Q. 134.2( 5 Votes )

Show that the rel

Answer :

We have

The relation “≥” on the set R of all real numbers.

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, let the relation having “≥” be P.

We can write

P = {(a, b): a ≥ b, a, b R}

Check for Reflexivity:

a R

If (a, a) P

a ≥ a, which is true.

Since, every real number is equal to itself.

So, a R, then (a, a) P.

P is reflexive.

P is reflexive.

Check for Symmetry:

a, b R

If (a, b) P

a ≥ b

Now, replace a by b and b by a. We get

b ≥ a, might or might not be true.

Let us check:

Take a = 7 and b = 5.

a ≥ b

7 ≥ 5, holds.

b ≥ a

5 ≥ 7, is not true as 5 < 7.

b ≥ a, is not true.

(b, a) P

So, if (a, b) P, then (b, a) P

a, b R

P is not symmetric.

P is not symmetric.

Check for Transitivity:

a, b, c R

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

a ≥ b and b ≥ c

a ≥ b ≥ c

a ≥ c

(a, c) P

So, if (a, b) P and (b, c) P, then (a, c) P

a, b, c R

P is transitive.

P is transitive.

Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.

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