Q. 18 C5.0( 2 Votes )

# Each of the follo

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

xy is the square of an integer. x, y N.

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): xy = a2, a = √(xy), a N} on N

Check for Reflexivity:

x N

We should have, (x, x) R

xx = a2, where a = √(xx)

x2 = a2, where a = √(x2)

which is true every time.

Take x = 1 and y = 4

xy = a2

1 × 4 = (√(1 × 4))2 [ a = √(xy)]

4 = (√4)2

4 = (2)2

4 = 4

So, x N, then (x, x) R

R is reflexive.

Check for Symmetry:

x, y N

If (x, y) R

xy = a2, where a = √(xy)

Now, replace x by y and y by x. We get

yx = a2, which is as same as xy = a2

where a = √(yx)

yx = a2

(y, x) R

So, if (x, y) R, and then (y, x) R x, y N

R is symmetric.

Check for Transitivity:

x, y, z N

If (x, y) R and (y, z) R

xy = a2 and yz = a2

xz = a2, may or may not be true.

Let us take x = 8, y = 2 and z = 50

xy = a2, where a = √(xy)

(8)(2) = (√(8 × 2))2

16 = (4)2

16 = 16, which is true.

yz = a2

(2)(50) = (√(2 × 50))2

100 = (10)2

100 = 100, which is true

xz = a2

(8)(50) = (√(8 × 50))2

400 = (20)2

400 = 400

We won’t be able to find a case to show a contradiction.

xz = a2

(x, z) R

So, if (x, y) R and (y, z) R, and then (x, z) R

x, y, z N

R is transitive.

Hence, the relation is symmetric and transitivity, but not reflexive.

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