First slide

Relations XII

Question

Let R be a relation over the set $N \times N$ and it is defined by $(a, b) R (c, d) \Rightarrow a + d = b + c .$ Then R is

Easy

A

Reflexive only
B

Symmetric only
C

Transitive only
D

An equivalence relation

Solution

We have $(a, b) R (a, b)$ for all $(a, b) \in N \times N$
Since $a + b = b + a .$ Hence, R is reflexive.
R is symmetric for we have $(a, b) R (c, d) \Rightarrow a + d = b + c$
$\Rightarrow d + a = c + b \Rightarrow c + b = d + a \Rightarrow (c, d) R (e, f) .$
Then by definition of R, we have
$a + d = b + c and c + f = d + e,$
whence by addition, we get
$a + d + c + f = b + c + d + e or a + f = b + e$
Hence, $(a, b) R (e, f)$
Thus, $(a, b) R (c, d) and (c, d) R (e, f) \Rightarrow (a, b) R (e, f) .$

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