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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.

Contents

1 Technique #1: deductive proof

2 Technique #2: proof by contradiction

3 Technique #3: mathematical induction

4 See also

5 Reference

Technique #1: deductive proof

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 $a \in R \ge 0$

Prove $a^2 \ge 0$

Proof:

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

$a \star a \ge a \star 0$

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

Technique #2: 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 Bolzano-Weirstrauss 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 $x n k$ is bounded and f(C) is bounded, thsi conclusion is wrong. F(x) must be bounded on [a,b].

Technique #3: mathematical induction

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

Prove $\sum_{k=1}^n\ 4k\ = \ 2(n+1)$ for all k and n.

Step 1, consider case $k = 1$ .

Right hand side (of equation) equals 4.

Left hand side equals 4.

Equation satisfied for n = 1.

Assume equation true for some n.

Step 2, consider case $n + 1$ .

$\sum_{k=1}^n\ 4k\ = \ 2(n+1)\ = \sum_{k=1}^n\ 2n(n+1)\ +\ 4(n+1) = \ 2(n+1)\ \star\ 2(n+1)(n+1+1)$

Simplify left and right hand sides.

Left hand side equals 2(n+1)(n+2)

Right hand side equals 2(n+1)(n+2)

Both sides are equivalent.

Based on cases n=1 and n=n+1, mathematical induction tells us equation true for all k,n, in N. $\mathcal{p}$

Reference

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

Categories: Mathematical analysis

Science Fair Project Encyclopedia Contents Page

10-26-2009 08:16:03
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