Proof techniques

In analytical calculus (often times known as advanced calculus, a specific subset of mathematical analysis), there are three important methods to determine that a given hypothesis is true or false.

Deductive Proof

A deductive proof, or direct proof, is very straightforward.

Given some hypothesis: There is a logical method to prove that the hypothesis is true through step by step operations and by using postulates, lemmas, and definitions in constructing the final conclusion.

Many mathematical hypothesis can be proven using deductive proof techniques.

Example

If $a\in R\geq 0$

Prove $a^{2}\geq 0$

Proof:

Since $a\geq 0$ then by multiplicative property of real number system,

$a\times a\geq a\times 0$

$a^{2}\geq 0{\mathcal {p}}$

Proof By Contradiction

Given some hypothesis: It is automatically assumed that the conclusion is wrong, and an antihypothesis must be proven. Then, deductive proof is used (stepwise operations and definitions) to reach a final conclusion. If this final conclusion contradicts the original hypothesis then it is said that the original conditions are proven to be true and correct.

More difficult mathematical hypothesis can be proved or disproved using proof by contradiction techniques.

Example

Extreme value theorem (proof by contradiction):

Assume $F(x)$ is unbounded on [a,b]. i.e., there exists ${x_{n}}\subset [a,b]$ such that $f(x_{n})\rightarrow \infty$.

Since $x_{n}$ is bounded, then the Bolzano-Weierstrass theorem says:

There exists ${x_{nk}}\rightarrow C\in [a,b]$ such that

$\lim _{n\to \infty }x_{nk}\rightarrow C$ and

$\lim _{n\to \infty }f(x_{nk})\rightarrow \infty$.

Since f is continuous at point C, then f(C) = $\lim _{k\to \infty }f{x_{k}}=\infty$

Because $x_{nk}$ is bounded and f(C) is bounded, this conclusion is wrong. F(x) must be bounded on [a,b].

Mathematical induction

Mathematical induction is very powerful two step process, and can only be used for the natural number $n\$. For example:

Prove $\sum _{k=1}^{n}\ 4k\ =\ 2n(n+1)$ for n > 0

Step 1, consider the base case $n=1$.

Right hand side equals 4.

Left hand side equals 4.

Equation satisfied for n = 1.

Step 2, assume the equation is true for some $n$ and consider case $n+1$. We want to prove:

$\sum _{k=1}^{n+1}\ 4k\ =\ 2(n+1)(n+2)$

Demonstrate that the desired n+1 result can be derived using the equation for n:

$\sum _{k=1}^{n+1}\ 4k\ =\sum _{k=1}^{n}\ 4k\ +\ 4(n+1)\ =\ 2n(n+1)\ +\ 4(n+1)\ =\ 2n^{2}\ +\ 6n+4\ =\ 2(n+1)(n+2)$

This proves the induction, that the equation is true for n+1 when it is true for n.

Since the equation is satisfied for n = 1 (step 1), then by step 2 it is satisfied for n = 2, and since it is satisfied for n = 2 then it is satisfied for n = 3, and so on, for n = 4, 5, ...

Mathematical induction tells us the equation is true for all integer n >= 1. ${\mathcal {p}}$

See also

• Proof theory
• Mathematical proof

Reference

1. Wade, William R. An Introduction to Analysis. Upper Saddle River, New Jersey: Pearson Prentice Hall, 2004.