Mathematical Induction

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A mathematical statement (or proposition) is a sentence that is either true or false, but not both. It must be a definite assertion that can be evaluated for its truth value. Example: '5 is an even n

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Let p(n) be a statement involving a natural number n. If: (i) p(1) is true (Base case), and (ii) For any k ≥ 1, if p(k) is true, then p(k+1) is also true (Inductive step) Then p(n) is true for all n

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Given: p(n): 2ⁿ > n + 1 Step 1: Replace n with 1 p(1): 2¹ > 1 + 1, which is 2 > 2 (False) Step 2: Replace n with k p(k): 2ᵏ > k + 1 Step 3: Replace n with k+1 p(k+1): 2ᵏ⁺¹ > (k+1) + 1, which is 2ᵏ⁺

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Step 1: BASE CASE Verify that p(1) is true (or p(a) where a is the starting value) Step 2: INDUCTIVE HYPOTHESIS Assume p(k) is true for some arbitrary natural number k ≥ 1 Step 3: INDUCTIVE STEP Usi

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Let p(n): 1 + 2 + 3 + ... + n = n(n+1)/2 Step 1 - Base Case: p(1) LHS = 1, RHS = 1(1+1)/2 = 1 Since LHS = RHS, p(1) is true. Step 2 - Inductive Hypothesis: Assume p(k) is true: 1 + 2 + ... + k = k(k

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Mathematical induction works because of the Well-Ordering Principle of natural numbers. Key insight: If p(1) is true and p(k) ⟹ p(k+1) for all k ≥ 1, then: • p(1) is true (given) • Since p(1) is tru

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Let p(n): 1² + 3² + 5² + ... + (2n-1)² = n(2n-1)(2n+1)/3 Step 1 - Base Case: p(1) LHS = 1² = 1 RHS = 1(2×1-1)(2×1+1)/3 = 1×1×3/3 = 1 p(1) is true. Step 2 - Inductive Hypothesis: Assume p(k): 1² + 3²

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To prove 'f(n) is divisible by d' using induction: Step 1: Verify f(1) is divisible by d Step 2: Assume f(k) is divisible by d This means f(k) = d × q for some integer q Step 3: Prove f(k+1) is div