Chapter 9 · IV · Geometry: logical reasoning

Propositions and Their Converses Worksheet

5 questions Propositions and truth Hypotheses and conclusions Converses and counterexamples Necessary and sufficient conditions

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Foundation

Converse · 1 practice mark

Question 1

The converse of “If n is divisible by 12, then n is divisible by 3” is:

  1. If n is divisible by 3, then it is divisible by 12
  2. If n is not divisible by 12, then it is divisible by 3
  3. Every integer is divisible by 12
  4. If n is divisible by 12, then it is not divisible by 3
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(a) If n is divisible by 3, then it is divisible by 12

  1. A converse swaps the hypothesis and conclusion.
  2. The swapped statement is false: n = 3 is a counterexample.

Practice

Truth and propositions · 3 practice marks

Question 2

For each sentence decide whether it is a proposition: (i) 7 is prime, (ii) Draw a triangle, (iii) 9 is prime. Explain why truth and proposition status differ.

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(i) True proposition; (ii) not a proposition; (iii) false proposition.

  1. A proposition is a declarative statement with a definite truth value.
  2. 7 is prime is true; 9 is prime is false because 9 = 3 × 3.
  3. An instruction such as Draw a triangle does not assert a true-or-false fact. A proposition need not be true.

Challenge

A false converse · 4 practice marks

Question 3

Prove that if an integer is divisible by 18 then it is divisible by both 3 and 6. Write the converse and refute it. What single divisibility condition characterises divisibility by both 3 and 6?

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The converse is false; being divisible by 6 characterises divisibility by both 3 and 6.

  1. Write n = 18k = 3(6k) = 6(3k) to prove the implication.
  2. Converse: if n is divisible by 3 and 6, then n is divisible by 18.
  3. n = 6 satisfies the hypothesis but is not divisible by 18.
  4. Every multiple of 6 is a multiple of 3. Thus “divisible by 3 and 6” is equivalent to “divisible by 6”, not by 18.

Necessary and sufficient · 4 practice marks

Question 4

For real x and y, determine whether x = y is necessary, sufficient, or both for x² = y². How does requiring x and y to be non-negative change the answer?

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Over all reals, sufficient but not necessary; over non-negative reals, both.

  1. Equality x = y permits squaring both sides, so it is sufficient.
  2. x = 2, y = −2 have equal squares but unequal values, disproving necessity over all reals.
  3. Factor x² − y² = (x − y)(x + y). Equal squares give x = y or x = −y.
  4. If both numbers are non-negative, the second possibility forces both to be zero. In this domain equal squares imply equal numbers.

Converse guided by a counterexample · 4 practice marks

Question 5

A learner claims “Equal diagonals imply a parallelogram.” Give a counterexample and add a condition that makes the conclusion valid.

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An isosceles trapezium is a counterexample; diagonals bisecting each other is sufficient.

  1. Take A(−2,0), B(2,0), C(1,2), D(−1,2). AC and BD each have length √13.
  2. AB and CD are parallel but AD and BC are not, so it is not a parallelogram.
  3. If the diagonals bisect each other, the parallelogram converse theorem applies.
  4. This bisection condition, not merely equal lengths, provides the required sufficient condition.