Question

Let A, B, C and D be sets such that | A| = |C| and |B| = |D|. Prove that |A × B| = |C × D|

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Darrell

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To prove that |A × B| = |C × D|, we need to show that there exists a bijective function between the two sets.

Let's consider the function f: A × B → C × D defined as f(a, b) = (c, d) where c is any element in C and d is any element in D. Since |A| = |C| and |B| = |D|, we know that there exists a bijective function g: A → C and a bijective function h: B → D.

Now, let's define a function F: C × D → A × B as F(c, d) = (g^(-1)(c), h^(-1)(d)), where g^(-1) and h^(-1) are the inverse functions of g and h, respectively.

We will prove that both f and F are bijections.

First, let's show that f is injective. Suppose (a1, b1) and (a2, b2) are two elements in A × B such that f(a1, b1) = f(a2, b2). This implies that (g(a1), h(b1)) = (g(a2), h(b2)). Since g and h are both injective functions, we conclude that a1 = a2 and b1 = b2. Therefore, f is injective.

Next, let's show that f is surjective. Let (c, d) be an element in C × D. Since g and h are both surjective functions, there exists a1 in A such that g(a1) = c, and there exists b1 in B such that h(b1) = d. Therefore, f(a1, b1) = (c, d). Hence, f is surjective.

Now, let's show that F is injective. Suppose (c1, d1) and (c2, d2) are two elements in C × D such that F(c1, d1) = F(c2, d2). This implies that (g^(-1)(c1), h^(-1)(d1)) = (g^(-1)(c2), h^(-1)(d2)). Since g^(-1) and h^(-1) are both injective functions, we conclude that c1 = c2 and d1 = d2. Therefore, F is injective.

Finally, let's show that F is surjective. Let (a, b) be an element in A × B. Since g and h are both surjective functions, there exists c1 in C such that g(a) = c1, and there exists d1 in D such that h(b) = d1. Therefore, F(c1, d1) = (g^(-1)(g(a)), h^(-1)(h(b))) = (a, b). Hence, F is surjective.

Since f is a bijection from A × B to C × D, and F is a bijection from C × D to A × B, we can conclude that |A × B| = |C × D|.

Answer: |A × B| = |C × D|

Let's consider the function f: A × B → C × D defined as f(a, b) = (c, d) where c is any element in C and d is any element in D. Since |A| = |C| and |B| = |D|, we know that there exists a bijective function g: A → C and a bijective function h: B → D.

Now, let's define a function F: C × D → A × B as F(c, d) = (g^(-1)(c), h^(-1)(d)), where g^(-1) and h^(-1) are the inverse functions of g and h, respectively.

We will prove that both f and F are bijections.

First, let's show that f is injective. Suppose (a1, b1) and (a2, b2) are two elements in A × B such that f(a1, b1) = f(a2, b2). This implies that (g(a1), h(b1)) = (g(a2), h(b2)). Since g and h are both injective functions, we conclude that a1 = a2 and b1 = b2. Therefore, f is injective.

Next, let's show that f is surjective. Let (c, d) be an element in C × D. Since g and h are both surjective functions, there exists a1 in A such that g(a1) = c, and there exists b1 in B such that h(b1) = d. Therefore, f(a1, b1) = (c, d). Hence, f is surjective.

Now, let's show that F is injective. Suppose (c1, d1) and (c2, d2) are two elements in C × D such that F(c1, d1) = F(c2, d2). This implies that (g^(-1)(c1), h^(-1)(d1)) = (g^(-1)(c2), h^(-1)(d2)). Since g^(-1) and h^(-1) are both injective functions, we conclude that c1 = c2 and d1 = d2. Therefore, F is injective.

Finally, let's show that F is surjective. Let (a, b) be an element in A × B. Since g and h are both surjective functions, there exists c1 in C such that g(a) = c1, and there exists d1 in D such that h(b) = d1. Therefore, F(c1, d1) = (g^(-1)(g(a)), h^(-1)(h(b))) = (a, b). Hence, F is surjective.

Since f is a bijection from A × B to C × D, and F is a bijection from C × D to A × B, we can conclude that |A × B| = |C × D|.

Answer: |A × B| = |C × D|

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