Chapter 2 · Sequences · Concept 5
2-24 Mathematical Induction for Proving Inequalities
2ⁿ eventually leaves 3n far behind — but not at first. At n = 3 it is actually smaller. So the claim is only true from a certain point on, and the proof has to know where that point is and why.
Guiding question
How do you prove an inequality for every n past a starting point?
Learning outcome. Apply the principle of mathematical induction to prove the validity of algebraic inequalities.
The same two parts, with one addition
Parts [1] and [2] are exactly as in lesson 2-23 — except part [1] now tests the starting value the question names, not necessarily n = 1.
and for anything with aⁿ in it
multiply both sides of the assumption by a
That turns ak into ak+1, which is what the target needs — and multiplying an inequality by a positive number keeps it pointing the same way.
But you are not finished there
With an equality, substituting gets you to the target. With an inequality it does not:
after multiplying: 2k+1 > 6k
but the target is: 2k+1 > 3(k + 1)
So you need a SECOND comparison.
Show that 6k > 3(k + 1) as well. Then chain them: if 2k+1 is bigger than 6k, and 6k is bigger than 3(k+1), then 2k+1 is bigger than 3(k+1). That chaining is the whole difficulty of this lesson.
Every one of these claims is false for small n and true from some point onward. That point is not decoration — it is what makes the statement true at all.
Warm Up · the book says n ≥ 4
| n | 2ⁿ | 3n | is it true? |
|---|---|---|---|
| 1 | 2 | 3 | FALSE ✗ |
| 2 | 4 | 6 | FALSE ✗ |
| 3 | 8 | 9 | FALSE ✗ |
| 4← starts here | 16 | 12 | true ✓ |
| 5 | 32 | 15 | true ✓ |
| 6 | 64 | 18 | true ✓ |
| 7 | 128 | 21 | true ✓ |
| 8 | 256 | 24 | true ✓ |
Look at the rows above n = 4. The claim is genuinely FALSE there — so "n ≥ 4" is not boilerplate, it is the whole reason the statement is true at all.
When n is a natural number greater than or equal to 4, prove that 2n > 3n.
- 1
Name the statement.
let (A) be 2n > 3nAs always.
Work with a partner. After multiplying the assumption, both of you write down the target separately. Compare them — if they are not the same, you still have a second comparison to do.
When h > 0 and n is a natural number greater than or equal to 3, prove that (1 + h)n > 1 + nh2.
Two letters this time, and one of them is not a whole number. The induction still runs on n alone — h just comes along for the ride.
In the Warm Up, where exactly did k ≥ 4 get used? And what would go wrong if the question had said n ≥ 1 instead?
Look at the second comparison, not the first.
Stopping after multiplying by a.
2ᵏ⁺¹ > 6k is not the target. You still have to show 6k > 3(k+1) and chain them.
Testing n = 1 when the question says n ≥ 4.
Part [1] must test the STARTING value. For 2ⁿ > 3n the claim is actually false at n = 1, 2 and 3.
Never using the condition k ≥ 4.
If your step never mentions it, you have almost certainly skipped the second comparison.
Writing the target from memory.
The target at k + 1 comes from substituting. For 3n + 1 that is 3(k+1) + 1 = 3k + 4, not 3k + 1.
Concluding "for all natural numbers".
If the proof started at 4, the conclusion is "for all n ≥ 4". Claiming more than you proved loses the mark.
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