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(b) Proposition. For each integer m, 5 divides (m-m). Proof. Let m e Z. We will prove that 5 divides (m³ - m) by proving that (m - m) =0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m = 0 (mod 5) and, hence, (m3 – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m = 1 (mod 5) and, hence, (ms -m) = (1 – 1) (mod 5), which means that (m3 - m) = 0 (mod 5). For the third case, if m = 2 (mod 5), then m = 32 (mod 5) and, hence, (m- m) = (32-2) (mod 5), which means that (m- m) = 0(mod 5).

(b) Proposition. For each integer m, 5 divides (m-m). Proof. Let m e Z. We will prove that 5 divides (m³ - m) by proving that (m - m) =0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m = 0 (mod 5) and, hence, (m3 – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m = 1 (mod 5) and, hence, (ms -m) = (1 – 1) (mod 5), which means that (m3 - m) = 0 (mod 5). For the third case, if m = 2 (mod 5), then m = 32 (mod 5) and, hence, (m- m) = (32-2) (mod 5), which means that (m- m) = 0(mod 5).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 46E
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I've been stuck on this one for a while. If the proposition is false, and the proof is incorrect, find the error in the proof and provide a counter example showing that it is false If the proposition is true, but the proof is wrong, fix the proof If everything is correct, nothing needs to be done
hence, (2a +b) = 0 (mod 3). (b) Proposition. For each integer m, 5 divides (m - m). Proof. Let m e Z. We will prove that 5 divides (m – m) by proving that (m-m) = 0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m5 = 0 (mod 5) and, hence, (m – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m = 1 (mod 5) and, hence, (m5- m) = (1– 1) (mod 5), which means that (ms- m) = 0 (mod 5). For the third case, if m = 2 (mod 5), then m5 hence, (m3 - m) = (32- 2) (mod 5), which means that (m - m) = 0 (mod 5). = 32 (mod 5) and, BY NC SA
Transcribed Image Text:hence, (2a +b) = 0 (mod 3). (b) Proposition. For each integer m, 5 divides (m - m). Proof. Let m e Z. We will prove that 5 divides (m – m) by proving that (m-m) = 0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m5 = 0 (mod 5) and, hence, (m – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m = 1 (mod 5) and, hence, (m5- m) = (1– 1) (mod 5), which means that (ms- m) = 0 (mod 5). For the third case, if m = 2 (mod 5), then m5 hence, (m3 - m) = (32- 2) (mod 5), which means that (m - m) = 0 (mod 5). = 32 (mod 5) and, BY NC SA
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