Question #5300

prove the relation a=L(mod m) is aequivalent relation?

Expert's answer

to prove we must

show that this relation is reflexive, symmetric and transitive

reflexive:

a=a( mod m) it is obvious statement by the definition of mod m

symmetric if

a=b(mod m) then b=a (mod m) to prove this we can write if a=b(mod m) then a have

the same remainder after division by m as b. so b have the same remainder after

division by m as a so b=a(mod m)

transitive: if a=b(mod m) and b=c(mod m)

then a=c(mod m) . if a=b (mod m) so a have reminder after division by m and its

equal to the same remainder of b , from b=c(mod m) we have that b have reminder

after division by m and its equal to the same remainder of c. so a and c have

the same remainder . so a=c(mod m)

show that this relation is reflexive, symmetric and transitive

reflexive:

a=a( mod m) it is obvious statement by the definition of mod m

symmetric if

a=b(mod m) then b=a (mod m) to prove this we can write if a=b(mod m) then a have

the same remainder after division by m as b. so b have the same remainder after

division by m as a so b=a(mod m)

transitive: if a=b(mod m) and b=c(mod m)

then a=c(mod m) . if a=b (mod m) so a have reminder after division by m and its

equal to the same remainder of b , from b=c(mod m) we have that b have reminder

after division by m and its equal to the same remainder of c. so a and c have

the same remainder . so a=c(mod m)

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