Proof Writing in Discrete Mathematics Test 1
Conditional Statement
If P then Q, denoted as P→Q Only false if P is T and Q is F
Direct Proof (often for proving universal)
1. Express the statement to be proven in a formalized format 2. Write the word Proof 3. Start proof by supposing x∈D and P(x) (hypothesis) - write in grammatically correct sentences, give a reason for each step in the proof 4. Show that the conclusion Q(x) is true by using definitions, previously established results, and the rule for logical inference
Definition of Rational
A number that can be expressed as a quotient of two integers with a nonzero denominator.
Irrational
A real number that is not rational.
Statement (Propositional) Form
Expression made up of statement variables and logical connectives
Proving Existential
Find one example which makes the statement true.
Definition of prime
For all positive integers r and s, if n=rs then either r=1 and s=n or r=n and s=1
Definition of composite
For some positive integers r and s such that n=rs and 1<r<n and 1<s<n
Existential Statement
Given one property that may or may not be true, there is at least one thing for which the property is true ∃y∈D, Q(y) only true if Q(y) is true for at least one y in the domain (D)
Disproving Existential
Must prove that it's negation (which will be an universal) is true
∼P
Not P, the negation of P
P∧Q
P and Q, the conjunction of P and Q only true when P is T and Q is T
Biconditional
P if and only if Q "iff"= if and only if P↔Q only true if P and Q are F or P and Q are T
Hypothesis
P in a conditional statement
P∨Q
P or Q, the disjunction of P and Q true except when P is F and Q is F
Conclusion
Q in a conditional statement
R
Set of all Real Numbers
Set
Set of all elements x in S x∈S
Z
Set of all integers
Q
Set of all rational numbers
De Morgan's Law 1
The negation of P∧Q is ∼P∨∼Q
De Morgan's Law 2
The negation of P∨Q is ∼P∧∼Q
Method of Generalization
To show that every element of a set satisfies a certain property, suppose x is a particular but arbitrarily chosen element of the set and show that x satisfies the property
Truth Values
When a proposition is True or False
Universal Statement
a certain property is true for all elements in a set ∀x∈D, Q(x) only true if Q(x) is true for all x in the domain (D)
Vacuously True
a conditional statement that is true by virtue of the fact that its hypothesis is false
Universal Existential Statement
a statement is universal because the first part says that a certain property is true for all objects of a given type, existential because it's second part exerts the existence of something ex. Every real number has an additive inverse.
Counterexample
a value for x in which Q(x) is false
Definition of even
an integer is even if it is equal to twice some integer even= 2k
Definition of odd
an integer that is one more than twice some integer odd= 2k+1
Truth Table
displays the truth values that correspond to all possible combinations of truth values for its component statement variables
Disproving Universal
find a counterexample and show how it does not satisfy the claim
Zero Product Property
if neither of two real numbers is not zero, then their product is also not zero.
Necessary Condition
if not r then not s if r does not occur, s cannot occur either Q
Sufficient Condition
if r then s r is enough to guarantee the occurrence of s P
Logically Equivalent
if two statements have the same truth values
Contrapositive of P→Q
negate and switch both terms ∼Q→∼P logically equivalent to P→Q
Inverse of P→Q
negate both sides ∼P→∼Q not logically equivalent to P→Q
Statement (Proposition)
sentence that is true or false but not both
Contradiction
statement form that is always false
Tautology
statement form that is always true
Existential Universal Statement
statement that is existential because its first part asserts that a certain object exists and its universal because the second part says that the object satisfies a certain property for all things of a certain kind Ex. there is a positive integer that is less than or equal to every positive integer
Converse of P→Q
switch terms P→Q not logically equivalent to P→Q
Negation of a Conditional
∼(P→Q) = P ∧∼Q
Negation of a Universal
∼(∀x∈D, Q(x)) = ∃x∈D, ∼Q(x)
Negation of an Existential
∼(∃x∈D, Q(x)) = ∀x∈D, ∼Q(x)
Order of Operations
∼, then ∧ and ∨, then →
