4.86. Evaluate the proposed proof of the following result. Result Let x, y E Z such that 3|x. If 3|(x +y), then 3 y. Proof Since 3x, it follows that a = 3a, where a E Z. Assume that 3 (x + y). Then x+y = 3b for some integer b. Hence, y = 36 – x = 3b – 3a = 3(b – a). Since b- a is an integer, 3 y. For the converse, assume that 3 y. Therefore, y = 3c, where c€ Z. Thus, x +y = 3a + 3c = 3(a + c). Since a + c is an integer, 3 (x +y).
4.86. Evaluate the proposed proof of the following result. Result Let x, y E Z such that 3|x. If 3|(x +y), then 3 y. Proof Since 3x, it follows that a = 3a, where a E Z. Assume that 3 (x + y). Then x+y = 3b for some integer b. Hence, y = 36 – x = 3b – 3a = 3(b – a). Since b- a is an integer, 3 y. For the converse, assume that 3 y. Therefore, y = 3c, where c€ Z. Thus, x +y = 3a + 3c = 3(a + c). Since a + c is an integer, 3 (x +y).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Any mistake in the proof or corrections needed to be made
Transcribed Image Text:4.86. Evaluate the proposed proof of the following result.
Result Let x, y € Z such that 3 x. If 3 (x + y), then 3 y.
Proof Since 3 x, it follows that x =
у — 36 — ӕ — 3b — За — 3(b — а).
Since b – a is an integer, 3|y.
For the converse, assume that 3|y. Therefore, y = 3c, where c € Z. Thus, x + y = 3a + 3c = 3(a + c). Since a + c is an
integer, 3|(x + y).
3a, where a E Z. Assume that 3|(x + y). Then x + y = 3b for some integer b. Hence,
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