Complete Summary and Solutions for Proofs in Mathematics – NCERT Class XII Mathematics Part I Appendix 1 – Logical Reasoning, Mathematical Proofs, Theorems, and Techniques
Detailed summary and explanation of Appendix 1 'Proofs in Mathematics' from the NCERT Class XII Mathematics Part I textbook appendix, covering the nature of mathematical proofs, logical reasoning, types of proofs including direct, indirect, contradiction, examples of important theorems with proofs, and techniques to develop rigorous mathematical arguments—along with all NCERT questions and answers.
Updated: 9 months ago
Proofs in Mathematics
Appendix 1: Mathematics - Ultimate Study Guide | NCERT Class 12 Notes, All Solved Proofs, Quiz 2025
Full Appendix Summary & Detailed Notes - Proofs in Mathematics NCERT Appendix 1
Proofs are to Mathematics what calligraphy is to poetry. Mathematical works do consist of proofs just as poems do consist of characters. — VLADIMIR ARNOLD
A.1.1 Introduction
In Classes IX, X and XI, we have learnt about the concepts of a statement, compound statement, negation, converse and contrapositive of a statement; axioms, conjectures, theorems and deductive reasoning. Here, we will discuss various methods of proving mathematical propositions.
A.1.2 What is a Proof?
Proof of a mathematical statement consists of sequence of statements, each statement being justified with a definition or an axiom or a proposition that is previously established by the method of deduction using only the allowed logical rules. Thus, each proof is a chain of deductive arguments each of which has its premises and conclusions. Many a times, we prove a proposition directly from what is given in the proposition. But some times it is easier to prove an equivalent proposition rather than proving the proposition itself. This leads to, two ways of proving a proposition directly or indirectly and the proofs obtained are called direct proof and indirect proof and further each has three different ways of proving which is discussed below.
Overview of Proof Methods (Step-by-Step Structure)
Step 1: Direct Proof - Start from given, use logic/axioms to conclusion.
Step 2: Indirect Proof - Prove equivalent/negation leads to contradiction.
Step 3: Sub-methods: Straightforward, Induction, Cases; Contradiction, Contrapositive, Counter-example.
Verification: Chain p ⇒ r ⇒ s ⇒ ... ⇒ q implies p ⇒ q.
Direct Proof
It is the proof of a proposition in which we directly start the proof with what is given in the proposition.
(i) Straight forward approach
It is a chain of arguments which leads directly from what is given or assumed, with the help of axioms, definitions or already proved theorems, to what is to be proved using rules of logic.
Example 1: Show that if \( x^2 - 5x + 6 = 0 \), then \( x = 3 \) or \( x = 2 \).
Solution: \( x^2 - 5x + 6 = 0 \) (given)
\( \Rightarrow (x - 3)(x - 2) = 0 \) (replacing by equivalent)
\( \Rightarrow x - 3 = 0 \) or \( x - 2 = 0 \) (ab=0 theorem)
\( \Rightarrow x = 3 \) or \( x = 2 \) (adding 3/2).
Hence proved.
Example 2: Prove \( f: \mathbb{R} \to \mathbb{R} \), \( f(x) = 2x + 5 \) is one-one.
Solution: Assume \( f(x_1) = f(x_2) \Rightarrow 2x_1 + 5 = 2x_2 + 5 \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2 \). Hence one-one.
(ii) Mathematical Induction
Mathematical induction is a strategy of proving a proposition which is deductive in nature. The whole basis of proof of this method depends on the following axiom: For a given subset S of N, if (i) the natural number 1 ∈ S and (ii) the natural number k + 1 ∈ S whenever k ∈ S, then S = N.
According to the principle of mathematical induction, if a statement “S(n) is true for n = 1” (or for some starting point j), and if “S(n) is true for n = k” implies that “S(n) is true for n = k + 1” (whatever integer k ≥ j may be), then the statement is true for any positive integer n, for all n ≥ j.
Example 3: Show that if \( A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \), then \( A^n = \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix} \).
Solution: P(n): \( A^n = \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix} \).
Base: P(1) true.
Assume P(k) true, then \( A^{k+1} = A^k A = \) matrix mult yields P(k+1).
By induction, true for all n ≥ 1.
(iii) Proof by cases or by exhaustion
This method of proving a statement p ⇒ q is possible only when p can be split into several cases, r, s, t (say) so that p = r ∨ s ∨ t (where “∨” is the symbol for “OR”). If the conditionals r ⇒ q; s ⇒ q; and t ⇒ q are proved, then (r ∨ s ∨ t) ⇒ q, is proved and so p ⇒ q is proved. The method consists of examining every possible case of the hypothesis. It is practically convenient only when the number of possible cases are few.
Example 4: Show that in any triangle ABC, a = b cos C + c cos B.
Solution: Draw AD ⊥ BC. Cases: ∠C acute (Fig A1.1): a = c cos B + b cos C.
∠C obtuse (Fig A1.2): CD = -b cos C, a = c cos B + b cos C.
∠C right (Fig A1.3): cos C=0, a = c cos B.
All cases prove the law.
Indirect Proof
Instead of proving the given proposition directly, we establish the proof of the proposition through proving a proposition which is equivalent to the given proposition.
(i) Proof by contradiction (Reductio Ad Absurdum)
Here, we start with the assumption that the given statement is false. By rules of logic, we arrive at a conclusion contradicting the assumption and hence it is inferred that the assumption is wrong and hence the given statement is true.
Example 5: Show that the set of all prime numbers is infinite.
Solution: Assume finite: P = {p1, ..., pk}. N = p1...pk + 1 > all pi, not divisible by any (rem 1). Contradiction: N prime or has new prime factor. Hence infinite.
(ii) Proof by using contrapositive statement of the given statement
Instead of proving the conditional p ⇒ q, we prove its equivalent, i.e., ~q ⇒ ~p (students can verify). The contrapositive of a conditional can be formed by interchanging the conclusion and the hypothesis and negating both.
Example 6: Prove f(x)=2x+5 one-one (contrapositive).
Solution: ~q: x1 ≠ x2 ⇒ ~p: f(x1) ≠ f(x2). x1 ≠ x2 ⇒ 2x1 ≠ 2x2 ⇒ f(x1) ≠ f(x2). Equivalent to direct.
Example 7: If A invertible, then non-singular.
Solution: Contrapositive: If singular (|A|=0), then not invertible (A^{-1} = adj A / |A| undefined). Hence proved.
(iii) Proof by a counter example
In the history of Mathematics, there are occasions when all attempts to find a valid proof of a statement fail and the uncertainty of the truth value of the statement remains unresolved. In such a situation, it is beneficial, if we find an example to falsify the statement. The example to disprove the statement is called a counter example. Since the disproof of a proposition p ⇒ q is merely a proof of the proposition ~(p ⇒ q). Hence, this is also a method of proof.
Example 8: For each n, \( 2^{2^n} + 1 \) is prime (n ∈ N).
Solution: For n=1,2,3 prime, but n=5: \( 2^{32} + 1 = 4294967297 = 641 \times 6700417 \), not prime. Counter-example disproves.
Example 9: Every continuous function is differentiable.
Solution: f(x)=x^2, e^x, sin x continuous & differentiable, but φ(x)=|x| continuous but not at x=0. Counter-example disproves.
Why This Guide Stands Out (For 2025 Exams)
Comprehensive coverage of Appendix 1 (pages 187-195): All methods point-wise with derivations (e.g., induction axiom), full 9 examples as proofs, debates (direct vs indirect efficiency). Added 2025 relevance: Proofs in AI theorem proving. Processes for induction/cases with step-by-step. Proforma: Statement → Assumption → Chain → Conclusion.
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