User:Popcrate/sandbox

= Topics to Review =

Logic - Statements

 * Converse
 * Inverse
 * Contrapositive

Consider a statement of the form: ∀x ∈ D, if P(x) then Q(x).
 * Its contrapositive is the statement: ∀x ∈ D,if ∼Q(x) then ∼P(x).
 * Its converse is the statement: ∀x ∈ D,if Q(x) then P(x).
 * Its inverse is the statement: ∀x ∈ D,if ∼P(x) then ∼Q(x).

More definitions
 * ∀x,r(x) is a sufficient condition for s(x)” means “∀x, if r(x) then s(x).”


 * “∀x,r(x) is a necessary condition for s(x)” means “∀x, if ∼r(x) then ∼s(x)” or, equivalently, “∀x, if s(x) then r(x).”


 * “∀x,r(x) only if s(x)” means “∀x, if ∼s(x) then ∼r(x)” or, equivalently, “∀x, if r(x) then s(x).”

Negation
Negation of a Universal Statement"∼(∀x ∈ D, Q(x)) ≡ ∃x ∈ D such that ∼Q(x)."Negation of an Existential Statement"∼(∃x ∈ D such that Q(x)) ≡ ∀x ∈ D,∼Q(x)."Negation of a Universal Conditional Statement"∼(∀x,if P(x) then Q(x)) ≡ ∃x such that P(x) and ∼Q(x)."

Definitions
If n is an integer, then:
 * n is even ⇔ ∃ an integer k such that n = 2k.
 * n is odd ⇔ ∃ an integer k such that n = 2k + 1.


 * n is prime ⇔ ∀ positive integers r and s, if n = rs then either r = 1 and s = n or r = n and s = 1.


 * n is composite ⇔ ∃ positive integers r and s such that n = rs and 1 < r < n and 1 < s < n.
 * r is rational ⇔ ∃ integers a and b such that $$r = \frac{a}{b}$$ and b $$\neq$$ 0.
 * Divisibility: d | n ⇔ ∃ an integer k such that n = dk.
 * "n is divisible by d"
 * "d is a factor of n"

Method of Direct Proof

 * 1) Express the statement to be proved in the form “∀x ∈ D, if P(x) then Q(x).”
 * 2) “Suppose x ∈ D and P(x).”
 * 3) Show that the conclusion Q(x) is true by using definitions, previously established results, and the rules for logical inference.

Theorems, Corollaries, and Lemmas

 * The sum of any two rational numbers is rational.
 * The double of a rational number is rational.
 * For all integers a and b, if a and b are positive and a divides b, then a ≤ b.
 * The only divisors of 1 are 1 and −1.
 * For all integers a, b, and c, if a divides b and b divides c, then a divides c.
 * Any integer n > 1 is divisible by a prime number.
 * The Parity Property: Any two consecutive integers have opposite parity.
 * The square of any odd integer has the form 8m + 1 for some integer m.
 * For all real numbers r,−|r| ≤ r ≤ |r|.
 * For all real numbers r, | − r| = |r|.
 * For all real numbers x and y, |x + y| ≤ |x| + |y|.

Unique Factorization of Integers Theorem
Given any integer n > 1, there exist a positive integer k, distinct prime numbers $$p_1, p_2, \ldots, p_k$$, and positive integers $$e_1, e_2, \ldots, e_k$$, such that"$n=p_{1}^{e_1} p_{2}^{e_2} p_{3}^{e_3} \ldots p_{k}^{e_k}$"and any other expression for n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written.

Given any integer n > 1, the standard factored form of n is an expression of the form"$n=p_{1}^{e_1} p_{2}^{e_2} p_{3}^{e_3} \ldots p_{k}^{e_k}$"where k is a positive integer; $$p_1, p_2, \ldots, p_k$$ are prime numbers; $$e_1, e_2, \ldots, e_k$$are positive integers; and $$p_1 < p_2 < \cdots < p_k$$.

The Quotient-Remainder Theorem
Given any integer n and positive integer d, there exist unique integers q and r such that"$n=dq+r$  and    $0 \leq r < d$"

MOD and DIV
Given an integer n and a positive integer d,"$n \text{ div } d$ = the integer quotient obtained when n is divided by d, and""$n \bmod d$ = the non-negative integer remainder obtained when n is divided by d."

Symbolically:

If n and d are integers and d > 0, then

'$$\begin{aligned} && n \text{ div } d = q && \text{and} && n \bmod d = r && \Leftrightarrow && n = dq + r && \end{aligned}$$'

where q and r are integers and 0 ≤ r < d.

Method of Proof by Division into Cases
To prove a statement of the form “If $$A_1$$ or $$A_2$$ or ... or $$A_n$$, then C,” prove all of the following:
 * If $$A_1$$, then C,


 * If $$A_2$$, then C,



This process shows that C is true regardless of which of $$A_1$$, $$A_2$$ ,..., $$A_n$$happens to be the case.
 * If $$A_n$$, then C

Equations
$$\sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r} $$