CBSE MATHEMATICS • SELECTED CENTRAL & STATE COMPETITIVE EXAMS
AIMSSchool of MathematicsAim through AIMS
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GANITA MANJARI · GRADE 9 · PART II · CHAPTER 9

AIMS Quick Revision

Think first. Use AI purposefully. Verify always.

Reconnect the chapter before attempting a mastery paper. Read the claim, identify its direction, and justify or give a counterexample.

Concept Map

  1. STATEWhat Is a Proposition?
  2. REVERSEUnderstanding the Converse
  3. TESTFinding Counterexamples
  4. COMPAREProposition and Converse Together
  5. REASONBuilding and Reading Implication Chains
  6. APPLYPropositions in Number Theory and Geometry
  7. PROVEConverse of the Baudhāyana–Pythagoras Theorem
  8. MASTERReasoning with Propositions and Converses

Key Ideas & Important Language

A proposition is a statement with a truth value. “If P, then Q”, “P implies Q” and “Q when P” express the same direction.

Proposition vs Converse

Original: P → Q. Converse: Q → P. Reversing exchanges conditions; it does not prove truth.

Counterexamples

One valid exception disproves an always-true claim. It must satisfy the starting condition and fail the conclusion.

Truth Relationship Comparison

  • True / false: square → rectangle.
  • True / true: positive integer square → odd positive factor count, with a separate pairing proof in each direction.
  • False / false: multiple of 4 → multiple of 10 (counterexamples 4 and 10).
  • These examples are not exhaustive: rectangle → square is false while its converse is true.

Implication Chain

For a positive integer, distinct factor partners contribute two factors each. An odd total therefore requires one factor to pair with itself, making the number a perfect square. Conversely, a perfect square has exactly one positive self-partner. All remaining factors pair distinctly, giving an odd total.

Baudhāyana–Pythagoras Converse

For positive triangle sides, a2 + b2 = c2 implies the angle opposite c is 90°. Construct perpendicular legs a,b, find hypotenuse c, then use SSS.

Given triangle ABC: BC equals a, CA equals b, AB equals c. Angle C is not yet proved right. Schematic, not to scale.CBAabcGiven · angle C to be provedConstructed triangle XYZ with right angle Z, YZ equals a, ZX equals b. XY is proved equal to c.ZYXabXY = cConstructed · angle Z = 90°
Original AIMS schematic, not to scale. Matching sides by SSS pairs A with X, B with Y and C with Z.

Common Mistakes

  • Calling a false statement “not a proposition”. Falsity is a truth value.
  • Adding “not” or changing the numbers instead of exchanging the two conditions.
  • Using a value that fails the starting condition, or treating five successful tests as a proof.
  • Using the same verdict for both directions without checking each.
  • Assuming a forward arrow justifies its reverse, or forgetting that 1 is a positive square with one factor.
  • Treating a diagram that looks correct as sufficient evidence, or forgetting the real-number domain when comparing squares.
  • Using the longest side as a leg, or marking the original angle right before proving it.
  • Confusing “I have not proved it” with “it is false”, or accepting a conclusion before checking its conditions.

Mini Examples

REVERSE · For positive integers: if n is a multiple of 14, then n is even.

Write n = 14k = 2(7k), so the original is true. Converse: if n is even, then n is a multiple of 14. The value 2 is even but is not a multiple of 14. The converse is false.

TEST · Claim: every positive multiple of 7 is odd.

Try 14, a permitted positive integer. 14 = 7 × 2, so it meets the starting condition. 14 is even, so the conclusion fails. The claim is false.

REASON · Use 49 to illustrate the factor-pair argument, then explain the general reason.

The positive factors of 49 are 1, 7 and 49: the pairs are (1,49) and (7,7). For any positive integer with an odd factor count, distinct pairs cannot account for every factor; a repeated factor f gives n = f2. Conversely, for n = f2 with positive f, f is the unique self-partner. All other factors pair distinctly, so the count is odd.

PROVE · Given: BC = a, CA = b, AB = c and a2 + b2 = c2. Prove ∠C = 90°.

Construct ΔXYZ with YZ = a, ZX = b and ∠Z = 90°. The forward theorem gives XY2 = a2 + b2 = c2. Since lengths are positive, XY = c. AB = XY, BC = YZ and CA = ZX, hence ΔABC ≅ ΔXYZ by SSS (A ↔ X, B ↔ Y, C ↔ Z). Therefore ∠C = ∠Z = 90°. The triangle is right-angled opposite c. Verify numerically: 82 + 152 = 64 + 225 = 289 = 172.

5-Question Quick Check

Selected original AIMS bank questions; revision, not an examination.

0 correct · 0 attempted

One question at a time. Think independently, check, use a hint and retry. Written reasoning remains important.

QUESTION 1 OF 5

Choose a false proposition.
Think independently before checking.

Fixed, reviewed answer keys; no live AI. Attempts are kept while this session is open. Saved completion is practice evidence, not an exam mark or mastery certificate.