Mathematical induction is a powerful technique used to prove statements or formulas that are true for all natural numbers. It’s a bit like a domino effect: if you can show that the first domino falls and that each falling domino will cause the next one to fall, then you’ve proven that all the dominos will fall. In mathematical terms, this method helps us show that a certain statement is true for every number in a sequence, starting from a base number and moving upwards.
What Is Mathematical Induction?
Mathematical induction involves two main steps:
- Base Case: Show that the statement is true for the initial value, often when n=1n = 1n=1.
- Inductive Step: Assume the statement is true for some arbitrary natural number kkk, and then show that if it’s true for kkk, it must also be true for k+1k + 1k+1.
If both steps are fruitful, you can conclude that the articulation is genuine for all normal numbers beginning from the base case.
Example 1: Entirety of the To begin with nnn Common Numbers
Let’s use mathematical induction to prove the formula for the sum of the first nnn natural numbers: S(n)=n(n+1)2S(n) = \frac{n(n + 1)}{2}S(n)=2n(n+1)
Step 1: Base Case
For n=1n = 1n=1, the sum of the first natural number is: S(1)=1S(1) = 1S(1)=1 Using the formula: 1⋅(1+1)2=22=1\frac{1 \cdot (1 + 1)}{2} = \frac{2}{2} = 121⋅(1+1)=22=1 So, the formula works for n=1n = 1n=1.
Step 2: Inductive Step
Assume the formula is true for some natural number kkk. It is: S(k)=k(k+1)2S(k) = \frac{k(k + 1)}{2}S(k)=2k(k+1)
We need to show that the formula holds for k+1k + 1k+1. According to the formula, the sum of the first k+1k + 1k+1 numbers is: S(k+1)=S(k)+(k+1)S(k + 1) = S(k) + (k + 1)S(k+1)=S(k)+(k+1) Using the induction hypothesis: S(k+1)=k(k+1)2+(k+1)S(k + 1) = \frac{k(k + 1)}{2} + (k + 1)S(k+1)=2k(k+1)+(k+1)
Combine the terms: S(k+1)=k(k+1)+2(k+1)2=k2+k+2k+22=(k+1)(k+2)2S(k + 1) = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{k^2 + k + 2k + 2}{2} = \frac{(k + 1)(k + 2)}{2}S(k+1)=2k(k+1)+2(k+1)=2k2+k+2k+2=2(k+1)(k+2)
This matches the formula for n=k+1n = k + 1n=k+1. So, the formula is correct for k+1k + 1k+1, and by induction, it’s true for all natural numbers.
Example 2: Total of the Initial N odd numbers
Let’s prove the sum of the first nnn odd numbers is n2n^2n2.
Step 1: Base Case
For n=1n = 1n=1, the sum of the first odd number is: 1=121 = 1^21=12 The formula works for n=1n = 1n=1.
Step 2: Inductive Step
Assume the formula is true for some natural number kkk. That is: Sum of first k odd numbers=k2\text{Sum of first } k \text{ odd numbers} = k^2Sum of first k odd numbers=k2
We need to show it’s true for k+1k + 1k+1. The sum of the first k+1k + 1k+1 odd numbers is: Sum of first (k+1) odd numbers=k2+(2k+1)\text{Sum of first } (k + 1) \text{ odd numbers} = k^2 + (2k + 1)Sum of first (k+1) odd numbers=k2+(2k+1) Using the induction hypothesis: k2+(2k+1)=(k+1)2k^2 + (2k + 1) = (k + 1)^2k2+(2k+1)=(k+1)2
So, the formula works for k+1k + 1k+1, and by induction, it’s true for all natural numbers.
Example 3: Divisibility by 3
We’ll prove that 2n−12^n – 12n−1 is divisible by 3 for all n≥1n \geq 1n≥1.
Step 1: Base Case
For n=1n = 1n=1: 21−1=12^1 – 1 = 121−1=1 1 is not divisible by 3, so let’s correct this to start from n=2n = 2n=2: For n=2n = 2n=2: 22−1=32^2 – 1 = 322−1=3 3 is divisible by 3. The base case works for n=2n = 2n=2.
Step 2: Inductive Step
Assume the statement is true for some natural number kkk. That is: 2k−1 is divisible by 32^k – 1 \text{ is divisible by 3}2k−1 is divisible by 3
We need to show that 2k+1−12^{k+1} – 12k+1−1 is also divisible by 3. Start with: 2k+1−1=2⋅2k−12^{k+1} – 1 = 2 \cdot 2^k – 12k+1−1=2⋅2k−1 Rewrite it using the induction hypothesis: 2⋅2k−1=2⋅(2k−1)+12 \cdot 2^k – 1 = 2 \cdot (2^k – 1) + 12⋅2k−1=2⋅(2k−1)+1 Since 2k−12^k – 12k−1 is divisible by 3, let’s say 2k−1=3m2^k – 1 = 3m2k−1=3m for some integer mmm. Then: 2⋅(2k−1)=2⋅3m=6m2 \cdot (2^k – 1) = 2 \cdot 3m = 6m2⋅(2k−1)=2⋅3m=6m So: 2⋅2k−1=6m+1−1=6m2 \cdot 2^k – 1 = 6m + 1 – 1 = 6m2⋅2k−1=6m+1−1=6m Which is clearly divisible by 3.
Thus, the statement is true for k+1k + 1k+1, and by induction, it’s true for all n≥2n \geq 2n≥2.
Example 4: Powers of 2
Let’s prove that 2n>n2^n > n2n>n for all n≥1n \geq 1n≥1.
Step 1: Base Case
For n=1n = 1n=1: 21=22^1 = 221=2 2 is greater than 1. The base case works.
Step 2: Inductive Step
Assume 2k>k2^k > k2k>k for some natural number kkk. We need to show 2k+1>k+12^{k+1} > k + 12k+1>k+1. Start with: 2k+1=2⋅2k2^{k+1} = 2 \cdot 2^k2k+1=2⋅2k Using the induction hypothesis: 2k+1=2⋅2k>2⋅k2^{k+1} = 2 \cdot 2^k > 2 \cdot k2k+1=2⋅2k>2⋅k We need 2⋅k>k+12 \cdot k > k + 12⋅k>k+1, which simplifies to: k>1k > 1k>1 This inequality holds for k≥2k \geq 2k≥2. For k=1k = 1k=1, the base case was verified.
So, the statement is true for k+1k + 1k+1, and by induction, it’s true for all n≥1n \geq 1n≥1.
Conclusion
Mathematical induction is a crucial tool in mathematics that helps us prove statements for an infinite number of cases. By following a structured approach with the base case and the inductive step, we can prove a wide range of mathematical propositions. Whether it’s sums, sequences, or inequalities, induction allows us to establish truth across all natural numbers systematically. Understanding this method opens doors to solving complex problems with confidence and clarity.