Let A,BA,B, and CC be sets such that |A|=11|A|=11, |B|=7|B|=7 and |C|=10|C|=10. For each element (x,y)∈A×A(x,y)∈A×A, we associate with it a one-to-one function f(x,y):B→Cf(x,y):B→C. Prove that there will be two distinct elements of A×AA×A whose associated functions have the same range.
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