Skip to content

Mathematics · Ch 6 — Application of Derivatives

Appendix 1 — Proofs in Mathematics

Appendix 1 — Proofs in Mathematics

The Purpose of Proofs

A proof transforms a guess or a pattern into an unshakable truth. It is a logical chain of reasoning that starts from accepted statements (axioms, definitions, or previously proven theorems) and, through valid steps, arrives at a new statement (the theorem).


Key Types of Proofs Covered

Three fundamental proof techniques, each a different tool for establishing truth.

1. Direct Proof

Assume the hypothesis (the "if" part) is true, then use logical deductions to show the conclusion (the "then" part) must also be true.

Structure: If PP is true, then QQ is true.

Example: Prove that the product of two even integers is even.

  • Let aa and bb be even integers, so a=2ma = 2m and b=2nb = 2n for some integers m,nm, n.
  • Their product is a×b=(2m)(2n)=4mn=2(2mn)a \times b = (2m)(2n) = 4mn = 2(2mn).
  • Since 2mn2mn is an integer, a×ba \times b is of the form 2×(integer)2 \times (\text{integer}), hence even.
2. Proof by Contradiction

Assume the opposite of what you want to prove (the conclusion is false) and show this leads to a contradiction. Since the assumption is impossible, the original statement must be true.

Structure: To prove P⇒QP \Rightarrow Q, assume PP true and QQ false, derive a contradiction; therefore QQ must be true.

Example: Prove that 2\sqrt{2} is irrational.

  • Assume 2\sqrt{2} is rational: 2=pq\sqrt{2} = \frac{p}{q} in lowest terms (p,qp, q integers with no common factor, q≠0q \neq 0).
  • Squaring: 2=p2q22 = \frac{p^2}{q^2}, so p2=2q2p^2 = 2q^2. Then p2p^2 is even, so pp is even; let p=2rp = 2r.
  • Substitute: 4r2=2q2⇒q2=2r24r^2 = 2q^2 \Rightarrow q^2 = 2r^2, so q2q^2 is even, hence qq is even.
  • But then pp and qq share the factor 22, contradicting "lowest terms."
  • Therefore the assumption is false, and 2\sqrt{2} is irrational.
3. Proof by Induction

Used to prove a statement claimed true for all natural numbers nn. It works like a row of dominoes: knock over the first, and let each one knock over the next.

Structure: To prove P(n)P(n) for all n∈Nn \in \mathbb{N}:

  1. Base Case: Show P(1)P(1) is true.
  2. Inductive Step: Assume P(k)P(k) is true (the induction hypothesis), and use it to prove P(k+1)P(k+1).

Example: Prove that P(n):1+2+3+⋯+n=n(n+1)2P(n): 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}. …