10 Problems with Functions, Sets, Recurrence Relations, Modular Operations, and Equivalence

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Question
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Solution
9)
We need to prove the relation is reflexive, symmetric and transitive.
For reflexivity, this means (a,b) R (a,b) This is true since a*b=b*a => R is reflexive.
For symmetry, this means if (a,b) R (c,d), then ad=bc.
On the other hand, cb=da (we reverted the order)=> (c,d) R (a,b)=> R is a symmetric relation.
For transitivity, we assume that (a,b) R (c,d) and also (c,d) R (e,f).
In this case it means that ad=bc and also cf=de
We multiply the two equalities side by side => adcf=bcde => adcf-bcde=0=> (af-be)cd=0...

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