Q. 5 C5.0( 3 Votes )

The following rel

Answer :

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

Check for Reflexivity:

For a

If aRa,

|a| ≤ a, is true

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.

|a| = positive.

& |a| ≤ a

Hence, |a| ≤ a.

aRa is true.

So, a , then aRa is true.

R is reflexive.

R is reflexive.

Check for Symmetry:

a, b

If aRb,

|a| ≤ b

Replace a by b and b by a, we get

|b| ≤ a, which might be true or not.

Let a = 2 and b = 3.

|a| ≤ b

|2| ≤ 3, is true

|b| ≤ a

|3| ≤ 2, is not true

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 and |b| ≤ c

|a| ≤ c or not.

Let us check.

If |a| ≤ b and |b| ≤ c

b ≠ |b|

Say, if b = –2

–2 ≠ |–2|

–2 ≠ 2

But, from |a| ≤ b

b ≥ 0 in every case otherwise the statement would not hold true.

b can only accept positive values including 0.

b is a whole number.

if |a| ≤ b and |b| ≤ c

|a| ≤ b, b ≤ c

|a| ≤ b ≤ c

|a| ≤ c

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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