Q. 18 C5.0( 2 Votes )

Each of the follo

Answer :

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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