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Strong vs Weak Induction Comparison cheat sheet - grade 10-12

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This cheat sheet compares weak induction and strong induction, two proof methods used to prove statements about positive integers or nonnegative integers. Students need it because induction proofs have a strict structure, and small wording errors can make a proof incomplete. It helps students choose the right form of induction and write clear base cases, hypotheses, and induction steps.

Weak induction assumes one previous case, usually P(k)P(k), to prove the next case P(k+1)P(k+1). Strong induction assumes all previous cases from the base case through P(k)P(k) to prove P(k+1)P(k+1). Both methods are logically equivalent, but strong induction is often easier when the next result depends on several earlier results.

Key Facts

  • Weak induction proves a statement P(n)P(n) by showing a base case such as P(1)P(1), then proving P(k)P(k+1)P(k) \Rightarrow P(k+1) for all k1k \ge 1.
  • Strong induction proves P(n)P(n) by showing base cases, then proving P(1),P(2),,P(k)P(k+1)P(1), P(2), \ldots, P(k) \Rightarrow P(k+1) for all kk in the required range.
  • The induction hypothesis in weak induction is usually a single assumption, such as assuming P(k)P(k) is true.
  • The induction hypothesis in strong induction assumes every earlier statement is true, such as P(b),P(b+1),,P(k)P(b), P(b+1), \ldots, P(k).
  • If a statement begins at n=bn = b, the base case should prove P(b)P(b), not automatically P(1)P(1).
  • Multiple base cases may be needed when the induction step uses results such as P(k1)P(k-1) or P(k2)P(k-2) to prove P(k+1)P(k+1).
  • Strong induction is useful for divisibility, recurrence relations, tiling problems, and proofs where P(k+1)P(k+1) depends on a smaller case not equal to P(k)P(k).
  • Weak and strong induction are logically equivalent because each can be used to justify the other, but they often lead to different proof styles.

Vocabulary

Mathematical induction
A proof method used to show that a statement P(n)P(n) is true for every integer nn in a specified range.
Base case
The first value or values of nn that are proven directly before using the induction step.
Induction hypothesis
The temporary assumption that earlier statement values, such as P(k)P(k) or P(b),,P(k)P(b), \ldots, P(k), are true.
Induction step
The part of an induction proof that uses the induction hypothesis to prove the next statement, usually P(k+1)P(k+1).
Weak induction
An induction method that assumes P(k)P(k) is true and uses it to prove P(k+1)P(k+1).
Strong induction
An induction method that assumes all statements from the base case through P(k)P(k) are true and uses them to prove P(k+1)P(k+1).

Common Mistakes to Avoid

  • Forgetting the base case makes the proof invalid because the induction step only shows how truth passes from one case to the next.
  • Using P(k+1)P(k+1) inside the induction hypothesis is wrong because P(k+1)P(k+1) is the statement that still needs to be proved.
  • Choosing weak induction when the proof needs P(k1)P(k-1) or another earlier case can leave a logical gap, so strong induction or extra base cases may be needed.
  • Proving only examples such as P(1)P(1), P(2)P(2), and P(3)P(3) is not induction because examples do not prove the statement for all integers.
  • Starting the base case at the wrong value is incorrect because the domain might begin at n=0n = 0, n=1n = 1, or another specified integer.

Practice Questions

  1. 1 Use weak induction to prove that 1+2++n=n(n+1)21 + 2 + \cdots + n = \frac{n(n+1)}{2} for all integers n1n \ge 1.
  2. 2 Use induction to prove that 33 divides 22n12^{2n} - 1 for all integers n1n \ge 1.
  3. 3 A sequence is defined by a1=1a_1 = 1, a2=1a_2 = 1, and an=an1+an2a_n = a_{n-1} + a_{n-2} for n3n \ge 3. Explain what base cases are needed to prove a property of ana_n by induction.
  4. 4 Decide whether weak induction or strong induction is more natural for proving that every integer n2n \ge 2 can be written as a product of primes, and explain why.