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C959 Flashcards (Master set) With Questions And 100% ALL CORRECT ANSWERS
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JACKLINEC959 Flashcards (Master set) With Questions And 100%
ALL CORRECT ANSWERS
Terms in this set (78)
Prove the statement by checking each element individually. Unlike proof by cases,
Proof by Exhaustion
exhaustive proof looks at every case in a way that is not general.
Used to disprove a universal statement. For example, to disprove the statement, "All
Proof by Counterexample prime numbers are odd" find one example where this statement is false—the number
2—which is both even and prime.
In the direct proof, we assume the hypothesis p is
true and we try to prove that q is true. Thus making
Direct Proof
p→q true.
A proof that makes use of the fact that p→q is
equivalent to its contrapositive ¬q→¬p. So we
Proof by Contrapositive
assume that ¬q is true and try to prove that ¬p is
true.
An indirect proof where if the theorem being
proven has the form p → q, then the beginning
Proof by Contradiction
assumption is p 𝖠 ¬q which is logically equivalent
to ¬(p → q).
In proof by case, we are looking at all the possible
cases that might arise in a theorem.
Proof by Cases
∴ Means "therefore".
Conditional Statements If p then q. p → q. Only not true when p is true and q is false. It breaks the contract.
Converse Converse of p → q is q → p.
Contrapositive Contrapositive of p → q is ¬q → ¬p.
Inverse Inverse of p → q is ¬p → ¬q.
p if and only if q. p ↔ q. True when p and q have the same truth value (including
Biconditional Statement
False ↔ False) and is false when p and q have different truth values.
p is logically equivalent to q. p ≡ q. Logically equivalent if they have the same truth
Logical Equivalence value regardless of the truth values of their individual propositions. So p ≡ ¬¬p or ¬p
≡ p → ¬q.
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, Conjunction — p 𝖠 q. True only if p and q are true.
Disjunction — p ∨ q. True when either of p or q, or both is true.
Logical Operator
Exclusive or — p ⊕ q. True if either one of p or q is true but not both.
Negation — ¬p. Negates value. If True then not True, if False then not False.
1. ¬ (not)
2. 𝖠 (and)
Order of Operations 3. ∨ (or)
4. →
5. ↔
Predicate Logical statement whose truth value is a function of one or more variables.
Variable is free to take on any value in the domain. In (∀x P(x)) 𝖠 Q(x), Q(x) is the free
Free Variable
variable.
Bound Variable Variable is bound to a quantifier. In (∀x P(x)) 𝖠 Q(x), P(x) is the bound variable.
¬∀ x F(x) as "Not every bird can fly." Which is logically equivalent (≡) to ∃ x ¬ F(x) as
"There exists a bird that cannot fly."
DeMorgan's Laws for Quantifiers ¬∃ x Q(x) as "It is not true that there is a child in the class who is absent today." Which
is logically equivalent (≡) to ∀ x ¬ Q(x) as "Every child in the class is not absent
today."
-∀x∀y M(x,y)
- ∃x∃y M(x,y)
Logical Symbol and Translations
- ∃x∀y M(x,y)
- ∀x∃y M(x,y)
DeMorgan's Law for Nested Qualifiers
The set of natural numbers: All integers greater than or equal to 0.
N
For example, 0, 1, 2, …
The set of all integers. For example, …, -2, -1, 0, 1, 2,
…
Z
The set of rational numbers: All real numbers that can be expressed as a/b, where a
Q and b are integers and b ≠ 0.
For example, 0, 1/2, 5.23, -5/3
The set of real numbers. Real numbers can be a continuous quantity.
R
For example, 0, 1/2, 5.23, -5/3, π, √2
Universal Set A = B if and only if A ⊆ B and B ⊆ A
Proper Subset An element of B that is not an element of A (i.e., A ≠ B). A ⊂ B.
f and g are two functions, where f: X → Y and g: Y → Z. The composition of g with f,
Composition of Functions
denoted g ο f, is the function (g ο f): X → Z, such that for all x ∈ X, (g ο f)(x) = g(f(x)).
If f: R+ → R+, f(x) = x³ and g: R+ → R+, g(x) = x + 2. Then, (f ο g)(x) = f(g(x)) = (x + 2)³
Composition of Functions (Example)
and (g ο f)(x) = g(f(x)) = x³ + 2.
Exponential Functions
Logarithmic Functions
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