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Proof Assume that 2 | (x² − 1). So x² − 1 = 2y for some integer y. Thus, x² = 2y + 1 is an odd integer. It then follows by Theorem 3.12 that x too is odd. Hence, x = 2z + 1 for some integer z. Then x² − 1 = (2z + 1)² − 1 = (4z² + 4z + 1) − 1 = 4z² + 4z = 4(z² + z). Since z² + z is an integer, 4 | (x² − 1).

Proof Assume that 2 | (x² − 1). So x² − 1 = 2y for some integer y. Thus, x² = 2y + 1 is an odd integer. It then follows by Theorem 3.12 that x too is odd. Hence, x = 2z + 1 for some integer z. Then x² − 1 = (2z + 1)² − 1 = (4z² + 4z + 1) − 1 = 4z² + 4z = 4(z² + z). Since z² + z is an integer, 4 | (x² − 1).

Advanced Engineering Mathematics
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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Please if able explain the following example in detail, I dont get the substitution and am very confused.

Result Proof Let x = Z. If 2 | (x² − 1), then 4 | (x² − 1). - Assume that 2 | (x² − 1). So x². - 1 = 2y for some integer y. Thus, x² = 2y + 1 is an odd integer. It then follows by Theorem 3.12 that x too is odd. Hence, x = 2z + 1 for some integer z. Then x² − 1 = (2z + 1)² − 1 = (4z² + 4z + 1) − 1 = 4z² + 4z = 4(z² + z). Since z² + z is an integer, 4 | (x² − 1).
Transcribed Image Text:Result Proof Let x = Z. If 2 | (x² − 1), then 4 | (x² − 1). - Assume that 2 | (x² − 1). So x². - 1 = 2y for some integer y. Thus, x² = 2y + 1 is an odd integer. It then follows by Theorem 3.12 that x too is odd. Hence, x = 2z + 1 for some integer z. Then x² − 1 = (2z + 1)² − 1 = (4z² + 4z + 1) − 1 = 4z² + 4z = 4(z² + z). Since z² + z is an integer, 4 | (x² − 1).
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