Question #284208

Show that the relation R on Z × Z defined by (a, b) R (c, d) if and only if a + d = b + c


is an equivalence relation.


Note: A relation on a set A is called an equivalence relation if it is reflexive, symmetric,


and transitive.


1
Expert's answer
2022-01-04T15:12:12-0500

A relation R on Z×ZZ \times Z defined by (a,b)R(c,d)(a, b) R(c, d) iff a+d=b+c.a+d=b+c.

we see that,

a+b=b+aa,bza+b=b+a \forall a, b \in z

Therefore (a,b)R(a,b)(a,b)R(a, b) R(a, b) \forall(a, b) \in R

therefore R is reflexive relation

Let (a, b) R(c, d)

a+d=b+cc+b=d+a(c,d)R(a,b)\begin{aligned} &\Rightarrow a+d=b+c \\ &\Rightarrow c+b=d+a \\ &\Rightarrow(c, d) R(a, b) \end{aligned}

Therefore R is symmetric relation.

Let, (a, b) R(c, d) and (c, d) R(a, f)a+d=b+candc+f=d+e(a+d)+(c+f)=(b+c)+(d+e)(a+f)+(d+c)=(b+e)+(d+c)a+f=b+e(a,b)R(e,f)\begin{aligned} &\text{Let, (a, b) R(c, d) and (c, d) R(a, f)}\\ &\Rightarrow a+d=b+c and c+f=d+e\\ &\Rightarrow(a+d)+(c+f)=(b+c)+(d+e)\\ &\Rightarrow(a+f)+(d+c)=(b+e)+(d+c)\\ &\Rightarrow a+f=b+e\\ &\Rightarrow(a, b) R(e, f)\\ \end{aligned}

Therefore R is transitive relation.

Therefore R is an equivalence relation.

since, ad=b+c\quad a \rightarrow d=b+c

d=a+b+c\Rightarrow d=-a+b+c

Therefore [(a,b)]={(c,a+b+c);cZ}[(a, b)]=\{(c,-a+b+c) ; c \in \mathbb{Z}\}


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