Q. 18 D3.7( 4 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


x + 4y = 10, x, y N


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


The relation can also be defined as


R = {(x, y): 4x + y = 10} on N


Check for Reflexivity:


x N


We should have, (x, x) R


4x + x = 10, which is obviously not true everytime.


Take x = 4.


4x + x = 10


16 + 4 = 10


20 = 10, which is not true.


That is 20 ≠ 10.


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


R is not reflexive.


Check for Symmetry:


x, y N


If (x, y) R


4x + y = 10


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


4y + x = 10, which may or may not be true.


Take x = 1 and y = 6


4x + y = 10


4(1) + 6 = 10


4 + 6 = 10


10 = 10


4y + x = 10


4(6) + 1 = 10


24 + 1 = 10


25 = 10, which is not true.


4y + x ≠ 10


(y, x) R


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


R is not symmetric.


Check for Transitivity:


x, y, z N


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


Then, (x, z) R


We have


4x + y = 10


y = 10 – 4x


Where x, y N


So, put x = 1


y = 10 – 4(1)


y = 10 – 4


y = 6


Put x = 2


y = 10 – 4(2)


y = 10 – 8


y = 2


We can’t take y >2, because if we put y = 3


y = 10 – 4(3)


y = 10 – 12


y = –2


But, y ≠ –2 as y N


So, only ordered pairs possible are


R = {(1, 6), (2, 2)}


This relation R can never be transitive.


Because if (a, b) R, then (b, c) R


R is not reflexive.


Hence, the relation is neither reflexive nor symmetric nor transitive.


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