Foundation Exam
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چکیده
First, we prove the function g is an injection (i.e., g is one-to-one). We need to prove that if g(m) = g(n) --(1), where m and n are two integers, then m = n --(2). Since g(m) (and also g(n)) can be defined by two different formulas depending whether the argument m (and n) is even or odd, we consider the following 3 cases: (Case 1) Both m and n are even. In this case, (1) implies 1 – m = 1 – n; thus, –m = –n, and m = n. So (2) is proved in this case. (Case 2) Both m and n are odd. In this case, (1) implies m + 3 = n + 3; thus, m = n is true. So (2) is proved in this case. (Case 3) One of the two numbers is even, the other is odd. We may assume m is even and n is odd (that is, we name the even number m and name the other n). In this case, (1) implies (using the definition of the function g) that 1 – m = n + 3. Thus, m + n = 1 – 3 = –2. However, since m is even and n is odd, m + n must be odd, so it cannot be equal to –2. That is, we proved that when m ≠ n and m is even and n is odd, g(m) = g(n) leads to a contradiction, which provided an indirect proof of (1) ⇒ (2) in this case. Therefore, we proved that (1) implies (2) in all cases, which proved the injection property of function g.
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