Question #130296
let A be the set of integers and C be the set of ordered pairs (x,y)£A×A such that y is not equal to zero define relation ~ on C(x,y)~(z,w) if yz=zw prove that defines an equivalence relation on C
1
Expert's answer
2020-08-27T16:21:03-0400

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

transitive.

I. Reflexive: Let (x,y)(x, y) be an ordered pair of integers, y0.y\not =0.To show C is reflexive we must show ((x,y),(x,y))C.((x, y),(x,y))\in C. Multiplication of integers is commutative, so xy=yx.xy=yx. Thus ((x,y),(x,y))C.((x, y),(x,y))\in C.


II. Symmetric: Let (x,y)(x, y)and (z,w)(z,w) be ordered pairs of integers such that ((x,y),(z,w))C.((x, y),(z,w))\in C. Then yz=xw.yz=xw. This equation is equivalent to wx=zy,wx=zy, so ((z,w),(x,y))C.((z, w),(x,y))\in C. This shows CC is symmetric.


III. Transitive: Let (x,y),(z,w),(x,y), (z,w), and (u,v)(u,v) be ordered pairs of integers such that ((x,y),(z,w))C((x, y),(z,w))\in C and ((z,w),(u,v))C((z, w),(u,v))\in C. Then yz=xwyz=xw and wu=zv.wu=zv. Thus, yzu=xwuyzu=xwu and xwu=xzv,xwu=xzv, which implies yzu=xzv.yzu=xzv. Since z0,z\not=0, we can cancel it from both sides of this equation to get yu=xv.yu=xv. This shows ((x,y),(u,v))C,((x, y),(u,v))\in C, and so CC is transitive.


Since CC is reflexive, symmetric, and transitive then CC is an equivalence relation.



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