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Inductive reasoning

A pattern has held in every case you have tried. Does that make it true, how do you hunt for the case that breaks it, and what is the pattern good for?

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Why this exists

Nobody finds a theorem by starting at the proof. Somebody first notices something: the angles of every triangle they have measured add to about 180^\circ, or the odd numbers keep adding up to square numbers. The noticing comes first, and it has a name.

Inductive reasoning is the step from “it happened in these cases” to “maybe it always happens”. It is how most mathematics is discovered. It is also, taken alone, unreliable, and the whole skill is knowing exactly how far it can be trusted.

In geometry the cases are usually drawings: you draw a few figures, notice something, and state it for every figure of that kind. But the figures you draw are the easy ones: symmetric, upright, acute. A claim can hold in all of them and still be false. So a geometric conjecture is tested by hunting for the least tidy figure the hypothesis allows. If one breaks the claim, the next question is how to repair it. If none does, and you can see why, that reason is the start of a proof. Inductive reasoning makes the guess, and this lesson follows the guess until it is broken, repaired or proved.

Definition

Inductive reasoning draws a general statement from particular cases. You look at examples, notice a pattern, and state that the pattern holds in every case, including the ones you have not looked at.

The statement it produces is a conjecture: a claim that looks true and has not been proved. Inductive reasoning cannot do more than that. However many cases agree, the conclusion is probable, never certain, because the next case has not been checked.

Compare deductive reasoning, which starts from statements already accepted and reaches a conclusion that must follow. The two work as a pair. Inductive reasoning proposes; deductive reasoning decides, either by proving the conjecture or by finding a counterexample that ends it.

Conjectures and counterexamples

A geometric conjecture nearly always has the shape: every figure with property H has property C. For example: if a quadrilateral has two pairs of equal sides, it is a parallelogram. Here H is the hypothesis and C the conclusion.

A counterexample figure is one particular figure that

  • has every property in H, and
  • does not have property C,

with both facts shown exactly, from coordinates, given lengths or a construction, never by eye. It is the geometric form of a counterexample.

To repair a false conjecture is to change it so its counterexamples no longer apply: add a hypothesis, which shrinks the set of figures the claim is about, or weaken the conclusion. A repaired claim is a new conjecture, with its own test and its own proof to come.

Intuition

Think of the cases you have checked as a few lit windows in a very long street. You can see a pattern in the lit ones, and it is sensible to guess the dark ones match. But a guess about the dark windows is not the same as having looked. And when the claim is about every whole number or every triangle, the street never ends: however many windows you light, infinitely many stay dark.

So inductive reasoning has three moves, and only the first two belong to it:

  • Observe. Collect cases, and make them as different from each other as you can.
  • Conjecture. State the pattern precisely enough that it could be wrong.
  • Settle. Find a proof that covers every case, or one case where it fails.

Choosing varied cases matters. A pattern seen only in squares may be a fact about squares, not about quadrilaterals. The more a test case differs from the ones that suggested the pattern, the more its agreement is worth, and the more likely it is to expose a false one.

Hunting for the figure that breaks it

In geometry the settling usually starts as a hunt. Every figure you draw has features the hypothesis never asked for. Draw “a quadrilateral with two pairs of equal sides” and you will probably put the equal sides opposite each other, which gives a parallelogram. The claim passes, but only because of the extra feature. The hunt removes such features one at a time:

What your drawing has that the claim did not ask forBreak it with
equal sides placed opposite each otherthe same sides placed next to each other
two equal sides, a line of symmetrya scalene triangle
every angle acutea right angle, then an obtuse one
every corner bulging outwarda concave shape such as a dart
the most special shape of its kindthe next one down the quadrilateral hierarchy

Put the figure on a grid when you can: then lengths come from the distance formula, angles from slopes, and the failure is exact.

Results

What inductive reasoning can give

Inductive reasoning produces conjectures, not proofs. No number of agreeing cases proves a claim about infinitely many.

One failing case disproves a conjecture. A single counterexample is enough. In geometry it is a counterexample figure, and it refutes the claim provided both halves, the hypothesis met and the conclusion failed, are shown exactly.

A general claim is proved only by a deductive argument that covers every case at once.

Any finite list of cases fits more than one rule. So “the pattern” in a list of values is never forced by the values alone.

Counterexample figures and repairs

A figure that misses a hypothesis proves nothing. A trapezoid cannot refute a claim about parallelograms, however close it looks.

Adding a hypothesis shrinks the claim. Excluding only the counterexample you found is not a repair: every figure like it is still there.

A proof of the repaired claim must use what was added. Otherwise it would prove the original claim too, and that was false.

Stop hunting when you can say why the hunt fails. Failed attempts are evidence, not proof. A reason that covers every attempt at once is the outline of a deductive proof.

Why it is true

Why agreeing cases prove nothing

A claim about the angle sum of every triangle is a claim about infinitely many triangles. Checking cases one at a time, you stop after finitely many, and every case you did not check is still open.

The result that any finite list fits more than one rule shows the danger concretely. The rules 2n and

2n + (n - 1)(n - 2)(n - 3)

both give 2, 4, 6 for n = 1, 2, 3, because the extra product is 0 at each of those values. At n = 4 the first gives 8 and the second gives 8 + 3 \cdot 2 \cdot 1 = 14. Multiplying more factors in the same way gives two rules that agree on as many cases as you like and then part. So agreement on the cases you checked can never tell you which rule you are looking at. Only an argument about why the pattern happens can.

Why one figure refutes, and why a repair must be used

“Every figure with H has C” is a universal statement. Its negation, “some figure has H and lacks C”, is proved by one such figure, and a figure without H is not one.

The repaired claim, “every figure with H and H' has C”, covers only figures covered before. A proof of it that never used H' would run on every figure with H alone and prove the old claim, which a figure has shown false.

Worked examples

Examples 1 to 4 go from cases to a conjecture and ask how far the cases can be trusted. Examples 5 to 10 test geometric conjectures by hunting for a counterexample figure. In Example 5 the hunt fails and a proof follows; in Examples 6 to 9 a figure breaks the claim and the claim is repaired; in Example 10 the hunt itself turns into the proof.

Example 1

From measurements to a conjecture

The angles of three triangles are measured, as in Figure 1. What conjecture do they suggest, and how strong is it?

60^\circ
70^\circ
50^\circ
60^\circ + 70^\circ + 50^\circ = 180^\circ
35^\circ
40^\circ
105^\circ
35^\circ + 40^\circ + 105^\circ = 180^\circ
90^\circ
35^\circ
55^\circ
90^\circ + 35^\circ + 55^\circ = 180^\circ
Figure 1Three very different triangles, one total. Three cases suggest a rule; they do not prove it.
Solution

Add the angles in each triangle:

\begin{aligned} 60 + 70 + 50 &= 180 \\ 35 + 40 + 105 &= 180 \\ 90 + 35 + 55 &= 180 \end{aligned}

The triangles were chosen to differ: one with all angles acute, one with an obtuse angle, one with a right angle. All three total 180^\circ, so the conjecture is: the angles of every triangle add to 180^\circ.

It is a strong conjecture, since three very different shapes agree. It is still only a conjecture. Real measurements are worse than this: a protractor reading is off by a degree or so, and three readings may total 179^\circ or 181^\circ. Measuring can never tell 180^\circ from 180.1^\circ. The deductive proof, which draws a line through one corner parallel to the opposite side, is in the lesson on the triangle angle sum, and it is what turns the conjecture into a theorem.

Example 2

Diagonals of a polygon

Count the diagonals of polygons with 3 to 7 sides and use the pattern to predict the number for 8 sides.

Solution

Drawing and counting gives this table.

sides n34567
diagonals025914

The differences between neighboring counts are 2, 3, 4, 5. If that continues, the next difference is 6, and an octagon has 14 + 6 = 20 diagonals.

A prediction like this is a conjecture about one case. Drawing an octagon and counting would test it, but a reason is better. From each corner, diagonals go to every corner except itself and its two neighbors, so each of the n corners starts n - 3 diagonals. That counts every diagonal twice, once from each end, so there are

\frac{n(n - 3)}{2}

diagonals. For n = 8 this is \displaystyle \frac{8 \cdot 5}{2} = 20, which confirms the prediction, and the argument is deductive: it holds for every n, not only the ones in the table.

Example 3

Odd numbers and squares

Add the first few odd numbers. State a conjecture, test it on the first ten odd numbers, and say what the test shows.

Solution
\begin{aligned} 1 &= 1 \\ 1 + 3 &= 4 \\ 1 + 3 + 5 &= 9 \\ 1 + 3 + 5 + 7 &= 16 \end{aligned}

The totals are 1^2, 2^2, 3^2, 4^2. Conjecture: the first n odd numbers add to n^2.

The tenth odd number is 19, and 1 + 3 + 5 + \dots + 19 = 100 = 10^2. The conjecture survives a test further out, which raises confidence. It proves nothing about n = 11, or about any case not yet added up.

The conjecture is true, but only an argument that covers every n at once can show it. That argument, a step from each case to the next, is written out in Conjecture, and why a pattern is not a proof.

Example 4

Points on a circle

Put points on a circle and join every pair with a chord, as in Figure 2. Place the points in general position, which here means that no three chords meet at one point inside the circle. Count the regions. Predict the count for 6 points, then check it.

1 region 2 regions 4 regions 8 regions 16 regions 31 regions
Figure 2Points in general position (no three chords through one inner point). The counts double four times in a row, and then give 31, not 32.
Solution

With 1, 2, 3, 4, 5 points the counts are 1, 2, 4, 8, 16. Each count doubles the one before, so the natural prediction for 6 points is 32.

Counting the last circle gives 31. The doubling pattern, after five agreeing cases, is false. Five agreements were a coincidence of small numbers. The true count is one region, plus one for each chord, plus one for each crossing; six points give 15 chords and 15 crossings, so 1 + 15 + 15 = 31. Why that count is right, and how it carries on past six points, is worked through in Conjecture, and why a pattern is not a proof.

The condition on the points matters. Space six points evenly and the three chords through opposite points, which are diameters, all meet at the center. The points are no longer in general position, and only 30 regions appear; the Boundaries below say why that 30 is no evidence against the 31.

Example 5

Choosing a test that could fail

Figure 3 joins the midpoints of the sides of a quadrilateral in order. Form a conjecture, and choose a test case likely to break it.

Figure 3Join the midpoints of the sides in order. Even the dart gives a parallelogram, though part of it lies outside the dart, past the dent. That is correct, not a drawing error.
Solution

In the irregular quadrilateral on the left, the inner shape has opposite sides that look equal and parallel: a parallelogram. Drawing more ordinary quadrilaterals gives the same thing. Conjecture: joining the midpoints of any quadrilateral’s sides in order gives a parallelogram.

A good test is the case least like the ones already drawn. Every quadrilateral so far bulged outward, so push one corner in and make a dart, as on the right. The inner shape is still a parallelogram, even though part of it now pokes out past the dent and lies outside the dart. The claim was about the shape of the inner quadrilateral, not about it staying inside, so the test is passed. A test that could easily have failed and did not is worth more than ten tests that were never at risk.

The conjecture is in fact true. Call the quadrilateral ABCD and draw the diagonal \overline{AC}. In triangle ABC the midsegment theorem makes the inner side joining the midpoints of \overline{AB} and \overline{BC} parallel to \overline{AC} and half its length. In triangle ADC it does the same for the inner side joining the midpoints of \overline{CD} and \overline{DA}. So one pair of opposite inner sides is both parallel and equal, which makes a parallelogram. Nothing in that argument asked the quadrilateral to bulge outward, which is why the dart obeys it. That is the deductive proof.

Example 6

Two pairs of equal sides

Test the conjecture: if a quadrilateral has two pairs of equal sides, it is a parallelogram.

Solution

The hunt now finds its figure. First drawings put each pair of equal sides opposite each other. The claim did not ask for that, so put each pair next to each other instead: A(0, 4), B(3, 0), C(0, -2), D(-3, 0).

A B C D
5
5
\displaystyle \sqrt{13}
\displaystyle \sqrt{13}
Figure 4Two pairs of equal sides, 5 and \displaystyle \sqrt{13}, but each pair meets at a corner. Not a parallelogram.

Hypothesis: by the distance formula,

\begin{aligned} AB = AD &= \sqrt{3^2 + 4^2} = 5 \\ CB = CD &= \sqrt{3^2 + 2^2} = \sqrt{13} \end{aligned}

so there are two pairs of equal sides. Conclusion: a parallelogram has its opposite sides parallel, but \overline{AB} has slope \displaystyle -\frac{4}{3} and the opposite side \overline{DC} has slope \displaystyle -\frac{2}{3}. So ABCD is not a parallelogram, and the conjecture is false. The figure is a kite.

Repair by saying where the equal sides sit: a quadrilateral whose opposite sides are equal in pairs is a parallelogram. That is true, and is proved in proving a quadrilateral is a parallelogram once triangle congruence is available.

Example 7

The angle bisector in a triangle

Test the conjecture: in every triangle, the bisector of an angle bisects the opposite side. Then repair it.

Solution

A first drawing is usually isosceles, with the bisected angle between the equal sides, and there the claim holds. Break the equal sides. Take the right triangle A(0, 0), B(6, 0), C(0, 3).

A B C D
\displaystyle 2\sqrt{5}
\displaystyle \sqrt{5}
Figure 5\overline{AD} bisects \angle A, and cuts \overline{BC} into pieces \displaystyle 2\sqrt{5} and \displaystyle \sqrt{5}.

\overline{AB} runs along the x-axis and \overline{AC} up the y-axis, so the line y = x makes 45^\circ with each and bisects \angle A. Side \overline{BC} lies on the line x + 2y = 6, and y = x meets it where 3x = 6, at D(2, 2). Then

\begin{aligned} BD &= \sqrt{4^2 + 2^2} = 2\sqrt{5} \\ DC &= \sqrt{2^2 + 1^2} = \sqrt{5} \end{aligned}

BD is twice DC, so the conjecture is false. The pieces are in the ratio of the legs, 6 : 3, which is the angle bisector proportion.

Repair it by adding back the feature the drawings had: in a triangle with AB = AC, the bisector of \angle A bisects \overline{BC}. Here there is a reason, so hunting further is pointless, though the proof itself has to wait for a later lesson. Triangles ABD and ACD have AB = AC (given), \angle BAD \cong \angle CAD (the bisector) and \overline{AD} in common. That is the if-part of SAS, one of the congruence criteria set out in a later lesson, which will make them congruent, so BD = DC; the proof is written out as Example 1 of proving the Base Angles Theorem. The plan uses both added facts, the equal sides and that the bisected angle lies between them; Exercise 20 shows the second cannot be dropped.

Example 8

Where the altitudes meet

Test the conjecture: the altitudes of a triangle meet at a point strictly inside it, on none of its sides.

Solution

In acute drawings it holds, so make one angle obtuse: A(0, 0), B(8, 0), C(-2, 4). The angle at A is obtuse, since \overrightarrow{AB} \cdot \overrightarrow{AC} = 8(-2) + 0(4) = -16 < 0.

A B C H
Figure 6In an obtuse triangle the altitudes meet at H, outside.

An altitude is the line through a vertex perpendicular to the line of the opposite side. The one from C is perpendicular to the x-axis, so it is x = -2. The one from B is perpendicular to \overline{AC}, of slope -2, so it has slope \displaystyle \frac{1}{2} and is \displaystyle y = \frac{1}{2}(x - 8). At x = -2 that gives y = -5, so these two meet at H(-2, -5). The one from A is perpendicular to \overline{BC}, of slope \displaystyle -\frac{2}{5}, so it is \displaystyle y = \frac{5}{2}x, and it passes through H too, since \displaystyle \frac{5}{2}(-2) = -5.

Every point of the triangle has y \ge 0, and H has y = -5. The altitudes meet outside, so the conjecture is false.

Repair: the altitudes of an acute triangle meet strictly inside it, which is true. So is the weaker “the altitudes of any triangle meet at one point”: that point is the orthocenter, and this figure shows it can lie outside.

Example 9

Two sides and the area

Test the conjecture: if two triangles have two pairs of equal sides and equal areas, they are congruent. Then find every triangle the hypothesis allows.

Solution

First drawings make both triangles acute and alike. Keep the sides and the area, and let the angle between the sides open past 90^\circ. With A(0, 0) and B(8, 0), take C(3, 4) for one triangle and C'(-3, 4) for the other.

A B C A B C′
8
8
5
5
\displaystyle \sqrt{41}
\displaystyle \sqrt{137}
4
4
Figure 7Sides 8 and 5, area 16, in both. The third sides differ.

Hypothesis: \overline{AB} is shared, and \displaystyle AC = AC' = \sqrt{3^2 + 4^2} = 5. Both triangles have base 8 and height 4, so both have area \displaystyle \frac{1}{2}(8)(4) = 16. Conclusion: the third sides are

\begin{aligned} BC &= \sqrt{5^2 + 4^2} = \sqrt{41} \\ BC' &= \sqrt{11^2 + 4^2} = \sqrt{137} \end{aligned}

Congruent triangles have all three sides equal, so these are not congruent. False.

Now make every try at once. Place any such triangle with its side of 8 on \overline{AB} and its side of 5 starting at A. Area 16 on base 8 means height 4, so the third vertex has y = \pm 4, and being 5 from A means x^2 + 16 = 25, so x = \pm 3. The points below the axis are mirror images of C and C', so there are only two triangles, and their angles at A add to 180^\circ (\overline{AC'} is the mirror image of \overline{AC} in the y-axis). The same steps with letters repair the claim by weakening its conclusion: two triangles with two pairs of equal sides and equal areas either are congruent or have the angles between those sides adding to 180^\circ.

Example 10

Equal diagonals: from hunt to proof

Test the conjecture: if a quadrilateral’s diagonals are equal, it is a rectangle. Repair it, and decide when to stop hunting.

Solution

Rectangles are what one draws first. Let two sides lean in by the same amount: the isosceles trapezoid A(-3, 0), B(3, 0), C(1, 2), D(-1, 2).

A B C D
Figure 8Equal diagonals, and no right angle anywhere.
\begin{aligned} AC &= \sqrt{4^2 + 2^2} = \sqrt{20} \\ BD &= \sqrt{4^2 + 2^2} = \sqrt{20} \end{aligned}

The diagonals are equal. But \overline{AD} has slope 1, so m\angle DAB = 45^\circ, not 90^\circ. False. Every isosceles trapezoid has equal diagonals.

Repair: a parallelogram with equal diagonals is a rectangle. Hunt again, with the lean as a letter so every try is made at once. Place a parallelogram with A(0, 0), B(b, 0) and D(s, h), where b > 0 and h > 0. Since \overline{DC} \parallel \overline{AB}, C is at height h; since \overline{BC} \parallel \overline{AD}, the step from B to C matches the step from A to D. So C(b + s, h), and

\begin{aligned} AC^2 &= (b + s)^2 + h^2 \\ BD^2 &= (s - b)^2 + h^2 \\ AC^2 - BD^2 &= (b + s)^2 - (s - b)^2 = 4bs \end{aligned}

Since b > 0, the diagonals are equal only when s = 0. Then D is straight above A, so \angle A is right, and since neighboring angles are same-side interior angles adding to 180^\circ, all four are right: ABCD is a rectangle.

That is where the hunt stops. The algebra did not try more parallelograms; it covered all of them, so it is the proof (another is in proving a parallelogram is a rectangle). It used “parallelogram” where it had to, in placing C.

Boundaries

Inductive reasoning is not mathematical induction. The names are close and the ideas are not. Mathematical induction is a deductive proof method: prove the first case, then prove that each case forces the next. Those two proofs together settle every whole number from the first case on, with nothing left unchecked. Inductive reasoning only suggests a claim. The first often proves what the second found, as it does for the odd numbers of Example 3.

It cannot prove a claim about infinitely many cases, however many agree. A claim can hold for a million cases and fail on the next. See why a pattern is not a proof.

When the cases are finite and every one is checked, it is no longer inductive reasoning. “Every whole number from 1 to 20 is a sum of at most four squares” can be proved by writing out all twenty. Checking every case of a finite claim is a deductive proof, a proof by exhaustion; the gap that makes inductive reasoning uncertain is the unchecked case, and here there is none.

It works only on cases that meet the claim’s conditions. A measured case with a slightly wrong reading, or a figure that is not quite what it claims to be, is not evidence either way. The circle count needs points in general position; six evenly spaced points break that condition, so their 30 says nothing against the 31.

A counterexample’s failure must be exact. Sides measured as 3.0 and 3.1 centimeters do not make a counterexample; the difference may be the pencil. This is what a diagram may assume, from given, prove and the diagram, turned round.

The figure must meet the hypotheses hidden in the definitions. A quadrilateral normally means one whose sides do not cross. A bow-tie can have its opposite sides equal in pairs without being a parallelogram, but it refutes nothing about the repair in Example 6.

A repair is a new claim. It can be false again (Exercise 21), and it is true only once proved. Enough added hypotheses make any claim true, so a good repair adds as little as it can.

A figure refutes a statement, not everything near it. The obtuse triangle of Example 8 kills “the altitudes meet inside”, not “the altitudes meet at one point”.

Counterexamples outside geometry are their own lesson. Searching for the one case that breaks a claim about numbers or sets, and the logic of what a single failure proves, is counterexample in general.

Where this goes next

Every theorem in geometric proof started life as somebody’s conjecture, and in a geometry course the cycle repeats: experiment with figures, state what you see, hunt for the figure that breaks it, then prove it. The theorems about parallelograms, isosceles triangles and altitudes all went through it. The proving is deductive reasoning. A claim that survives moves on to given and prove and a two-column proof; its converse starts a new hunt. Example 9 has a famous cousin, that two sides and a non-included angle fix a triangle, whose counterexample figure is in why SSA is not a criterion. Number patterns lead to proof by induction, the tool built for claims about every whole number, and how sure a long run of agreeing cases can make you is the subject of Conjecture, and why a pattern is not a proof.

Exercises

Forming a conjecture

  1. 11^2 = 121, 111^2 = 12321 and 1111^2 = 1234321. Conjecture the value of 11111^2, then check it by multiplying.
  2. 9 \times 1 + 2 = 11, 9 \times 12 + 3 = 111 and 9 \times 123 + 4 = 1111. Write the next two lines of the pattern and check them.
  3. A triangle’s angles add to 180^\circ, a quadrilateral’s to 360^\circ and a pentagon’s to 540^\circ. Conjecture the angle sum of a hexagon and of a 12-sided polygon.
  4. Use the formula from Example 2 to find the number of diagonals of a 10-sided polygon, and check that it continues the pattern of differences.
  5. 1 \times 2 = 2, 2 \times 3 = 6, 3 \times 4 = 12 and 4 \times 5 = 20. State a conjecture about the product of two consecutive whole numbers.

Testing a conjecture

  1. Conjecture: n^2 + n + 17 is prime for every whole number n. Test n = 1, 2, 3, 4, 5, then n = 16. For the last one, write 16^2 + 16 as 16 \times 17.
  2. Conjecture: n^3 - n is a multiple of 6 for every whole number n. Test n = 2, 3, 4, 5. Does the test prove the conjecture?
  3. Does the pattern of Exercise 1 still hold for nine ones? Find 111111111^2.
  4. Does it hold for ten ones? Find 1111111111^2.
  5. Seven points are placed on a circle in general position, so no three chords meet at a point inside it. They give 21 chords and 35 crossings. How many regions are there, and what would doubling have predicted?
  6. A student measures the four angles of a quadrilateral as 88^\circ, 93^\circ, 91^\circ and 87^\circ. Find the total, and say what conjecture is reasonable.
  7. Find 4 + 5 + 6, 10 + 11 + 12 and 29 + 30 + 31. What do you notice about the sum of three consecutive whole numbers?
  8. Find 1 + 2 + 3 + 4, 2 + 3 + 4 + 5 and 3 + 4 + 5 + 6. Is the sum of four consecutive whole numbers always a multiple of 4? What else do the sums have in common?

Does the figure count?

  1. Claim: a quadrilateral with four equal sides is a square. Which of these is a counterexample? (a) a square of side 3; (b) the quadrilateral (0, 0), (5, 0), (8, 4), (3, 4); (c) a 3 by 5 rectangle.
  2. Claim: a quadrilateral with a pair of parallel sides is a parallelogram. Is the quadrilateral (0, 0), (6, 0), (4, 3), (1, 3) a counterexample?
  3. Claim (Example 8): the altitudes of a triangle meet at a point strictly inside it, on none of its sides. Is the right triangle (0, 0), (4, 0), (0, 3) a counterexample?

Find a counterexample figure

Give a figure by coordinates or lengths, and show exactly that it breaks the claim.

  1. A quadrilateral with one pair of opposite sides equal is a parallelogram.
  2. A quadrilateral whose diagonals bisect each other is a rectangle.
  3. The diagonals of a kite bisect each other.
  4. Triangle A(0, 4), B(-3, 0), C(3, 0) has AB = AC = 5. The bisector of the base angle \angle ABC meets \overline{AC} at \displaystyle D\left(\frac{15}{11}, \frac{24}{11}\right). Find AD and DC, and say what this shows about the repair in Example 7.

Repair the claim

  1. Claim: a quadrilateral with perpendicular diagonals is a rhombus. The kite of Example 6, whose diagonals lie on the axes, breaks it, so a student repairs it to: a quadrilateral with perpendicular diagonals that is not a kite is a rhombus. Test the repair on A(0, 3), B(2, 0), C(0, -1), D(-1, 0).
  2. Claim: a quadrilateral with four right angles is a square. Give a counterexample and two different repairs.
  3. Claim: if two angles of a triangle are equal, the triangle is equilateral. Give a counterexample by its angles, then repair the claim by weakening the conclusion.
  4. An altitude of length 6 cuts a triangle into two right triangles, and its foot splits the base into pieces 4 and 9. The top angle looks right. Decide with the converse of the Pythagorean theorem. Then decide again with pieces 4 and 8.

Counterexample or proof?

Decide whether each claim is true, and give a counterexample figure or a proof.

  1. Every triangle has an angle of at least 60^\circ.
  2. A quadrilateral whose diagonals are perpendicular and bisect each other is a rhombus.
  3. An exterior angle of a triangle is larger than every interior angle of the triangle. If this is false, repair it.

Reasoning

  1. Prove the conjecture from Exercise 12 by writing the three numbers as n - 1, n and n + 1.
  2. Prove that the sum of four consecutive whole numbers is always even and never a multiple of 4.
  3. A computer checks a claim about whole numbers for every n from 1 to 1{,}000{,}000 and finds no failure. Has the claim been proved? Explain.
  4. Explain the difference between inductive reasoning and proof by induction.
  5. Show that 2^n and n^2 - n + 2 give the same values for n = 1, 2, 3, and find where they first differ.
  6. In Example 5, why is testing the dart worth more than testing a rectangle?
  7. A student draws a quadrilateral with perpendicular diagonals, measures its sides as 3.1, 3.0, 3.0 and 3.1 centimeters, and calls it a counterexample to the claim of Exercise 21. What is missing, and how would you fix it?
  8. Explain why a proof of a repaired claim must use the hypothesis that was added.
  9. In Example 10 the hunt among parallelograms kept failing. Explain what made it reasonable to stop hunting, and how the reason became the proof.
Answers
  1. 11111^2 = 123454321. The pattern counts up to the number of ones and back down, and multiplying confirms it.
  2. 9 \times 1234 + 5 = 11111 and 9 \times 12345 + 6 = 111111. Both check: 11106 + 5 and 111105 + 6.
  3. Hexagon 720^\circ, 12-sided polygon 1800^\circ. Each extra side adds 180^\circ, so the conjecture is {(n - 2) \cdot 180^\circ} for n sides. The proof, which splits the polygon into triangles, is in the polygon angle sum.
  4. 35, since 10 \cdot 7 \div 2 = 35. The pattern gives 14, 20, 27, 35 for 7, 8, 9, 10 sides, with differences 6, 7, 8.
  5. The product of two consecutive whole numbers is always even. One of any two consecutive whole numbers is even, which is why; 7 \times 8 = 56 fits.
  6. False. n = 1, 2, 3, 4, 5 give 19, 23, 29, 37, 47, all prime, and so does every n from 0 to 15. But n = 16 gives 16^2 + 16 + 17 = 16 \times 17 + 17 = 17 \times 17 = 289. Sixteen agreeing cases, then a failure.
  7. n = 2, 3, 4, 5 give 6, 24, 60, 120, which are 6 \times 1, 6 \times 4, 6 \times 10 and 6 \times 20, so the conjecture survives. The test does not prove it: four cases say nothing about n = 6. (It is true, for a reason the test never saw; the proof is in Conjecture, and why a pattern is not a proof.)
  8. Yes. 111111111^2 = 12345678987654321.
  9. No. 1111111111^2 = 1234567900987654321. Carrying spoils the pattern once a digit would reach 10.
  10. 57 regions: 1 + 21 + 35 = 57. Doubling from 31 would give 62, and doubling all the way from 1 would give 64.
  11. The total is 359^\circ. The reasonable conjecture is 360^\circ, since each reading can be off by a degree; a measured 359^\circ neither confirms nor breaks it.
  12. 15, 33, 90. Each is a multiple of 3, in fact 3 times the middle number.
  13. 10, 14, 18. Not a multiple of 4, since 10 is not. All three are even, and each is 2 more than a multiple of 4.
  14. Only (b). Its sides are all 5: two run level and are 5 long, and the two slanted ones are \displaystyle \sqrt{3^2 + 4^2} = 5. At (0, 0) the sides go along (5, 0) and (3, 4), whose dot product is 15 \ne 0, so the angle is not right and it is not a square. (a) meets the conclusion; (c) misses the hypothesis.
  15. Yes. The sides from (0, 0) to (6, 0) and from (1, 3) to (4, 3) are both horizontal, so parallel. The other two have slopes \displaystyle -\frac{3}{2} and 3, so they are not parallel, and it is not a parallelogram.
  16. Yes. The altitude from (4, 0) is perpendicular to the leg on the y-axis, so it is the x-axis; the altitude from (0, 3) is the y-axis. They meet at (0, 0), where the third altitude starts. That point is a vertex, on two sides, so not strictly inside.
  17. (0, 0), (6, 0), (4, 2), (2, 2). Both slanted sides are \displaystyle \sqrt{2^2 + 2^2} = \sqrt{8}, but the other pair are 6 and 2, and a parallelogram’s opposite sides are equal.
  18. (0, 0), (4, 0), (5, 2), (1, 2). Both diagonals have midpoint \displaystyle \left(\frac{5}{2}, 1\right), but at (0, 0) the sides go along (4, 0) and (1, 2), with dot product 4 \ne 0, so the angle is not right.
  19. The kite of Example 6. \overline{AC} has midpoint (0, 1) and \overline{BD} has midpoint (0, 0). The midpoints differ, so the diagonals do not bisect each other.
  20. \displaystyle AD = \frac{25}{11} and \displaystyle DC = \frac{30}{11}, from \displaystyle \sqrt{15^2 + 20^2} = 25 and \displaystyle \sqrt{18^2 + 24^2} = 30, each over 11. They are unequal, so in an isosceles triangle the bisector of a base angle does not bisect the opposite side. The repair must name the angle between the equal sides.
  21. It fails. The diagonals lie on the axes, so they are perpendicular, but the sides are \displaystyle AB = \sqrt{13}, \displaystyle BC = \sqrt{5}, \displaystyle CD = \sqrt{2} and \displaystyle DA = \sqrt{10}. No two are equal, so it is not a kite and not a rhombus. Excluding kites removed one counterexample, not the reason.
  22. A 2 by 5 rectangle. Repairs: add that two adjacent sides are equal, or that the diagonals are perpendicular. Each one forces a square.
  23. Angles 40^\circ, 40^\circ, 100^\circ: two are equal and the triangle is not equilateral. Repair: then the triangle is isosceles (the converse of the Base Angles Theorem).
  24. With 4 and 9: the slanted sides are \displaystyle \sqrt{4^2 + 6^2} = \sqrt{52} and \displaystyle \sqrt{9^2 + 6^2} = \sqrt{117}, the base is 13, and 52 + 117 = 169 = 13^2, so it is right. With 4 and 8: the sides are \displaystyle \sqrt{52} and 10, the base is 12, and 52 + 100 = 152 \ne 144, so it is not right.
  25. True. If every angle were less than 60^\circ, the three would add to less than 180^\circ, but they add to exactly 180^\circ (the triangle angle sum).
  26. True. Put the crossing point at the origin with one diagonal on the y-axis; the other is perpendicular to it, so it lies on the x-axis. Each is cut in half there, so the corners are (0, a), (b, 0), (0, -a), (-b, 0) with a, b > 0, and by the distance formula every side is \displaystyle \sqrt{a^2 + b^2}.
  27. False. In a triangle with angles 30^\circ, 30^\circ, 120^\circ, the exterior angle at the 120^\circ corner is 60^\circ. Repair: an exterior angle is larger than each of the two interior angles not next to it. That is true, since it equals their sum (the exterior angle theorem) and each is positive.
  28. (n - 1) + n + (n + 1) = 3n, which is 3 times a whole number, so a multiple of 3.
  29. The sum is n + (n + 1) + (n + 2) + (n + 3) = 4n + 6. Since 4n + 6 = 2(2n + 3), it is even, and since 4n + 6 = 4(n + 1) + 2, it leaves remainder 2 when divided by 4.
  30. No. The claim covers infinitely many whole numbers, and the check covers finitely many. It is strong evidence for a conjecture, but the number 1{,}000{,}001 is still unchecked.
  31. Inductive reasoning goes from examples to a conjecture and never proves it. Proof by induction is a deductive method: it proves the first case and that each case forces the next, which proves the claim for every whole number from the first case on.
  32. Both give 2, 4, 8 for n = 1, 2, 3. At n = 4, 2^4 = 16 but 16 - 4 + 2 = 14.
  33. The conjecture came from quadrilaterals that bulge outward, and a rectangle is another of those. The dart differs in exactly the way the earlier cases did not, so it was the likeliest to fail, and its agreement says more.
  34. Measurements that close cannot tell equal from unequal, so nothing has been shown to fail. Give the figure exactly instead, for example the kite of Example 6: its diagonals lie on the axes, and its sides are 5 and \displaystyle \sqrt{13}.
  35. A proof that never used the added hypothesis would run on every figure with the old hypotheses alone, and so would prove the old claim. A counterexample has shown that claim false, so such a proof has a mistake in it.
  36. Writing the lean as a letter gave AC^2 - BD^2 = 4bs, which says why every lean fails: any slide makes one diagonal longer than the other. That covers every parallelogram at once, so it is already the proof: equal diagonals force s = 0, a right angle.
kg1128 · updated 2026-10-02 · CC-BY-SA-4.0