2.6: Conjectures and Counterexamples
- Page ID
- 2141
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Educated guesses and examples that disprove them.
A conjecture is an “educated guess” that is based on examples in a pattern. A counterexample is an example that disproves a conjecture.
Suppose you were given a mathematical pattern like \(h = \dfrac{−16}{t^2}\). What if you wanted to make an educated guess, or conjecture, about \(h\)?
Use the following information for Examples 1 and 2:
A car salesman sold 5 used cars to five different couples. He noticed that each couple was under 30 years old. The following day, he sold a new, luxury car to a couple in their 60’s. The salesman determined that only younger couples by used cars.
Example \(\PageIndex{1}\)
Is the salesman’s conjecture logical? Why or why not?
Solution
It is logical based on his experiences, but is not true.
Example \(\PageIndex{2}\)
Can you think of a counterexample?
Solution
A counterexample would be a couple that is 30 years old or older buying a used car.
Example \(\PageIndex{3}\)
Here’s an algebraic equation and a table of values for \(n\) and \(t\).
\(t=(n−1)(n−2)(n−3)\)
\(n\) | \((n−1)(n−2)(n−3)\) | \(t\) |
---|---|---|
1 | \((0)(−1)(−2)\) | 0 |
2 | \((1)(0)(−1)\) | 0 |
3 | \((2)(1)(0)\) | 0 |
Solution
After looking at the table, Pablo makes this conjecture:
The value of \((n−1)(n−2)(n−3)\) is 0 for any number n.
Is this a true conjecture?
This is not a valid conjecture. If Pablo were to continue the table to n=4, he would have see that \((n−1)(n−2)(n−3)=(4−1)(4−2)(4−3)=(3)(2)(1)=6\)
In this example \(n=4\) is the counterexample.
Example \(\PageIndex{4}\)
Arthur is making figures for an art project. He drew polygons and some of their diagonals.
From these examples, Arthur made this conjecture:
If a convex polygon has \(n\) sides, then there are \(n−2\) triangles formed when diagonals are drawn from any vertex of the polygon.
Is Arthur’s conjecture correct? Or, can you find a counterexample?
Solution
The conjecture appears to be correct. If Arthur draws other polygons, in every case he will be able to draw \(n−2\) triangles if the polygon has n sides.
Notice that we have not proved Arthur’s conjecture, but only found several examples that hold true. So, at this point, we say that the conjecture is true.
Example \(\PageIndex{5}\)
Give a counterexample to this statement: Every prime number is an odd number.
Solution
The only counterexample is the number 2: an even number (not odd) that is prime.
Review
Give a counterexample for each of the following statements.
- If \(n\) is a whole number, then \(n^2 >n\).
- All numbers that end in 1 are prime numbers.
- All positive fractions are between 0 and 1.
- Any three points that are coplanar are also collinear.
- All girls like ice cream.
- All high school students are in choir.
- For any angle there exists a complementary angle.
- All teenagers can drive.
- If \(n\) is an integer, then \(n>0\).
- All equations have integer solutions.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.5.
Resources
Vocabulary
Term | Definition |
---|---|
conjecture | A conjecture is an educated guess that is based on examples in a pattern. |
counterexample | A counterexample is an example that disproves a conjecture. |
fraction | A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number. |
Additional Resources
Interactive Element
Video: Inductive Reasoning
Activities: Conjectures and Counterexamples Discussion Questions
Study Aids: Types of Reasoning Study Guide
Practice: Conjectures and Counterexamples
Real World: Conjectures And Counterexamples