For each integer n ≥ 0, 1 + 3n ≤ 4". Proof: Let the property P(n) be the inequality 1 + 3n ≤ 4". We will prove that P(n) is true for every integer n ≥ 0 0 Show that P 0 is true: P(0) is true because its left-hand side is 1 + 3 . 0 = 1 and its right-hand side is 4 = 1, and 1 ≤ 1. Show that for every ✓ integer k≥0 if ✓ P(k) is true then P 0 is true: Let k be any integer with k ≥ 0 and suppose 1+ 3k ≤ 4k (inductive hypothesis). We must show that 1 + 3 k+1 ≤ 4k+4 Multiplying both sides of the inequality in the inductive hypothesis by 4 gives 4. 4 + ( 3k k≤ 4k+4 (Inequality 1). Now 1 + 3 (k + 1) = 4+ (k+1) ≤ 4 + 12k because 3k ≤ 12k since k ≥0 Thus, 1+3 k+3 ≤ 412k (Inequality 2). k+4 ≤ 4 • k+1 or, equivalently, It follows from Inequality 1 and Inequality 2 and the transitive property of order that 1 +3 k+3 ≤ 4k+4 which is what was to be shown. '
For each integer n ≥ 0, 1 + 3n ≤ 4". Proof: Let the property P(n) be the inequality 1 + 3n ≤ 4". We will prove that P(n) is true for every integer n ≥ 0 0 Show that P 0 is true: P(0) is true because its left-hand side is 1 + 3 . 0 = 1 and its right-hand side is 4 = 1, and 1 ≤ 1. Show that for every ✓ integer k≥0 if ✓ P(k) is true then P 0 is true: Let k be any integer with k ≥ 0 and suppose 1+ 3k ≤ 4k (inductive hypothesis). We must show that 1 + 3 k+1 ≤ 4k+4 Multiplying both sides of the inequality in the inductive hypothesis by 4 gives 4. 4 + ( 3k k≤ 4k+4 (Inequality 1). Now 1 + 3 (k + 1) = 4+ (k+1) ≤ 4 + 12k because 3k ≤ 12k since k ≥0 Thus, 1+3 k+3 ≤ 412k (Inequality 2). k+4 ≤ 4 • k+1 or, equivalently, It follows from Inequality 1 and Inequality 2 and the transitive property of order that 1 +3 k+3 ≤ 4k+4 which is what was to be shown. '
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:For each integer n ≥ 0, 1 + 3n ≤ 4".
Proof: Let the property P(n) be the inequality 1 + 3n ≤ 4".
We will prove that P(n) is true for every integer n ≥ 0
0
Show that P 0
is true: P(0) is true because its left-hand side is 1 + 3 .
0
= 1 and its right-hand side is 4
= 1, and 1 ≤ 1.
Show that for every
✓ integer k≥0
if
✓ P(k) is true then
P
0
is true: Let k be any integer with k ≥ 0
and suppose 1+ 3k ≤ 4k (inductive hypothesis).
We must show that 1 + 3
k+1
≤ 4k+4
Multiplying both sides of the inequality in the inductive hypothesis by 4 gives 4.
4 +
(
3k
k≤ 4k+4
(Inequality 1).
Now 1 + 3 (k + 1) = 4+ (k+1)
≤ 4 + 12k because 3k ≤ 12k since k ≥0
Thus, 1+3
k+3
≤ 412k (Inequality 2).
k+4
≤ 4 •
k+1
or, equivalently,
It follows from Inequality 1 and Inequality 2 and the transitive property of order that 1 +3
k+3
≤ 4k+4
which is what was to be shown.
'
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