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Mathematical Reasoning - New Statements from Old (Negation, Compound Statements)

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Negation of a Statement: The denial of a statement is called its negation. If pp is a statement, its negation is denoted by p\sim p (read as 'not pp'). For example, if pp is 'Every natural number is an integer', then p\sim p is 'It is false that every natural number is an integer' or 'There exists at least one natural number which is not an integer'. Visually, if statement pp represents a region in a Venn diagram, p\sim p represents everything outside that specific region.

Compound Statements: A compound statement is a statement which is made up of two or more simple statements, known as component statements. These are joined by connecting words like 'and', 'or', 'if-then', and 'if and only if'. For instance, 'The sun is a star and the earth is a planet' is a compound statement where the connective 'and' joins two distinct facts.

The Connective 'And' (\wedge): A compound statement with 'and' is true only if all its component statements are true. If even one component is false, the entire compound statement is false. In a set-theoretic visual, the 'And' operation corresponds to the intersection (\cap) of two sets, representing only the overlapping area where both conditions are satisfied simultaneously.

The Connective 'Or' (\vee): A compound statement with 'or' is true if at least one of the component statements is true. It is false only when both components are false. In logic, we usually use the 'inclusive or' (either one or both). Visually, this corresponds to the union (\cup) of two sets, covering the entire area occupied by either or both circles in a Venn diagram.

Conditional Statements (Implies): A statement of the form 'If pp, then qq' is called a conditional statement, denoted by pqp \Rightarrow q. Here, pp is the hypothesis and qq is the conclusion. This can be visualized as a directed flow or an arrow pointing from the cause to the effect. The statement is false only in the specific case where pp is true but qq is false.

Bi-conditional Statements (If and only if): A statement of the form 'pp if and only if qq' is denoted by pqp \Leftrightarrow q. It means both pqp \Rightarrow q and qpq \Rightarrow p are true. Visually, this represents a perfect equivalence or a state where two sets or conditions are exactly the same and coincide perfectly with each other.

Quantifiers: These are phrases like 'There exists' (existential quantifier, denoted by \exists) and 'For all' or 'For every' (universal quantifier, denoted by \forall). Negating a statement with 'For all' results in a statement containing 'There exists', and vice versa. For example, the negation of 'All squares are rectangles' is 'There exists a square that is not a rectangle'.

Contradiction and Tautology: A compound statement that is always true regardless of the truth values of its components is a Tautology. Conversely, a statement that is always false is a Contradiction. On a truth table, a Tautology would show a column of all 'T's, while a Contradiction would show all 'F's.

📐Formulae

Negation: (p)p\sim(\sim p) \equiv p

De Morgan's Law 1: (pq)(p)(q)\sim(p \wedge q) \equiv (\sim p) \vee (\sim q)

De Morgan's Law 2: (pq)(p)(q)\sim(p \vee q) \equiv (\sim p) \wedge (\sim q)

Conditional Negation: (pq)p(q)\sim(p \Rightarrow q) \equiv p \wedge (\sim q)

Contrapositive: pq(q)(p)p \Rightarrow q \equiv (\sim q) \Rightarrow (\sim p)

Converse: The converse of pqp \Rightarrow q is qpq \Rightarrow p

Inverse: The inverse of pqp \Rightarrow q is (p)(q)(\sim p) \Rightarrow (\sim q)

💡Examples

Problem 1:

Write the negation of the compound statement: 'The number 2 is even and the number 7 is a prime number.'

Solution:

Step 1: Identify the component statements. Let pp be 'The number 2 is even' and qq be 'The number 7 is a prime number'. Step 2: Identify the connective. The connective is 'and' (\wedge). Step 3: Apply De Morgan's Law for negation: (pq)(p)(q)\sim(p \wedge q) \equiv (\sim p) \vee (\sim q). Step 4: Formulate the negation in words. p\sim p is 'The number 2 is not even' and q\sim q is 'The number 7 is not a prime number'. Step 5: Combine using 'or'. The result is: 'The number 2 is not even or the number 7 is not a prime number.'

Explanation:

To negate a conjunction (And statement), we negate both components and change the connective to a disjunction (Or statement) following De Morgan's laws.

Problem 2:

Identify the component statements and find the truth value of: 'If 3 is a prime number, then 3+2=53 + 2 = 5.'

Solution:

Step 1: Identify components. Let pp: '3 is a prime number' and qq: '3+2=53 + 2 = 5'. Step 2: Determine truth values of components. pp is True (TT) because 3 has only two factors. qq is True (TT) because 3+23 + 2 indeed equals 5. Step 3: Check the conditional rule pqp \Rightarrow q. Here we have TTT \Rightarrow T. Step 4: According to the truth table for conditional statements, TTT \Rightarrow T results in True.

Explanation:

A conditional statement 'If pp, then qq' is only false if the first part is true and the second part is false. Since both parts are true here, the whole statement is true.