Show That the Relation \(R\) on \(N\times N\) is Reflexive, Symmetric and Transitive
Question
Let \(R\) be a relation on \(N\times N\) defined by
\[ (a,b)R(c,d)\iff a+d=b+c \]
for all \[ (a,b),(c,d)\in N\times N \]
Show that:
(i) \[ (a,b)R(a,b) \] for all \[ (a,b)\in N\times N \]
(ii) \[ (a,b)R(c,d)\Rightarrow(c,d)R(a,b) \]
(iii) \[ (a,b)R(c,d)\text{ and }(c,d)R(e,f)\Rightarrow(a,b)R(e,f) \]
Solution
(i) Reflexive
For \[ (a,b)R(a,b), \] we need
\[ a+b=b+a \]
which is true.
Hence,
\[ \boxed{ (a,b)R(a,b) } \] for all \[ (a,b)\in N\times N \]
(ii) Symmetric
Suppose
\[ (a,b)R(c,d) \]
Then,
\[ a+d=b+c \]
Rearranging,
\[ c+b=d+a \]
Therefore,
\[ (c,d)R(a,b) \]
Hence,
\[ \boxed{ (a,b)R(c,d)\Rightarrow(c,d)R(a,b) } \]
(iii) Transitive
Suppose
\[ (a,b)R(c,d) \] and \[ (c,d)R(e,f) \]
Then,
\[ a+d=b+c \]
and
\[ c+f=d+e \]
Adding,
\[ a+d+c+f=b+c+d+e \]
Cancelling \[ c+d \] from both sides,
\[ a+f=b+e \]
Therefore,
\[ (a,b)R(e,f) \]
Hence,
\[ \boxed{ (a,b)R(c,d)\text{ and }(c,d)R(e,f) \Rightarrow (a,b)R(e,f) } \]