Q. 5 B5.0( 2 Votes )

# The following relations are defined on the set of real numbers :aRb if 1 + ab > 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 1 + ab > 0

Check for Reflexivity:

For a

If aRa,

1 + aa > 0

1 + a2 > 0

If a is a real number.

[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]

This means, even if ‘a’ was negative.

a2 = positive.

a2 + 1 = positive

And any positive number is greater than 0.

Hence, 1 + a2 > 0

aRa is true.

So, a , then aRa is true.

R is reflexive.

R is reflexive.

Check for Symmetry:

a, b

If aRb,

1 + ab > 0

Replace a by b and b by a, we get

1 + ba > 0

Whether we write ab or ba, it is equal.

ab = ba

So, 1 + ba > 0

bRa is true.

So, if aRb is true, then bRa is true.

a, b

R is symmetric.

R is symmetric.

Check for Transitivity:

a, b, c

If aRb and bRc.

1 + ab > 0 and 1 + bc > 0

1 + ac > 0 or not.

Let us check.

1 + ab > 0 means ab > –1.

1 + bc > 0 means bc > –1.

1 + ac > 0 means ac > –1.

If ab > –1 and bc > –1.

ac > –1 should be true.

Take a = –1, b = 0.9 and c = 1

ab > –1

(–1)(0.9) > –1

–0.9 > –1, is true on the number line.

bc > –1

(0.9)(1) > –1

0.9 > –1, is true on the number line.

ac > –1

(–1)(1) > –1

–1 > –1, is not true as –1 cannot be greater than itself.

ac > –1 is not true.

1 + ac > 0 is not true.

aRc is not true.

So, if aRb is true and bRc is true, then aRc is not true.

a, b, c

R is not transitive.

R is not transitive.

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