Let R be a relation defined on \[\mathbb{N}\] as \[a\mathrm{R}b\] if 2a+3b is a multiple of \[5,a,b\in \mathbb{N}\]. Then R is
Solutions · 1
Official solution
a R b if 2a + 3b = 5m, m \[\in\] l
(1) \[(a,a) \in R\] as \[2a + 3a = 5a,a \in N\]
Hence, R is reflexive
(2) If \[(a,b) \in R\] then 2a+3=5m
Now, 5(a+b)=5n3a+2b+2a+3b=5n
\[\therefore\] 3a+2b=5(n−m)
\[\therefore\] \[(b,a) \in R\]
\[\therefore\] R is symmetric
(3) If \[(a,b) \in R\] and \[(b,c) \in R\] then
2a+3b=5m,2b+3c=5n
\[ \Rightarrow 2a + 5b + 3c = 5(m + n)\]
\[ \Rightarrow 2a + 3c = 5(m = n - b)\]
\[\therefore\] \[(a,c) \in R\]
\[\therefore\] R is transitive
Hence, R is equivalence relation.
Option (1) is correct.