Q. 5 A

# The following relations are defined on the set of real numbers :aRb if a – b > 0Find whether these relations are reflexive, symmetric or transitive.

Let set of real numbers be .

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.

We have

aRb if a – b > 0

Check for Reflexivity:

For a

If aRa,

a – a > 0

0 > 0

But 0 > 0 is not possible.

Hence, aRa is not true.

So, a , then aRa is not true.

R is not reflexive.

R is not reflexive.

Check for Symmetry:

a, b

If aRb,

a – b > 0

Replace a by b and b by a, we get

b – a > 0

[Take a = 12 and b = 6.

a – b > 0

12 – 6 > 0

6 > 0, which is a true statement.

Now, b – a > 0

6 – 12 > 0

–6 > 0, which is not a true statement as –6 is not greater than 0.]

bRa is not true.

So, if aRb is true, then bRa is not true.

a, b

R is not symmetric.

R is not symmetric.

Check for Transitivity:

a, b, c

If aRb and bRc.

a – b > 0 and b – c > 0

a – c > 0 or not.

Let us check.

a – b > 0 means a > b.

b – c > 0 means b > c.

a – c > 0 means a > c.

If a > b and b > c,

a > b, b > c

a > b > c

a > c

Hence, aRc is true.

So, if aRb is true and bRc is true, then aRc is true.

a, b, c

R is transitive.

R is transitive.

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