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Mathematical Reasoning - Validating Statements

Grade 11CBSE

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

🔑Concepts

Direct Method of Validation: This method is used to prove a conditional statement of the form p    qp \implies q. We assume that the antecedent pp is true and use logical reasoning, axioms, and previously proven theorems to show that the consequent qq must also be true. Visually, imagine a directed path starting at point pp and moving forward step-by-step through valid logical gates until reaching destination qq.

Method of Contrapositive: This method relies on the logical equivalence between a statement and its contrapositive: p    qq    pp \implies q \equiv \sim q \implies \sim p. To prove the statement, we assume that qq is false (i.e., q\sim q is true) and demonstrate that pp must therefore be false (i.e., p\sim p is true). This can be visualized as looking at the logical flow in a mirror; if the reflection (negation) of the result leads back to the reflection of the start, the original path is valid.

Method of Contradiction (Reductio ad Absurdum): To prove a statement pp is true, we begin by assuming the negation p\sim p is true. We then follow a chain of logical deductions until we reach a contradiction (a statement that is clearly false or conflicts with our initial assumption). This implies that our assumption p\sim p was wrong, hence pp must be true. Visually, this is like exploring a branch in a decision tree that leads to a 'Dead End' or a logical 'Wall,' forcing us back to the only other possible branch.

Validation by Counter-example: This method is used to disprove a statement that is claimed to be true for all cases (e.g., 'For all xS,p(x)x \in S, p(x)'). To invalidate such a statement, we only need to find a single instance where the condition does not hold. Visually, if a property is claimed to cover an entire shaded region, a counter-example is a single 'outlier' point located outside that shaded boundary.

Validating 'If and Only If' (Bi-conditional) Statements: To validate a statement of the form p    qp \iff q (p if and only if q), we must prove two parts: the necessity (p    qp \implies q) and the sufficiency (q    pq \implies p). Both directions must be validated independently for the bi-conditional to hold. Visually, this represents a two-way street where traffic (logic) can flow freely in both directions between pp and qq.

Validating Statements with 'And' and 'Or': A compound statement joined by 'And' (pqp \land q) is validated by showing both pp and qq are true. A statement joined by 'Or' (pqp \lor q) is validated by showing at least one of them is true. For 'Or' statements, we often assume pp is false and prove qq must be true. Visually, 'And' is like a series circuit where all switches must be closed, while 'Or' is like a parallel circuit where any one path allows the current to flow.

Quantifiers and Validity: Statements containing 'For every' (Universal Quantifier, \forall) require a general proof that covers every element in the domain. Statements containing 'There exists' (Existential Quantifier, \exists) only require finding or constructing one specific element that satisfies the condition. Visually, \forall refers to the entire set (the whole circle), while \exists refers to at least one specific dot within that circle.

📐Formulae

Contrapositive: p    qq    p\text{Contrapositive: } p \implies q \equiv \sim q \implies \sim p

Converse: q    p\text{Converse: } q \implies p

Inverse: p    q\text{Inverse: } \sim p \implies \sim q

De Morgansˊ Law 1: (pq)pq\text{De Morgan\'s Law 1: } \sim(p \land q) \equiv \sim p \lor \sim q

De Morgansˊ Law 2: (pq)pq\text{De Morgan\'s Law 2: } \sim(p \lor q) \equiv \sim p \land \sim q

Bi-conditional: p    q(p    q)(q    p)\text{Bi-conditional: } p \iff q \equiv (p \implies q) \land (q \implies p)

💡Examples

Problem 1:

Check the validity of the statement: 'If nn is an integer and n2n^2 is even, then nn is even' using the method of contrapositive.

Solution:

  1. Let pp be the statement 'n2n^2 is even' and qq be the statement 'nn is even'. We need to prove p    qp \implies q.
  2. The contrapositive is q    p\sim q \implies \sim p, which is: 'If nn is not even (i.e., nn is odd), then n2n^2 is not even (i.e., n2n^2 is odd)'.
  3. Assume q\sim q is true, so nn is odd. Thus, we can write n=2k+1n = 2k + 1 for some integer kk.
  4. Squaring both sides: n2=(2k+1)2=4k2+4k+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1.
  5. Simplify: n2=2(2k2+2k)+1n^2 = 2(2k^2 + 2k) + 1.
  6. Let m=2k2+2km = 2k^2 + 2k. Then n2=2m+1n^2 = 2m + 1, which is the form of an odd integer.
  7. Since n2n^2 is odd, p\sim p is true. Thus, q    p\sim q \implies \sim p is valid.
  8. Therefore, the original statement p    qp \implies q is true.

Explanation:

We use the contrapositive method because proving that the square of an odd number is odd is algebraically simpler than working directly with square roots of even numbers. Since the contrapositive is logically equivalent to the original statement, validating it validates the original.

Problem 2:

Prove by contradiction that 2\sqrt{2} is an irrational number.

Solution:

  1. Assume the negation: p:2p: \sqrt{2} is rational.
  2. By definition of rational numbers, 2=ab\sqrt{2} = \frac{a}{b}, where aa and bb are integers with no common factors other than 1 (b0b \neq 0).
  3. Squaring both sides: 2=a2b2    a2=2b22 = \frac{a^2}{b^2} \implies a^2 = 2b^2.
  4. This means a2a^2 is even, so aa must also be even. Let a=2ka = 2k.
  5. Substitute a=2ka = 2k into the equation: (2k)2=2b2    4k2=2b2    2k2=b2(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies 2k^2 = b^2.
  6. This means b2b^2 is even, so bb must also be even.
  7. If aa and bb are both even, they have a common factor of 2. This contradicts our assumption that aa and bb have no common factors.
  8. Therefore, the assumption that 2\sqrt{2} is rational is false. Hence, 2\sqrt{2} is irrational.

Explanation:

The contradiction method is effective here because it shows that assuming the statement is rational leads to a logical impossibility (the numbers being both coprime and having a common factor), forcing the conclusion that the number must be irrational.