Q. 18 D3.7( 3 Votes )

# Each of the following defines a relation on N :x + 4y = 10, x, y ∈ NDetermine which of the above relations are reflexive, symmetric and transitive.

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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