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Mathematical Reasoning - Statements

Grade 11CBSE

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

🔑Concepts

A mathematical statement is a sentence which is either true or false, but not both. Sentences involving variables like 'He', 'She', or 'It', or relative terms like 'today' or 'here', are generally not statements unless the context fixes them. Visualize a decision tree where a sentence only qualifies as a statement if it follows a single path to either a 'True' or 'False' leaf node, without any ambiguity.

Negation of a statement involves denying the statement. If pp is a statement, its negation is denoted by p\sim p (read as 'not pp'). If pp is true, p\sim p is false, and vice versa. This can be visualized using a two-row truth table where the value in the second column is the exact opposite of the first.

Compound statements are formed by combining two or more simple statements using logical connectives such as 'and' (conjunction, denoted by \wedge) and 'or' (disjunction, denoted by \vee). Think of 'and' as a series electrical circuit where the bulb glows only if both switches are closed, and 'or' as a parallel circuit where the bulb glows if at least one switch is closed.

Quantifiers are phrases like 'There exists' (existential quantifier, \exists) and 'For all' (universal quantifier, \forall). They define the scope of the statement. Visually, 'For all' covers every single point within a defined boundary (like a shaded circle), whereas 'There exists' highlights at least one specific point within that boundary.

Implications or Conditional statements are of the form 'If pp, then qq', denoted by pqp \Rightarrow q. Here, pp is the hypothesis and qq is the conclusion. This relationship is like a one-way street: the truth of pp forces the truth of qq, but the truth of qq does not necessarily mean pp is true.

The Contrapositive and Converse are variations of the conditional statement pqp \Rightarrow q. The Contrapositive is qp\sim q \Rightarrow \sim p, which is logically equivalent to the original statement. The Converse is qpq \Rightarrow p, which may or may not be true. Visually, if pqp \Rightarrow q represents 'All squares are rectangles', the contrapositive is 'If it is not a rectangle, it is not a square'.

A Bi-conditional statement is of the form 'pp if and only if qq', denoted by pqp \Leftrightarrow q. This means both pqp \Rightarrow q and qpq \Rightarrow p are true. This can be visualized as a two-way arrow, indicating that pp and qq are logically identical in terms of truth value.

Validating statements can be done through direct proof, contrapositive proof, or proof by contradiction. In a proof by contradiction, we assume the negation of the statement is true and work towards a logical impossibility (e.g., 1=01 = 0), thereby proving the original statement must be true.

📐Formulae

Negation: (p)=p\sim(\sim p) = p

Conjunction ('and'): pqp \wedge q is True only if both pp and qq are True.

Disjunction ('or'): pqp \vee q is False only if both pp and qq are False.

Conditional: pqp \Rightarrow q is False only when pp is True and qq is False.

Bi-conditional: pqp \Leftrightarrow q is True only when both pp and qq have the same truth value.

Contrapositive of pqp \Rightarrow q is qp\sim q \Rightarrow \sim p

Converse of pqp \Rightarrow q is qpq \Rightarrow p

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

💡Examples

Problem 1:

Check whether the following compound statement is true or false: 'Number 2525 is a multiple of 55 and 88'.

Solution:

  1. Let pp be 'Number 2525 is a multiple of 55'. This statement is True because 5×5=255 \times 5 = 25.
  2. Let qq be 'Number 2525 is a multiple of 88'. This statement is False because 2525 cannot be divided by 88 without a remainder.
  3. The compound statement is pqp \wedge q.
  4. Since pp is True and qq is False, the conjunction pqp \wedge q is False.

Explanation:

In an 'and' statement, both component parts must be true for the whole statement to be true. Since one part is false, the entire compound statement is false.

Problem 2:

Write the contrapositive and converse of the following statement: 'If a number nn is even, then n2n^2 is divisible by 44'.

Solution:

Let pp: 'a number nn is even' and qq: 'n2n^2 is divisible by 44'.

  1. Contrapositive (qp\sim q \Rightarrow \sim p): 'If n2n^2 is not divisible by 44, then the number nn is not even'.
  2. Converse (qpq \Rightarrow p): 'If n2n^2 is divisible by 44, then the number nn is even'.

Explanation:

The contrapositive is formed by negating both statements and switching their positions. The converse is formed by simply switching the positions of the hypothesis and conclusion.