krit.club logo

Mathematical Reasoning - Validating Statements

Grade 11ICSE

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

🔑Concepts

•

A Mathematical Statement is a declarative sentence that is either true or false, but not both simultaneously. Visualizing this as a toggle switch that can only be in one of two positions (On/Off) helps distinguish statements from exclamatory or interrogative sentences.

•

Negation of a Statement involves the denial of the statement, denoted by ∼p\sim p. If statement pp is represented as a shaded region in a Venn diagram, its negation ∼p\sim p is represented by everything outside that shaded region within the universal set.

•

Compound Statements are formed by combining two or more simple statements using logical connectives like 'And' and 'Or'. A compound statement with 'And' (conjunction) is true only if both component statements are true, similar to two switches in a series circuit where both must be closed for current to flow.

•

The 'Or' connective (disjunction) creates a compound statement that is true if at least one of the component statements is true. This functions like a parallel circuit where the flow of logic is maintained if either path is open.

•

The Contrapositive method of validation states that to prove the conditional statement 'If pp, then qq' (p  ⟹  qp \implies q), one can instead prove 'If not qq, then not pp' (∼q  ⟹  ∼p\sim q \implies \sim p). These two are logically equivalent and can be visualized as two sides of the same coin.

•

The Method of Contradiction is a powerful validation technique where to prove a statement pp is true, we initially assume pp is false (assume ∼p\sim p is true). If this assumption leads to a logical absurdity or a conflict with known facts, the original statement pp must be true.

•

Quantifiers like 'There exists' (∃\exists) and 'For all' (∀\forall) are used to validate the scope of a statement. 'For all' requires every single element in a set to satisfy a property, while 'There exists' only requires finding one single example (a counter-example search in reverse) to be true.

•

Validating Biconditional Statements of the form 'p if and only if q' (p  ⟺  qp \iff q) requires proving both directions: if pp then qq, and if qq then pp. This represents a perfect logical symmetry between two statements.

📐Formulae

∼(p∧q)≡∼p∨∼q\sim(p \land q) \equiv \sim p \lor \sim q (De Morgan's Law for Conjunction)

∼(p∨q)≡∼p∧∼q\sim(p \lor q) \equiv \sim p \land \sim q (De Morgan's Law for Disjunction)

p  ⟹  q≡∼q  ⟹  ∼pp \implies q \equiv \sim q \implies \sim p (Contrapositive Equivalence)

∼(p  ⟹  q)≡p∧∼q\sim(p \implies q) \equiv p \land \sim q (Negation of Implication)

p  ⟺  q≡(p  ⟹  q)∧(q  ⟹  p)p \iff q \equiv (p \implies q) \land (q \implies p) (Biconditional Equivalence)

💡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 'n2n^2 is even' and qq be '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. Let n=2k+1n = 2k + 1 for some integer kk.
  4. Then n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1.
  5. Since n2n^2 is of the form 2m+12m + 1 where m=2k2+2km = 2k^2 + 2k, n2n^2 is odd.
  6. Thus, ∼q  ⟹  ∼p\sim q \implies \sim p is true, which validates the original statement p  ⟹  qp \implies q.

Explanation:

The contrapositive method allows us to prove a statement by proving its logically equivalent negative reverse, which is often algebraically simpler.

Problem 2:

Validate the statement '7\sqrt{7} is irrational' using the method of contradiction.

Solution:

  1. Assume the negation is true: Suppose 7\sqrt{7} is rational.
  2. Then 7=ab\sqrt{7} = \frac{a}{b} where a,ba, b are integers, b≠0b \neq 0, and gcd⁡(a,b)=1\gcd(a, b) = 1 (they are in simplest form).
  3. Squaring both sides: 7=a2b2  ⟹  a2=7b27 = \frac{a^2}{b^2} \implies a^2 = 7b^2.
  4. This means a2a^2 is divisible by 77, so aa must be divisible by 77. Let a=7ka = 7k.
  5. Substituting aa: (7k)2=7b2  ⟹  49k2=7b2  ⟹  7k2=b2(7k)^2 = 7b^2 \implies 49k^2 = 7b^2 \implies 7k^2 = b^2.
  6. This means b2b^2 is divisible by 77, so bb must be divisible by 77.
  7. Both aa and bb have a common factor 77, which contradicts the assumption that gcd⁡(a,b)=1\gcd(a, b) = 1.
  8. Therefore, the assumption is false, and 7\sqrt{7} is irrational.

Explanation:

By showing that the assumption of being rational leads to a mathematical impossibility (a contradiction), we prove that the original statement must be true.