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Thinking Mathematically Compound Statements and Connectives.

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Presentation on theme: "Thinking Mathematically Compound Statements and Connectives."— Presentation transcript:

1 Thinking Mathematically Compound Statements and Connectives

2 “Compound” Statements Simple statements can be connected with “and”, “Either … or”, “If … then”, or “if and only if.” These more complicated statements are called “compound.” Examples “Miami is a city in Florida” is a true statement. “Atlanta is a city in Florida” is a false statement. “Either Miami is a city in Florida or Atlanta is a city in Florida” is a compound statement that is true. “Miami is a city in Florida and Atlanta is a city in Florida” is a compound statement that is false.

3 “And” Statements When two statements are represented by p and q the compound “and” statement is p /\ q. p: Harvard is a college. q: Disney World is a college. p/\q: Harvard is a college and Disney World is a college. p/\~q: Harvard is a college and Disney World is not a college.

4 “Either... or” Statements When two statements are represented by p and q the compound “Either... or” statement is p\/q. p: The bill receives majority approval. q: The bill becomes a law. p\/q: The bill receives majority approval or the bill becomes a law. p\/~q: The bill receives majority approval or the bill does not become a law.

5 “If... then” Statements When two statements are represented by p and q the compound “If... then” statement is: p  q. p: Ed is a poet. q: Ed is a writer. p  q: If Ed is a poet, then Ed is a writer. q  p: If Ed is a writer, then Ed is a poet. ~q  ~p: If Ed is not a writer, then Ed is not a Poet

6 “If and only if” Statements When two statements are represented by p and q the compound “if and only if” statement is: p  q. p: The word is set. q: The word has 464 meanings. p  q: The word is set if and only if the word has 464 meanings. ~q  ~p: The word does not have 464 meanings if and only if the word is not set.

7 Symbolic Logic Statements of Logic NameSymbolic Form Negation~p Conjunctionp/\q Disjunctionp\/q Conditionalp  q Biconditionalp  q


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