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Let A and B be sets. Show that A x B and B x A have equal cardinality by constructing an explicit bijection between the two sets. Then use the following proposition to prove that multiplication is commutative. (Let n, m be natural numbers. Then nxm=mxn)

Proposition: Cardinal arithmetic

a) Let X be a finite set, and let x be an object which is not an element of X. Then X U (union) {x} is finite and #(X U {x})= #(X)+1

b) Let X and Y be finite sets. Then X U Y is finite and # (X U Y) is less than or equal to #(X) + #(Y). If in addition X and Y are disjoint (i.e., X intersection Y = the empty set), then #(X U Y)= #(X) + #(Y)

c) Let X be a finite set, and let Y be a subset of X. Then Y is finite, and #(Y) is less than or equal to #(X), If in addition Y does not equal X (i.e. Y is a proper subset of X), then we have #(Y) is less than #(X)

d) If X is a finite set, and f:X-->Y is a function, then f(X) is a finite set with #(f(X)) less than or equal to #(X). If in addition f is one to one, then #(f(X)) = #(X)

e) Let X and Y be finite sets. Then cartesian product X x Y is finite and # (X x Y) = #(X) x #(Y)

f) Let X and Y be finite sets. Then the set Y^X is finite and #(Y^X)= #(Y)^#(X)

#### Solution Preview

Consider the mapping f:AxB-->BxA which sends (a,b) to (b,a). This mapping is injective. Indeed, let f(a,b)=f(a',b'). We have f(a,b)=(b,a), and f(a',b')=(b',a'). Therefore (b,a)=(b',a'). This implies that b=b' and a=a'. Thus (a,b)=(a',b'). ...

#### Solution Summary

The proof about cardinality is examined.

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