Proof

Proof Methods

Proof by deduction

Use algebra to define a logical arguement to prove or disprove.

Example 1:

$n^2 - 4n + 5$ is positive for any integer.

$n^2 - 4n + 5 = (n-2)^2 + 1$

because $x^2 \ge 0$, and $+1$ doesnt change it:

$(n-2)^2 + 1 > 0$

$\therefore$ $n^2 - 4n + 5 > 0$

So this is true.

Proof by exhaustion

Substitute all possible values into the conjecture (only works with limited number of possible values).

Example 1:

n is an integer

$2 \le n \le 7$

$p = n^2 + 2 \not= 4x$

with x as an integer

All results are false $\therefore$

$p = n^2 + 2 \not= 4x$ is true

Disproof by counter example

using the methods above, but finding a result which doesnt fit with the statement

Example 1:

$\sqrt{x^2 + 9} \equiv x + 3$

$x^2 + 9 \equiv (x + 3)^2$

$x^2 + 9 \equiv x^2 + 6x + 9$

$0 \equiv 6x$

when x = 2:

$0 \equiv 6(2)$

$0 \equiv 12$

$\therefore$ this statement is false

Proof of irrationality of $\sqrt{2}$

$\sqrt{2} = \frac{p}{q}$

with $\frac{p}{q}$ in simplest form.

$2 = \frac{p^2}{q^2}$

$2q^2 = p^2$

$\therefore$ $p^2$ must be even

$\therefore$ $p$ must be even

replace $p$ with $2k$, where k is an integer

$2q^2 = (2k)^2$

$2q^2 = 4k^2$

$q^2 = 2k^2$

$\therefore$ $q^2$ must be even

$\therefore$ $q$ must be even

if $p$ and $q$ are both even, is contradicts our statement of $\frac{p}{q}$ in simplest form as they have a common factor of 2

$\therefore$ $\sqrt{2}$ is irrational

Proof of infinite primes

complete list of primes:

$p_1, p_2, p_3, \dots, p_n$

$Q = p_1 \times p_2 \times p_3 \times \dots \times p_n$

$p_1 \times p_2 \times p_3 \times \dots \times p_n$ is the only way to write Q as a product of primes

if we add 1 to Q

$Q = p_1 \times p_2 \times p_3 \times \dots \times p_n + 1$

cant create product of primes because any of the primes will have the multiple $p_1 \times p_2 \times p_3 \times \dots \times p_n$

eg. $p_2(p_1 \times p_3 \times \dots \times p_n) = Q - 1$

and since we cant add 1 in any way

this must be a new prime and the list is not complete contradicting our statement.

$\therefore$ there are infinite primes.