Does 7 = 5? (not a mathematical statement because you cannot assign true or false to it)
Natural Numbers denoted by
Set of integers where negative, positive, or zero are denoted by
Set of rational numbers denoted by contains all the numbers of the form where a is an integer and b is a non-zero integer
Set of real numbers denoted by contains all the numbers in decimal form
Negation - Finding the inverse meaning of a statement
Suppose A is a true statement, A would be false
( A) = A
Quantified Statements contains 4 parts:
A quantifier
A variable (any symbol representing a quantity or mathematical object)
a domain (any set)
an open sentence involving the variable
Open Sentence - A sentence that contains a variable, where the truth of the sentence is determined by the value of the variable chosen from the domain of the variable
Quantifiers - Describe “how many” elements of the domain are claimed to make the open sentence true
- Universal Quantifier
- Existential Quantifier
Universally Quantified Statement - x S, P(x)
This statement is read as “For all x in S, P(x) is true”
Existentially Quantified Statement - x S, P(x)
This statement is read as “There exists at least one value of x in S for which P(x) is true”
Nested Quantifiers - Quantifiers in a statement containing more than one quantifier
s , t , t > s
For all real numbers s, there exists a real number t such that t > s
x X, y Y, z Z, R(x,y,z)
Can be rewritten as:
x X, P(x), where P(x) is y Y, Q(x, y), where Q(x, y) is z Z, R(x, y, z)
We prove a statement when we demonstrate its true and the argument to do is called a proof
You can prove an inequality by proving all of it’s constraints (eg: x 1 and x 3)
We disprove a statement when we demonstrate it’s false
To disprove a universally quantified statement find a counter-example. For example if the statement is true for all domains find a value for which it isn’t
To disprove x S, P(x) prove it’s negation x S, P(x)
When proving an identity we do NOT start by assuming the truth of that identity
Be on the look out for extraneous solutions (solutions not in the domain created by acts such as rearranging the equation)
k , x , x + 2kx + k = 0 x , k , x + 2kx + k = 0
Changing the order of nested quantifiers can change the truth value of a quantified statement
To prove the implication x S, P(x) Q(x)
Assume x is an arbitrary element
Assume P(x) is true
Use that assumption to show Q(x) is true
We assume an integer is even if it can be written as 2k (k = an integer). Otherwise an integer is written in the form 2k + 1 where we say the integer is odd
3 6 exists since 6 = 3 * 2, exists an integer k such that 6 = 3k
-3 6 exists since 6 = (-3) * (-2), exists an integer k such that 6 = -3k
5 6 since no integer k exists so that 6 = k * 5
For all integers a, a 0 0 = 0 * a
For all non-zero integers a, a 0 since no integer k so that k * 0 = a
For all integers b, 1 b since b can be written as b = b * 1 and -1 b since b can be written b = (-b) * (-1)
Transfer of Divisibility (TD) - For all integers a, b, and c, if a b and b c, then a c
Proposition 8: For all integers if a, b and c, if a b or a c, then a bc
If a b, then a bc
Assume a b, there exists an integer k so that b = ka. Substitute for b in product bc, we get
bc = (ka)c = kac = (kc)a
kc is an integer we can conclude a bc
If a c, then a bc
Exists integer m so that c = ma. Substituting this equation we get
bc = b(ma) = bma = (bm)a
bm is an integer we conclude a bc
These proofs can be re-written as if a c, then a bc is similar, and is omitted
Divisibility of Integer Combinations (DIC) - For all integers a, b and c, if a b and a c, then for all integers x and y, a (bx + cy)
To prove an implication using contrapositive (A B) replace it with it’s contrapositive (B) (A)
Anytime we come across an argument claiming A and A are true we say there must be a contradiction in the argument
Axiom is a statement that is assumed to be true. No proof given
Axiom 1 - Principle of Mathematical Induction (POMI)
Let P(n) be an open sentence depends on n
If statements 1 and 2 are both true:
P(1)
For all k , if P(k), then P(k + 1)
then statement 3 is true:
For all n , P(n)
Axiom 2 - Principle of Strong Induction (POSI)
Let P(n) be an open sentence that depends on n
If statements 1 and 2 are both true
P(1)
For all k , if P(1) P(2) P(k) then P(k + 1)
then statement 3 is true:
3. For all n , P(n)
If you can prove the inductive conclusion P(k + 1) by assuming only P(k) then use POMI but if you need to assume P(i) for one or more i with i < k, then use POSI
Number of elements in a finite set is called cardinality denoted by | S |
Universe of Discourse - contains all the objects we might encounter in a given situation
{x : P(x)}
x is an element of
P(x) is true
{f(x) : x }
containing all objects of the form f(x) such that x is an element of
{f(x) : x , P(x)} or {f(x) : P(x), x }
x is an element of
P(x) is true
Union of two sets S and T written S T is the set of all elements belonging to either set S or set T(or both)
S T = {x: x S OR x T} = {x: (x S) (x T)}
The intersection of two sets S and T, written S T is the set of all elements belonging to both set S and set T
S T = {x: x S AND x T} = {x: (x S) (x T)}
Set difference of two sets S and T is written as S - T (or S \ T) is the set of all elements belonging to S but not T
S - T = {x : x S AND x T} = {x : (x S) (x T)} = {x : (x S) ((x T))}
Complement of set S, written is set of all elements not in S
= {x : x S}
= -
= -
Two sets are said to be disjoint when S T =
A set S is called a subset of set T, written symbolically as S T, when every element of S belongs to T
A set S is called a proper subset of a set T, written symbolically as S T, when S is a subset of T and there exists an element in T which does not belong in S
When S is not a subset of T, we can write it as S T
When we have S T we also say T is a superset of S
When we have S T we also say T is a proper subset of S
Diophantine Equation - Both coefficients and variables are integers, this equation is called linear if each term in the equation is a constant or a constant times a single variable
Linear Diophantine Equation Theorem, Part 1, (LDET 1)
For all integers a, b, and c, with a and b both non zero, the linear Diophantine Equation
ax + by = c
(in variables x and y) has an integer solution if and only if d c, where d = gcd(a, b)
Linear Diophantine Equation Theorem, Part 2, (LDET 2)
Let a, b and c be integers with a and b both non zero, and define d = gcd(a, b). If x = x and y = y is one particular integer solution to the linear Diophantine Equation ax + by = c, then the set of all solutions is given by