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Show that 32n 1 is always a multiple of 8 for all n = N = {1, 2, 3,...}. We say one integer A is a multiple of 8 means there exists an integer B such that A=8B. For example, 16 is a multiple of 8 since 16 = 8 × 2. Proof: Let P(n)= 32 - 1. 1) Base (n=1): P(1) = 32 - 1 = 8× Thus, P(1) is a multiple of 8. 2) Induction hypothesis: Assume P(k) = 32k - 1 is always a multiple of 8 for any k Є N = {1, 2, 3, . . . }. 3) Induction step: We want to show that P(k) is true P(k+1) is true where P(k+1)= Since 32k - 1 is always a multiple of 8, we can say there exists an integer x such that 32k - 1 = 8x. Thus, we can write P(k+1)=32kx -1 =32k × (8+ =32k x 8+ =32k x 8+ (by induction hypothesis) = 8(32k + Since the number inside the parentheses above is an integer, we have proved that P(k + 1) is a multiple of 8, i.e. P(k+1) is true.

Show that 32n 1 is always a multiple of 8 for all n = N = {1, 2, 3,...}. We say one integer A is a multiple of 8 means there exists an integer B such that A=8B. For example, 16 is a multiple of 8 since 16 = 8 × 2. Proof: Let P(n)= 32 - 1. 1) Base (n=1): P(1) = 32 - 1 = 8× Thus, P(1) is a multiple of 8. 2) Induction hypothesis: Assume P(k) = 32k - 1 is always a multiple of 8 for any k Є N = {1, 2, 3, . . . }. 3) Induction step: We want to show that P(k) is true P(k+1) is true where P(k+1)= Since 32k - 1 is always a multiple of 8, we can say there exists an integer x such that 32k - 1 = 8x. Thus, we can write P(k+1)=32kx -1 =32k × (8+ =32k x 8+ =32k x 8+ (by induction hypothesis) = 8(32k + Since the number inside the parentheses above is an integer, we have proved that P(k + 1) is a multiple of 8, i.e. P(k+1) is true.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Show that 32n 1 is always a multiple of 8 for all n = N = {1, 2, 3,...}. We say one integer A is a multiple of 8 means there exists an integer B such that A=8B. For example, 16 is a multiple of 8 since 16 = 8 × 2. Proof: Let P(n)= 32 - 1. 1) Base (n=1): P(1) = 32 - 1 = 8× Thus, P(1) is a multiple of 8. 2) Induction hypothesis: Assume P(k) = 32k - 1 is always a multiple of 8 for any k Є N = {1, 2, 3, . . . }. 3) Induction step: We want to show that P(k) is true P(k+1) is true where P(k+1)= Since 32k - 1 is always a multiple of 8, we can say there exists an integer x such that 32k - 1 = 8x. Thus, we can write P(k+1)=32kx -1 =32k × (8+ =32k x 8+ =32k x 8+ (by induction hypothesis) = 8(32k + Since the number inside the parentheses above is an integer, we have proved that P(k + 1) is a multiple of 8, i.e. P(k+1) is true.
Transcribed Image Text:Show that 32n 1 is always a multiple of 8 for all n = N = {1, 2, 3,...}. We say one integer A is a multiple of 8 means there exists an integer B such that A=8B. For example, 16 is a multiple of 8 since 16 = 8 × 2. Proof: Let P(n)= 32 - 1. 1) Base (n=1): P(1) = 32 - 1 = 8× Thus, P(1) is a multiple of 8. 2) Induction hypothesis: Assume P(k) = 32k - 1 is always a multiple of 8 for any k Є N = {1, 2, 3, . . . }. 3) Induction step: We want to show that P(k) is true P(k+1) is true where P(k+1)= Since 32k - 1 is always a multiple of 8, we can say there exists an integer x such that 32k - 1 = 8x. Thus, we can write P(k+1)=32kx -1 =32k × (8+ =32k x 8+ =32k x 8+ (by induction hypothesis) = 8(32k + Since the number inside the parentheses above is an integer, we have proved that P(k + 1) is a multiple of 8, i.e. P(k+1) is true.
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