Proof Techniques

Technique #1: Deductive Proof

$If\backslash \; a\; \backslash in\; R\; \backslash ge\; 0$

$Prove\backslash$ $a^2\; \backslash ge\; 0$

Proof:

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

$a\; \backslash star\; a\; \backslash ge\; a\; \backslash star\; 0$

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

Technique #2: Proof By Contradiction

$Assume\backslash \; f(x)\backslash \; is\backslash \; UNbounded\backslash \; on\backslash a,b\backslash$

$(i.e.\backslash \; there\backslash \; exists\backslash \; \{x\_n\}\backslash subseta,b\backslash \; such\backslash \; that\backslash \; f(x\_n)\backslash rightarrow\backslash infty.)$

$Since\backslash \; x\_n\backslash \; is\backslash \; bounded,\backslash \; then\backslash \; Bolzano-Weirstrauss\backslash \; Theorem\backslash \; says:\backslash$

$There\backslash \; exists\backslash \; \{x\_\{nk\}\}\backslash rightarrow\; C\; \backslash ina,b$

$such\backslash \; that\backslash \; \backslash lim\_\{n\; \backslash to\; \backslash infty\}x\; \_\{nk\}\; \backslash rightarrow\; c,\backslash \; and\backslash \; \backslash lim\_\{n\; \backslash to\; \backslash infty\}f\; (x\; \_\{nk\})\; \backslash rightarrow\; \backslash infty.$

$Since\backslash \; f\backslash \; is\backslash \; continuous\backslash \; at\backslash \; point\backslash \; C,\backslash \; then\backslash \; f(C)\backslash \; =\backslash \; \backslash lim\_\{k\; \backslash to\; \backslash infty\}f(x\_k)\backslash \; =\backslash \; \backslash infty.$

$Because\backslash \; x\; \_\{nk\}\backslash \; is\backslash \; bounded,\backslash \; and\backslash \; f(C)\backslash \; is\backslash \; bounded,\backslash \; this\backslash \; conclusion\backslash \; is\backslash \; wrong.$

$Therefore\backslash \; f(x)\backslash \; IS\backslash \; BOUNDED\backslash \; on\backslash a,b.\backslash \; \backslash mathcal\{p\}$

Technique #3: Mathematical Induction

$Prove\backslash \; \backslash sum\_\{k=1\}^n\backslash \; 4k\backslash \; =\; \backslash \; 2(n+1)\backslash \; for\backslash \; all\backslash \; k\backslash \; and\backslash \; n.$

$Step\backslash \; 1:\backslash \; Consider\backslash \; case\backslash \; k\backslash \; =\backslash \; 1.$

Right hand side equals 4.

Left hand side equals 4.

Equation satisfied for n = 1.

Assume equation true for some n.

$Step\backslash \; 2:\backslash \; Consider\backslash \; case\backslash \; n+1.$

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

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. $\backslash mathcal\{p\}$

See Also

1. proof theory
2. Mathematical proof

For Further Reference / Selected Bibliography

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