Mathematical Induction

🎯 Learning Objectives

• State the principle of mathematical induction (weak form)
• Apply induction to prove formulas, inequalities, and divisibility
• Use strong induction when multiple previous cases are needed
• Recognise which form of induction is appropriate
• Construct complete inductive proofs with base case and inductive step

4.1 Intuition: The Domino Effect

1. Base case: Knock over the first domino.
2. Inductive step: If domino $k$ falls, domino $k+1$ falls.
3. Conclusion: All dominoes fall.

🔑 Key Insight: Induction proves statements about infinitely many natural numbers using only two steps.

4.2 Weak (Simple) Induction

Principle

1. Base case: Prove $P(n_0)$ is true.
2. Inductive hypothesis: Assume $P(k)$ is true for some $k \geq n_0$.
3. Inductive step: Show $P(k) \to P(k+1)$.
4. Conclusion: By induction, $P(n)$ is true for all $n \geq n_0$.

Example 1: Sum of First $n$ Natural Numbers

• Base case ($n=1$): LHS = 1, RHS = $\frac{1(2)}{2} = 1$. ✓
• Inductive hypothesis: Assume $1 + 2 + \dots + k = \frac{k(k+1)}{2}$.
• Inductive step: Show for $k+1$:

Example 2: Sum of Squares

• Base ($n=1$): $1 = \frac{1 \cdot 2 \cdot 3}{6} = 1$. ✓
• IH: Assume for $n=k$.
• Step: $\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$

Example 3: Divisibility

• Base ($n=1$): $7^1 - 1 = 6$, divisible by 6. ✓
• IH: Assume $7^k - 1$ is divisible by 6, i.e., $7^k = 6m + 1$.
• Step: $7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 7 - 1 = 42m + 6 = 6(7m + 1)$. Thus $7^{k+1} - 1$ is divisible by 6. $\square$

Example 4: Inequality

• Base ($n=1$): $2^1 = 2 > 1$. ✓
• IH: Assume $2^k > k$.
• Step: $2^{k+1} = 2 \cdot 2^k > 2 \cdot k = k + k \geq k + 1$ (since $k \geq 1$). Thus $2^{k+1} > k + 1$. $\square$

Example 5: Sum of Odd Numbers

• Base ($n=1$): $1 = 1^2$. ✓
• IH: Assume $1 + 3 + 5 + \dots + (2k-1) = k^2$.
• Step: Add the next odd number $(2k+1)$:

4.3 Strong Induction

Principle

1. Base case(s): Prove $P(n_0), P(n_0+1), \dots, P(m)$.
2. Inductive hypothesis: Assume $P(j)$ for all $j$ with $n_0 \leq j \leq k$.
3. Inductive step: Use this to prove $P(k+1)$.

When to Use Strong Induction

Example 1: Prime Factorization

• Base ($n=2$): $2$ itself is prime. ✓
• IH: Assume every integer from 2 up to $k$ has a prime factorization.
• Step: Consider $n = k+1$.
  • If $k+1$ is prime, done.
  • If $k+1$ is composite, then $k+1 = ab$ where $2 \leq a, b \leq k$.
  • By IH, $a$ and $b$ have prime factorizations. Their product is a prime factorization of $k+1$. $\square$

Example 2: Fibonacci Numbers

• Base ($n=1,2$): $F_1 = 1 \geq (\frac{3}{2})^{-1} = \frac{2}{3}$ ✓; $F_2 = 1 \geq (\frac{3}{2})^0 = 1$ ✓.
• IH: Assume true for all $j \leq k$.
• Step: For $n = k+1$:

Example 3: Binary Representation

• Base ($n=1$): $1 = 2^0$. ✓
• IH: True for all $j < n$.
• Step: Let $2^k$ be the largest power of 2 with $2^k \leq n$.
  • If $n = 2^k$, done.
  • If $n > 2^k$, write $n = 2^k + r$, where $r < 2^k$ and $r < n$.
  • By IH, $r$ is a sum of distinct powers of 2, none exceeding $2^{k-1}$ (since $r < 2^k$).
  • Together with $2^k$, we get the representation. $\square$

4.4 Induction Pitfalls

📊 Formula Summary

✅ Practice Questions

Solution

Base ($n=1$): $1! = 1 \geq 2^{0} = 1$. ✓ IH: $k! \geq 2^{k-1}$. Step: $(k+1)! = (k+1)k! \geq (k+1)2^{k-1} \geq 2 \cdot 2^{k-1} = 2^k$ (since $k+1 \geq 2$ for $k \geq 1$). Thus $(k+1)! \geq 2^k$. $\square$ Q3: Prove $2^n \geq n^2$ for $n \geq 4$. Solution

Base ($n=4$): $2^4 = 16 \geq 4^2 = 16$. ✓ IH: $2^k \geq k^2$ for $k \geq 4$. Step: $2^{k+1} = 2 \cdot 2^k \geq 2k^2$. Need to show $2k^2 \geq (k+1)^2 = k^2 + 2k + 1$. $2k^2 - (k^2 + 2k + 1) = k^2 - 2k - 1 = (k-1)^2 - 2$. For $k \geq 4$, $(k-1)^2 - 2 \geq 9 - 2 = 7 > 0$. So $2k^2 > (k+1)^2$ and $2^{k+1} \geq 2k^2 \geq (k+1)^2$. $\square$ Q4: Prove that any amount of postage $\geq 8$ cents can be formed using 3-cent and 5-cent stamps. Solution

Base: $8 = 3 + 5$, $9 = 3 + 3 + 3$, $10 = 5 + 5$. ✓ IH: True for all values from 8 to $k$. Step: For $k+1 \geq 11$, consider $(k+1) - 3 = k - 2 \geq 8$. By IH, $k-2$ can be formed. Add one 3-cent stamp. $\square$ Q5: Prove that $F_n < 2^n$ for all $n \geq 1$ (Fibonacci numbers). Solution

Base: $F_1 = 1 < 2^1$, $F_2 = 1 < 2^2$. ✓ IH: Assume $F_k < 2^k$ and $F_{k-1} < 2^{k-1}$. Step: $F_{k+1} = F_k + F_{k-1} < 2^k + 2^{k-1} = 3 \cdot 2^{k-1} = \frac{3}{2} \cdot 2^k < 2 \cdot 2^k = 2^{k+1}$. $\square$ Join Discord PreviousLogical FallaciesNextAdvanced Proofs