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Definitions, Conjectures, and Counterexamples: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 What kind of support?

Classify each item as a definition, accepted postulate, proved result, or experimental observation:

Item A: a diameter of a circle in a plane is a segment through its center with both endpoints on the circle.

Item B: if B is between A and C, then $AB+BC=AC$ .

Item C: a diameter has length $d=2r$ for positive radius r.

Item D: a protractor reading of an angle is about 38 degrees.
1. Hint 1
  Ask whether each item fixes a meaning, supplies a starting fact, gives a justified conclusion, or reports a measurement.
2. Hint 2
  Segment addition is an accepted starting fact. A protractor reading has limited precision.
Answer

In order: definition; accepted postulate; proved result; experimental observation.

Full solution

Item A: The diameter statement fixes the meaning of the term, so it is a definition.

Item B: The segment-addition statement is an accepted postulate, including its betweenness condition.

Item C: The diameter-length result was proved in 1.4: the center is between the endpoints, and segment addition gives $r+r=2r$ .

Item D: The protractor reading reports a particular measurement, so it is an experimental observation.

Answer

In order: definition; accepted postulate; proved result; experimental observation.

Key idea

Track how a fact is supported before using it as a reason.
Problem 2 Classify by conditions

Use these definitions: a rectangle is a quadrilateral with four right angles, and a square is a rectangle with four congruent sides.

Figure A is a quadrilateral with four right angles and side lengths 10, 10, 10, 10 centimeters.

Figure B is a quadrilateral with four right angles and neighboring side lengths 6 and 13 centimeters.

Give every applicable name, square and rectangle, for each figure.

Figure A has four right angles and four side lengths of 10 cm. Figure B has four right angles and neighboring lengths of 6 cm and 13 cm. Classify from the stated conditions.

Text description of this figure

Figure A has four right angles and four side lengths of 10 cm. Figure B has four right angles and neighboring lengths of 6 cm and 13 cm. Classify from the stated conditions.
1. Hint 1
  Check the right-angle condition separately from the equal-side condition.
2. Hint 2
  A figure can satisfy both definitions; the names are inclusive.
Answer

A is both a square and a rectangle. B is a rectangle but not a square.

Full solution

Figure A meets the rectangle condition because it has four right angles.
Its four equal side lengths also make it a square.

Figure B meets the rectangle condition.
Its unequal neighboring side lengths prevent all four sides from being congruent, so it is not a square.

Answer

A is both a square and a rectangle. B is a rectangle but not a square.

Key idea

A more specific family can lie inside another family.
Problem 3 A claim about equal sides

A claim says that every quadrilateral with four congruent sides is a square.

Consider a quadrilateral with four side lengths of 7 centimeters and angle measures 60, 120, 60, and 120 degrees.

Decide whether this figure refutes the claim.

Justify your decision by checking the hypothesis and conclusion.
1. Hint 1
  A counterexample must satisfy the hypothesis but fail the conclusion.
2. Hint 2
  Compare the supplied angles with the right-angle requirement in the square definition.
Answer

Yes. It has four congruent sides but is not a square.

Full solution

The four equal side lengths satisfy the hypothesis.

A square must have four right angles.
The supplied angles are 60 or 120 degrees, not 90 degrees, so this figure fails the conclusion.

It is therefore a counterexample to the universal claim.

Answer

Yes. It has four congruent sides but is not a square.

Key idea

A counterexample belongs to the stated starting family and fails the promised result.
Problem 4 Separate the two parts

Rewrite “Every point on a circle centered at O with radius 16 cm is 16 cm from O” as an if-then statement. Identify its hypothesis and conclusion.
1. Hint 1
  Start with one arbitrary point P covered by the sentence.
2. Hint 2
  The membership condition belongs after “if”; the distance conclusion belongs after “then.”
Answer

If P is on the circle centered at O with radius 16 cm, then $OP=16$ cm. Hypothesis: P is on that circle. Conclusion: $OP=16$ cm.

Full solution

The original statement restricts attention to points on the specified circle.
Use that restriction as the hypothesis.

The promised result is the distance from O, giving the conditional: if P is on that circle, then $OP=16$ cm.

Answer

If P is on the circle centered at O with radius 16 cm, then $OP=16$ cm. Hypothesis: P is on that circle. Conclusion: $OP=16$ cm.

Key idea

An if-then statement separates the condition from the result it guarantees.
Problem 5 Change a conditional carefully

Let $p$ mean “segments $\overline{JK}$ and $\overline{LM}$ are congruent” and $q$ mean “ $JK=LM$ .” Write the converse, inverse, and contrapositive of “if $p$ , then $q$ ” in words.
1. Hint 1
  The converse swaps the hypothesis and conclusion; the inverse keeps their order and negates each one.
2. Hint 2
  The contrapositive makes both of those changes.
Answer

Converse: if $JK=LM$ , then $\overline{JK}$ and $\overline{LM}$ are congruent. Inverse: if those segments are not congruent, then $JK\ne LM$ . Contrapositive: if $JK\ne LM$ , then those segments are not congruent.

Full solution

The converse exchanges the two parts, so equality of the lengths becomes the hypothesis.

The inverse keeps the original order and negates both the segment-congruence condition and the length-equality conclusion.

The contrapositive exchanges the parts and negates both, so it begins with unequal lengths and concludes that the segments are not congruent.

Answer

Converse: if $JK=LM$ , then $\overline{JK}$ and $\overline{LM}$ are congruent. Inverse: if those segments are not congruent, then $JK\ne LM$ . Contrapositive: if $JK\ne LM$ , then those segments are not congruent.

Key idea

Order and negation are separate changes to a conditional.
Problem 6 Match truth without retesting

For a particular conditional about a family of figures, the original has been justified but its converse has a counterexample.

State whether its inverse and contrapositive are true or false, and explain how you know.
1. Hint 1
  Which form has the same logical promise as the original?
2. Hint 2
  The converse and inverse also form an equivalent pair.
Answer

The inverse is false; the contrapositive is true.

Full solution

The inverse is logically equivalent to the converse.
Since the converse is false, the inverse is false.

The contrapositive is logically equivalent to the original.
Since the original is true, the contrapositive is true.

Answer

The inverse is false; the contrapositive is true.

Key idea

Equivalent forms share a truth value, even when other related forms do not.
Problem 7 Test a midpoint biconditional

A statement says, “For any point $M$ , $M$ is the midpoint of segment $\overline{AB}$ if and only if $AM=MB$ ,” where $A$ and $B$ are distinct.

A point $M$ outside the segment is 5 cm from both $A$ and $B$ .

Decide whether the biconditional is true.

If not, write a corrected biconditional and justify both directions.
1. Hint 1
  Test the direction from equal distances to midpoint using the supplied point.
2. Hint 2
  A midpoint definition includes a location condition as well as equal lengths.
Answer

The biconditional is false. A correct version is: $M$ is the midpoint of $\overline{AB}$ if and only if $M$ lies on $\overline{AB}$ and $AM=MB$ .

Full solution

The supplied point has $AM=MB$ but is outside the segment, so it is not a midpoint.
This defeats the reverse direction of the proposed biconditional.

If $M$ is the midpoint of $\overline{AB}$ , its definition gives both membership in the segment and $AM=MB$ .

Conversely, if $M$ lies on $\overline{AB}$ and $AM=MB$ , it satisfies every condition of the midpoint definition.

Point M lies off segment AB, although AM and BM both have length 5 cm. Equal distances alone do not put M on the segment.

Answer

The biconditional is false. A correct version is: $M$ is the midpoint of $\overline{AB}$ if and only if $M$ lies on $\overline{AB}$ and $AM=MB$ .

Key idea

Both directions of a definition need all of its conditions.
Problem 8 A particular figure

Every square is a rectangle.

A card says that figure K is a rectangle.

Does the information establish that K is a square?

Does it establish that K is not a square?

Explain what the information does and does not establish.
1. Hint 1
  The given fact matches which part of the conditional?
2. Hint 2
  Compare a square rectangle with a rectangle having unequal neighboring sides.
Answer

Neither conclusion is established; K may or may not be a square.

Full solution

The given is the conclusion of the square-to-rectangle conditional, so detachment does not apply in reverse.

A square and a rectangle with unequal neighboring sides both fit the card.
The information leaves K’s square classification undecided.

Answer

Neither conclusion is established; K may or may not be a square.

Key idea

Knowing a conditional’s conclusion does not establish or deny its hypothesis.
Problem 9 Test an angle conjecture

Selected angles measure 40, 100, and 150 degrees. Each is bisected. Find the measure of each part in the three cases.

These examples suggest the conjecture “Every part of every bisected angle is acute.” Classify the reasoning that suggests this claim from the examples, then decide whether the conjecture is true.

Justify your decision.
1. Hint 1
  A bisector gives two equal parts whose measures add to the selected angle measure.
2. Hint 2
  Check an allowed angle outside the three examples, such as a straight angle.
Answer

Parts, in order: $20^\circ$ , $50^\circ$ , $75^\circ$ . Inductive reasoning; false conjecture. A straight angle or any reflex opening with $180^\circ<m<360^\circ$ is a valid counterexample.

Full solution

For the $40^\circ$ angle, each part has measure
$\frac{40^\circ}{2}=20^\circ.$

For the $100^\circ$ angle, each part has measure
$\frac{100^\circ}{2}=50^\circ.$

For the $150^\circ$ angle, each part has measure
$\frac{150^\circ}{2}=75^\circ.$

All three part measures are less than 90 degrees, so these examples support the proposed pattern.
Suggesting a general claim from examples is inductive reasoning.

A straight angle is an allowed case.
Each part after bisecting has measure
$\frac{180^\circ}{2}=90^\circ.$

A right angle is not acute.
This case satisfies the bisected-angle condition but fails the conjecture, so it refutes the claim.

A reflex opening with measure $m$ satisfying $180^\circ<m<360^\circ$ gives two parts each greater than $90^\circ$ and less than $180^\circ$ .
Those parts are obtuse, so any such opening also refutes the conjecture.

Answer

Parts, in order: $20^\circ$ , $50^\circ$ , $75^\circ$ . Inductive reasoning; false conjecture. A straight angle or any reflex opening with $180^\circ<m<360^\circ$ is a valid counterexample.

Key idea

Successful examples suggest a conjecture; one allowed counterexample refutes it.
Problem 10 Follow the whole chain

Use these two facts: if a selected angle is acute, then its measure is less than 90 degrees; if its measure is less than 90 degrees, then it is not obtuse.

State the conditional obtained by syllogism.

Angle V is acute.

State the conclusion about V obtained by detachment.

A different angle X is given to be not obtuse. Decide whether this information establishes that X is acute, and justify your decision.
1. Hint 1
  Find the matching middle condition in the two forward implications.
2. Hint 2
  For X, compare an acute angle with a right angle. Both are not obtuse.
Answer

Syllogism: acute implies not obtuse. Detachment: V is not obtuse. Whether X is acute is undecided; a right, straight, or reflex angle is not obtuse and not acute.

Full solution

The measure condition, less than 90 degrees, links the two conditionals.
Syllogism therefore gives acute implies not obtuse.

The given that V is acute matches the hypothesis of that combined conditional.
Detachment gives that V is not obtuse.

The given about X matches the conclusion of the combined conditional.
The forward implication does not establish its hypothesis.

An acute angle and a 90-degree right angle both satisfy “not obtuse.” The right angle is not acute, so the given does not decide whether X is acute.

A straight angle or a reflex angle is also a valid example that is neither obtuse nor acute.

Answer

Syllogism: acute implies not obtuse. Detachment: V is not obtuse. Whether X is acute is undecided; a right, straight, or reflex angle is not obtuse and not acute.

Key idea

A valid chain connects matching forward links and still needs a matching given for a particular conclusion.

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