MATH1011 Chap.4 Mathematical Induction in Two Forms
Mathematical Induction in Two Forms
A proof with a starting point and a reusable step
Mathematical induction proves a statement for every positive integer without checking those integers one at a time. The first principle has two obligations. Verify the statement at the first allowed value, then assume it at an arbitrary integer n and derive the statement at n + 1. The base case starts the chain; the implication carries truth along it.
A page of successful numerical checks supplies evidence for a pattern, but it does not replace either obligation.
Write P(n) as a complete mathematical statement before doing algebra. In a sum identity, P(n) includes the full sum and the claimed closed form. In a divisibility claim, convert the hypothesis into an equation such as an expression equals a fixed divisor times an integer.
That equation is what the next-case calculation can actually use. A correct proof shows the exact line at which the induction hypothesis enters.
When one base case is not enough
The second principle in this course is the two-step form. It verifies P(1) and P(2), assumes two consecutive statements P(n) and P(n + 1), and proves P(n + 2).
This structure matches a recurrence whose next term depends on the previous two. One base case would start only one parity chain and leave the assumed pair unavailable, so the recurrence itself tells you how many starting values the proof needs.
Keep this course-specific form separate from the broader strong-induction principle found in other texts. The question is not asking for every earlier case as an assumption.
It asks for two neighbouring cases and one step of length two. State the version you use instead of writing the vague phrase by induction.
How to avoid a circular inductive step
Open with the left side at the new index and work in one direction until it has the target form. Do not begin by writing the equation you are meant to prove and manipulate both sides as if it were already true.
For divisibility, expose a visible factor of the required divisor. For an identity, isolate the old partial sum so the hypothesis can replace it. Finish by naming the principle and the integer range, because the algebra alone does not announce why every case follows.
What this chapter covers
- 01
The base case and the induction hypothesis
- 02
Deriving P(n + 1) from P(n)
- 03
Two consecutive bases for a two-step recurrence
- 04
Divisibility proofs with an integer multiplier
- 05
Sum identities and isolating the old partial sum
- 06
Circular steps and missing conclusions
Worked example · free
Prove a divisibility pattern without assuming the target
- 1Check the first positive integer and express the result as a multiple of 7.
- 2Assume the claim at n by introducing an integer multiplier.
- 3Reach the case n + 1 and close by naming the principle.
Key terms
- Base Case
- The Base Case verifies the first value from which an inductive chain is meant to start.
- Induction Hypothesis
- The Induction Hypothesis is the temporary assumption that P(n), or the required consecutive cases, already holds at an arbitrary index.
- Inductive Step
- The Inductive Step derives the next required case from the explicitly stated hypothesis.
- First Principle
- The First Principle combines P(1) with the implication from P(n) to P(n + 1).
- Second Principle
- The Second Principle used here combines two base cases with a step from two consecutive statements to the case two places ahead.
Mathematical Induction in Two Forms FAQ
Why do several checked examples not prove an induction claim?
A finite list cannot rule out a later counterexample. Induction needs a base case to start the chain and a general implication that transports the statement from an arbitrary index to the next required index. The checked values may suggest the formula, but the implication is what reaches every positive integer.
When should I use the second principle in this course?
Use it when the next case naturally depends on two consecutive earlier cases. Verify both starting values, assume both neighbouring statements, and derive the case two places forward. Do not call it strong induction unless the question actually permits all earlier cases as assumptions.
How can I tell whether my inductive step is circular?
Point to the first line where the hypothesis is substituted. If no such line exists, or if the step begins by assuming the complete n + 1 equation and merely rearranges it, the proof is circular. Start from one side of the new case and derive the other.
Exam move
For each tutorial induction problem, write four labelled lines before doing any detailed algebra: statement, base, hypothesis and target. Then circle the exact substitution that uses the hypothesis. Mix sum identities, divisibility and two-step recurrences, because the surface algebra changes while the logical skeleton remains stable. Finish every proof with the named principle and its integer range.
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