Q. 3 C3.5( 4 Votes )

# Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive :R3 on R defined by (a, b) ϵ R3⇔ a2 – 4 ab + 3b2 = 0.

Here, R1, R2, R3, and R4 are the binary relations.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x A, xRx.

R is symmetric if for all x, y A, if xRy, then yRx.

R is transitive if for all x, y, z A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R3 on R defined by (a, b) R3 a2 – 4ab + 3b2 = 0

Check for Reflexivity:

a R,

(a, a) R3 needs to be proved for reflexivity.

If (a, b) R3, then we have

a2 – 4ab + 3b2 = 0

Replace b by a, we get

a2 – 4aa + 3a2 = 0

a2 – 4a2 + 3a2 = 0

–3a2 + 3a2 = 0

0 = 0, which is true.

(a, a) R3

So, a R, (a, a) R3

R3 is reflexive.

Check for Symmetry:

a, b R

If (a, b) R3, then we have

a2 – 4ab + 3b2 = 0

a2 – 3ab – ab + 3b2 = 0

a (a – 3b) – b (a – 3b) = 0

(a – b) (a – 3b) = 0

(a – b) = 0 or (a – 3b) = 0

a = b or a = 3b …(1)

Replace a by b and b by a in equation (1), we get

b = a or b = 3a …(2)

Results (1) and (2) does not match.

(b, a) R3

R3 is not symmetric.

Check for Transitivity:

a, b, c R

If (a, b) R3 and (b, c) R3

a2 – 4ab + 3b2 = 0 and b2 – 4bc + 3c2 = 0

a2 – 3ab – ab + 3b2 = 0 and b2 – 3bc – bc + 3c2

a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0

(a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0

(a – b) = 0 or (a – 3b) = 0

And (b – c) = 0 or (b – 3c) = 0

a = b or a = 3b And b = c or b = 3c

What we need to prove here is that, a = c or a = 3c

Take a = b and b = c

Clearly implies that a = c.

[ if a = b, just substitute a in place of b in b = c. We get, a = c]

Now, take a = 3b and b = 3c

If a = 3b Substitute in b = 3c. We get a = 9c, which is not the desired result.

(a, c) R3

R3 is not transitive.

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