Proofs in Mathematics — Telangana (SSC) Class 9 Mathematics Solutions (Free)
Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Proofs in Mathematics" — important questions with detailed answers, download PDF for board exam preparation.
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Proofs in Mathematics — Textbook Questions Answered (8)
Q1: State Euclid's proof that the sum of angles in a triangle equals 180°.
Step 1: Given triangle ABC
Step 2: Draw a line through C parallel to AB
Step 3: Since AB ∥ to the line through C:
Alternate interior angles are equal
Step 4: ∠BAC = ∠ACE (where E is on the parallel line, on one side of C)
Step 5: ∠ABC = ∠DCB (where D is on the parallel line, on other side of C)
Step 6: The angle on the parallel line = ∠ACD + ∠DCB + ∠ACE
Step 7: But angles on a straight line sum to 180°
Step 8: ∠ACE + ∠ACB + ∠BCD = 180°
Step 9: ∠BAC + ∠ACB + ∠ABC = 180°
Final Answer: Sum of angles in a triangle = 180°
Q2: Prove that the base angles of an isosceles triangle are equal.
Step 1: Let triangle ABC where AB = AC (equal sides)
Step 2: We need to prove ∠ABC = ∠ACB (base angles)
Step 3: Construct the angle bisector of ∠BAC meeting BC at D
Step 4: In triangles ABD and ACD:
AB = AC (given)
∠BAD = ∠CAD (AD is angle bisector)
AD = AD (common side)
Step 5: By SAS congruence: △ABD ≅ △ACD
Step 6: Therefore ∠ABD = ∠ACD
Step 7: Which means ∠ABC = ∠ACB
Final Answer: Base angles of isosceles triangle are equal
Q3: Prove using the contrapositive that if n² is even, then n is even.
Step 1: Direct statement: If n² is even, then n is even
Step 2: Contrapositive: If n is odd, then n² is odd
Step 3: Assume n is odd, so n = 2k + 1 for some integer k
Step 4: n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1
Step 5: n² = 2m + 1 where m = 2k² + 2k
Step 6: Therefore n² is odd
Step 7: Contrapositive is true, so original statement is true
Final Answer: If n² is even, then n is even (proved by contrapositive)
Q4: Prove that there are infinitely many prime numbers.
Step 1: Assume there are finitely many primes: p₁, p₂, p₃, ..., pₙ
Step 2: Consider the number N = p₁ × p₂ × p₃ × ... × pₙ + 1
Step 3: N is not divisible by p₁ (leaves remainder 1)
Step 4: N is not divisible by p₂ (leaves remainder 1)
Step 5: Similarly, N is not divisible by any of the listed primes
Step 6: Therefore, N is either prime itself, or has a prime factor not in our list
Step 7: This contradicts our assumption that p₁, ..., pₙ are ALL primes
Step 8: Therefore, there must be infinitely many primes
Final Answer: There are infinitely many primes (proved by contradiction)
Q5: Define axiom, postulate, theorem, and corollary.
Step 1: Axiom: A fundamental statement accepted as true without proof, universal to all mathematics. Example: Things equal to the same thing are equal to each other.
Step 2: Postulate: A fundamental assumption specific to a domain (like geometry), accepted without proof. Example: A line can be drawn between any two points.
Step 3: Theorem: A statement that is proved using axioms, postulates, and previously proved theorems. Example: Sum of angles in a triangle = 180°.
Step 4: Corollary: A theorem that follows immediately from another theorem with little or no additional proof. Example: Exterior angle of triangle = sum of remote interior angles (follows from angle sum theorem).
Final Answer: Axiom (universal truth), Postulate (domain-specific), Theorem (proved), Corollary (follows from theorem)
Q6: Prove using mathematical induction that 1 + 2 + 3 + ... + n = n(n+1)/2.
Step 1: Base case: n = 1
LHS = 1
RHS = 1(1+1)/2 = 1
Base case verified ✓
Step 2: Inductive step: Assume true for n = k
1 + 2 + 3 + ... + k = k(k+1)/2 ... (1)
Step 3: Prove for n = k+1
LHS = 1 + 2 + 3 + ... + k + (k+1)
Step 4: = k(k+1)/2 + (k+1) [using (1)]
Step 5: = (k+1)[k/2 + 1]
Step 6: = (k+1)[(k+2)/2]
Step 7: = (k+1)(k+2)/2 = RHS for n = k+1
Step 8: By mathematical induction, the formula holds for all positive integers n
Final Answer: 1 + 2 + 3 + ... + n = n(n+1)/2 (proved by induction)
Q7: Prove that the opposite angles of a cyclic quadrilateral are supplementary.
Step 1: Let ABCD be a cyclic quadrilateral inscribed in a circle
Step 2: Angle subtended by arc BCD at A = ∠DAB
Step 3: Angle subtended by arc BCD at center = 2∠DAB (by inscribed angle theorem)
Step 4: Arc BCD + Arc DAB = complete circle = 360°
Step 5: Angle subtended by arc DAB at C = 2 × (180° - ∠DAB) [from circle property]
Step 6: ∠BCD = 180° - ∠DAB
Step 7: Therefore ∠DAB + ∠BCD = 180°
Step 8: Similarly, ∠ABC + ∠CDA = 180°
Final Answer: Opposite angles of cyclic quadrilateral are supplementary
Q8: Distinguish between a theorem and its converse with an example.
Step 1: Theorem: A statement that has been proved
Example: If a triangle is isosceles, then its base angles are equal
Step 2: Converse: The statement obtained by interchanging the hypothesis and conclusion
Example: If a triangle's base angles are equal, then it is isosceles
Step 3: A theorem may be true, but its converse may not be true
Example: Theorem - If n = 2, then n² = 4 (TRUE)
Converse - If n² = 4, then n = 2 (FALSE, because n could be -2)
Step 4: In geometry, many converses are true (like the isosceles triangle example)
But in general, we cannot assume converse is true without proof
Final Answer: Theorem is proved; Converse reverses hypothesis-conclusion; Converse must be separately proved
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