Evaluate the proof by induction of the following statement: For all positive integers n. 4 divides...

 

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Evaluate the proof by induction of the following statement: For all positive integers n. 4 divides (32n +7). Proof: Let P (n) be the statement: 4 divides (32n+7). Basis step: We need to show that P(1) is true. P(1) is true since 32 + 7 = 16 is divisible by 4. Inductive step: For the inductive step, we assume that P(k) is true for an arbitrary k. Thus, we assume that divides (32k+7), that is, 32k+7 = 4m for some integer m. We must show that whenever the inductive hypothesis P(k) is true, then so is P(k+1). P(x + 1) is the statement 4 divides (32(k+1)+7). 32(k+1)+7 = = = = 9(32k) +7 8(32k) + 32k +7 8(32k) + 4m 4(2(32k) + m) Thus, 4 divides (32(k+1)+7), and P(k+1) is true. Thus, P(k) implies P(k+1). Therefore, by the Principle of Mathematical Induction, we conclude that P(n) is true for all positive integers. O a. None of the other alternatives. O b. The theorem is false and the proof incorrectly shows it is true. O c. The theorem is true but the proof contains arithmetic mistakes, which makes the proof incorrect. O d. The proof correctly shows that the theorem is true. Evaluate the proof by induction of the following statement: For all positive integers n. 4 divides (32n +7). Proof: Let P (n) be the statement: 4 divides (32n+7). Basis step: We need to show that P(1) is true. P(1) is true since 32 + 7 = 16 is divisible by 4. Inductive step: For the inductive step, we assume that P(k) is true for an arbitrary k. Thus, we assume that divides (32k+7), that is, 32k+7 = 4m for some integer m. We must show that whenever the inductive hypothesis P(k) is true, then so is P(k+1). P(x + 1) is the statement 4 divides (32(k+1)+7). 32(k+1)+7 = = = = 9(32k) +7 8(32k) + 32k +7 8(32k) + 4m 4(2(32k) + m) Thus, 4 divides (32(k+1)+7), and P(k+1) is true. Thus, P(k) implies P(k+1). Therefore, by the Principle of Mathematical Induction, we conclude that P(n) is true for all positive integers. O a. None of the other alternatives. O b. The theorem is false and the proof incorrectly shows it is true. O c. The theorem is true but the proof contains arithmetic mistakes, which makes the proof incorrect. O d. The proof correctly shows that the theorem is true.

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