I'm sure that this is true, because A and B have the same cardinality (number of elements in the set). So there has to be the same number of ordered pairs with C in both cases. But I'm not sure how to exactly prove it since we just started with cardinality.
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$\begingroup$ See also this post and other questions linked there. $\endgroup$– Martin SleziakDec 7, 2016 at 5:19
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2 Answers
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You just need a bijection between $A \times C$ and $B \times C$ given a bijection between $A$ and $B$.
Let $f: A \rightarrow B$ be a bijection. Then define $g: A \times C \rightarrow B \times C$ given by $g(a,c)=(f(a),c).$
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Ac= Ac ( for any value)
Now given that a=b Now we can replace the right hand side "a" of this given equation with its value B, Then equation will be
AC = BC (proved)
