, so (5,2)∈R(5, 2) \in R

Check (a,c)(a,c): 100≤23100 \leq 2^3

100≤8100 \leq 8 is false, so (100,2)∉R(100, 2) \notin R

RR is not transitive.


RR is neither reflexive, nor symmetric, nor transitive.

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