Notes/Discrete Structures/Mathematical Proofs (Induction).md

type
theoretical

Proving Equivalences

To prove that two statements P and Q are equivalent (P \iff Q):

1. Chain of Equivalences: Show that P(x_1, \ldots, x_n) \iff \ldots \iff Q(x_1, \ldots, x_n), transforming P into Q step by step.

2. Bi-conditional¹ Decomposition²: Use the fact that:

   
   (P \iff Q) \iff ((P \implies Q) \land (Q \implies P)).
   

   • Prove P \implies Q (first direction).
   • Prove Q \implies P (second direction).

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Proving Universal Statements

To prove a statement of the form \forall x\, P(x):

1. Proof by Exhaustion³:

   • If the domain of x is finite, verify P(x) for each possible value.

2. Proof by Universal Generalization⁴:

   • Let c be an arbitrary element of the domain.
   • Prove P(c).
   • Conclude that \forall x\, P(x) holds.

Example:

• For all integers n, if n is even, then n^2 is even.
• Proof:
  1. Let n be an arbitrary integer.
  2. Assume n is even. Then n = 2k for some integer k.
  3. Compute n^2 = (2k)^2 = 4k^2 = 2(2k^2), which is even.

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Proving Existential Statements

To prove a statement of the form \exists x\, P(x):

1. Constructive Proof⁵: Find a specific c such that P(c) is true.

2. Non-constructive Proof⁶:

   • Assume \forall x\, \neg P(x) (negation of existence).
   • Derive a contradiction.
   • Conclude that \exists x\, P(x) must be true.

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Proof by Contraposition

Instead of proving P \implies Q, prove its contrapositive⁷:

\neg Q \implies \neg P

Example:

• For all n \in \mathbb{N}, if n^2 is odd, then n is odd.
• Proof:
  1. Prove the contrapositive: If n is even, then n^2 is even.
  2. Assume n = 2k. Then n^2 = (2k)^2 = 4k^2 = 2(2k^2), which is even.

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Proof by Contradiction

To prove P:

1. Assume \neg P.
2. Derive a contradiction⁸.
3. Conclude that P must be true.

Example:

• \sqrt{2} is irrational.
• Proof:
  1. Assume \sqrt{2} is rational, so \sqrt{2} = \frac{p}{q}, where p, q \in \mathbb{Z}, q \neq 0, and \frac{p}{q} is in lowest terms.
  2. Square both sides: 2 = \frac{p^2}{q^2}, so 2q^2 = p^2.
  3. p^2 is even, so p is even (p = 2k).
  4. Substitute p = 2k: 2q^2 = (2k)^2 = 4k^2, so q^2 = 2k^2.
  5. q^2 is even, so q is even.
  6. Contradiction: p and q cannot both be even since \frac{p}{q} is in lowest terms.
  7. Conclude \sqrt{2} is irrational.

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Mathematical Induction

⁹ Generalized in Proofs (FP)

To prove a statement of the form \forall n \geq n_0, P(n):

1. Base Case: Prove P(n_0) is true.

2. Inductive Hypothesis - IH: Assume P(k) is true for some k \geq n_0.

3. Inductive Step: Prove P(k + 1) is true using P(k).

Example:

• 1 + 2 + \dots + n = \frac{n(n + 1)}{2} for all n \geq 1.
• Proof:
  1. Base Case: For n = 1, LHS = 1, RHS = \frac{1(1 + 1)}{2} = 1. Holds true.
  2. IH: Assume 1 + 2 + \dots + k = \frac{k(k + 1)}{2}.
  3. Inductive Step: Show 1 + 2 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2}.

     
     \text{LHS} = (1 + 2 + \dots + k) + (k + 1) = \frac{k(k + 1)}{2} + (k + 1).
     

     Simplify:

     
     \frac{k(k + 1)}{2} + (k + 1) = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2}.
     

     Matches RHS.

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Strong¹⁰ Mathematical Induction

To prove \forall n \geq n_0, P(n):

1. Base Cases: Prove P(n_0), P(n_0 + 1), \ldots, P(n_0 + m) are true.

2. Inductive Hypothesis: Assume P(i) is true for all n_0 \leq i \leq k.

3. Inductive Step: Prove P(k + 1) is true using the assumption P(i) for all i \leq k.

Example:

• Every n \geq 2 can be factored into primes.
• Proof:
  1. Base Case: n = 2 is a prime.
  2. IH: Assume every n \leq k can be factored into primes.
  3. Inductive Step: For n = k + 1:
     • If k + 1 is prime, done.
     • If composite, k + 1 = a \cdot b, where 2 \leq a, b \leq k.
     • By hypothesis, a and b can be factored into primes.
     • Combine the prime factors of a and b to get $k + 1$'s factorization.

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1. A statement where both directions are true, like "A if and only if B," meaning A leads to B, and B leads to A. ↩︎

2. Breaking a complex problem into smaller, simpler parts to solve each one step by step ↩︎

3. Checking every possible case individually to prove something is true for all of them. ↩︎

4. Showing something is true for all cases by proving it for an arbitrary or random example. Similar to formal proof for introducing the \forall quantifier ↩︎

5. Directly finding an example to show something exists or is true. ↩︎

6. Proving something exists or is true without giving a specific example, usually by ruling out the possibility of it not being true. ↩︎

7. Instead of directly proving "If A, then B," you prove "If not B, then not A," which means the same thing logically. ↩︎

8. A way of proving something is true by assuming it is false and showing this leads to a logical impossibility, which, as we know, really messes with everything. ↩︎

9. Proof that works by showing it works for the smallest case, then assuming it works for one number and proving it works for the next ↩︎

10. Like induction, but you assume everything is true up to a certain point to prove the next case ↩︎