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Each of the following defines a relation on N:

(i) x is greater than y, x, y \in N\\

(ii) x + y = 10, x, y \in N\\

(iii) x y is square of an integer x, y \in N\\

(iv) x + 4y = 10 x, y \in N\\

Determine which of the above relations are reflexive, symmetric, and transitive.
 

Answers (1)

(i)Here, x is greater than y; x, y \in N\\

If (x, x) \in R, then x > x, that does not satisfy for any x \in N.\\

Therefore, R is not reflexive.

Say,(x, y) \in R

\\Or, xRy\\ Or, x > y

\Rightarrow y > x \\ For any x, y \in N, the above condition is not true.

Hence, R is not symmetric.

Again, xRy and yRz

\\Or, x > y and y > z\\ Or, x > z\\ Or, xRz\\

Hence, R is transitive.

(ii) x + y = 10;x, y \in N

Thus,

\\R = \{ (x, y); x + y = 10, x, y \in N \} \\ R = \{ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \} \\

Therefore, (1, 1) \notin R\\

So, R is not reflexive.

Again, (x, y) \in R \Rightarrow (y, x) \in R\\

Therefore, R is symmetric.

And, (1, 9) \in R, (9, 1) \in R, but (1, 1) \notin R\\

Therefore, R is not transitive.

(iii)Here, xy is square of an integer x, y \in N

R = \{ (x, y) : \text{xy is a square of an integer} x, y \in N \} \\

So, (x, x) \in R, \forall x \in N\\

For any x \in N, x\textsuperscript{2 } is an integer.

Thus, R is reflexive.

If (x, y) \in R \Rightarrow (y, x) \in R\\

So, R is symmetric.

Again, if xy and yz both are square of an integer.

Then, xy = m\textsuperscript{2} \: \: and \: \: yz = n\textsuperscript{2} \text{for some m, n} \in Z\\

\\x = m\textsuperscript{2} /y and z = n\textsuperscript{2} /y\\ xz = m\textsuperscript{2} n\textsuperscript{2} / y\textsuperscript{2}, this must be the square of an integer.

Therefore, R is transitive.

(iv) x + 4y = 10; x, y \in N\\

\\R = \{ (x, y): x + 4y = 10; x, y \in N \} \\ R = \{ (2, 2), (6, 1) \} \\ Here, (1, 1) \notin R\\

Hence, R is not symmetric.

\\(x, y) \in R \Rightarrow x + 4y = 10\\ And, (y, z) \in R \Rightarrow y + 4z = 10\\ Or,\ x - 16z = -30\\ Or,\ (x, z) \notin R\\

Therefore, R is not transitive.
 

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