††affiliationtext: 1Facultad de Ingeniería y Ciencias, Universidad Adolfo
Ibáñez , Diagonal Las Torres 2640, Peñalolén , Santiago , 7941169, Chile

Education in Engineering– Higher Education– Incorrect Problems– Mathematics
Education– Teaching Techniques

#  Learning Mathematics through incorrect Problems

Hugo Caerols-Palma1 Katia Vogt-Geisse${{}^{1}}^{*}$<EMAIL_ADDRESS>
###### Abstract

In this article we describe special type of mathematical problems that may
help develop teaching methods that motivate students to explore patterns,
formulate conjectures and find solutions without only memorizing formulas and
procedures. These are problems that either their solutions do not make sense
in a real life context and/or provide a contradiction during the solution
process. In this article we call these problems incorrect problems. We show
several examples that can be applied in undergraduate mathematics courses and
provide possible ways these examples can be used to motivate critical
mathematical thinking. We also discuss the results of exposing a group of 168
undergraduate engineering students to an incorrect problem in a Differential
Equations course. This experience provided us with important insight on how
well prepared our students are to “out of the box” thinking and on the
importance of previous mathematical skills in order to master further
mathematical analysis to solve such a problem.

##  1 Introduction

Solving exercises and problems is fundamental for learning Mathematics and are
the base for teaching this subject. According to Halmos et. al, problem-
solving is the hearth of Mathematics and of a Mathematician’s work [2].
Teaching mathematics oriented towards problem-solving and developing
strategies that help students improve their problem-solving abilities has been
of great importance since the 1970s [7].

Schoenfeld et al. in [9] explain the importance that problem-solving methods
have had to teach mathematics in the last decades of the 20th century. The
article comments on the main focus of learning based on problem-solving while
developing mathematical thinking. The authors suggest to make changes in the
curriculum and in the teaching methods in order to develop the students’
mathematical thinking abilities. They propose that teachers assign the
following tasks to students:

  1. 1.

Exploring patterns, instead of memorizing formulas;

  2. 2.

Formulating conjectures, not only doing exercises;

  3. 3.

Finding solutions, instead of memorizing exercises or procedures.

The following is an example of a problem that illustrates a way to explain how
to achieve the previously listed tasks:

Given two initial values $x_{0}=a,x_{1}=b,$ determine if the mathematical
sequence

$x_{n+2}=\frac{x_{n+1}+x_{n}}{2},\;n\geq 0$ (1)

converges.

To explore patterns, a student could study the convergence of the sequence by
creating a table starting with the initial values and obtaining
$x_{2},x_{3},x_{4},...$ recursively, and this way exploring to which value the
sequence converges. Students could then formulate a conjecture of the value of
convergence that depends on the initial values chosen. Then, the conjecture
should be proven. How to show the conjecture depends on the knowledge of the
student and the mathematical tools each student manages, these problems can be
approached differently and may lead to distinct proofs. For instance, a
student could prove the convergence using an inductive reasoning, whereas a
student who knows how to deduce a direct formula for this type of recurrences
can find a more direct and different way to prove it.

In this article we describe a special type of problems that may help to
develop a teaching method that makes students explore patterns, formulate
conjectures and find solutions without only memorizing formulas and
procedures. These are problems that when solving them generally their
solutions do not make sense in a real life context or provide a contradiction.
In this article we will call those problems incorrect problems and will
provide and discuss a few examples.

We will start presenting a simple incorrect problem that could be used in a
basic Algebra course during the first years of undergraduate math studies when
studying the Field of Real Numbers or even at a high school level mathematics
course. Consider the following chain of reasoning:

Suppose that $a=b$ therefore $a^{2}=ab$ then $a^{2}-b^{2}=ab-b^{2}$ from the
previous $(a+b)(a-b)=b(a-b)$. Dividing by $(a-b)$ on both sides of the
equation we obtain $a+b=b$, but since $a=b$ we have $2a=a$ dividing by $a$,
gives us $2=1$.

The first time students are confronted with this problem they usually feel
puzzled when realizing that they have reached a contradiction: $2=1$ is
definitely not right. By proposing these types of exercises, professors expect
that their students become more careful in each step when using deductions to
solve them. Some, who at some point have faced this problem and found the
reason for the contradiction seldom forget it: there are dangers when dividing
by zero. Exercises such as the one just shown may be used to motivate and
explain the importance of a deeper understanding of the implied axioms
underlying a theory, in this case the axioms of the Field of Real Numbers, as
Lakatos et al. discuss in [5].

The objectives of this work are: first, to detect learning opportunities
obtained from analyzing the solutions of incorrect problems; second, to
explore how a group of students– that are used to be confronted with direct
“correct” problems– perform when being exposed to an incorrect problem; and
third, to discuss the initial perception of the professors– which was that the
use of incorrect problems will have positive effects on learning outcomes– is
confirmed in the case of the specific example and group of students analyzed,
aiming to give important insight which may help when dealing with similar
exercises in different undergraduate mathematics courses.

In Sec. 2 we present and discuss examples of incorrect problems. Particularly,
in Subsec. 2.1 we give an example that has appeared naturally while teaching
the course Multivariable Calculus and in Subsec. 2.2 and 2.3 we provide two
problems that are relevant in a Differential Equations course for
undergraduate engineering students. In Sec. 3 we show the results of assigning
an incorrect problem that uses Newton’s Law of cooling to 168 undergraduate
engineering students in 7 sections of the Differential Equations course at
Universidad Adolfo Ibáñez in Santiago, Chile. Finally, in Sec. 4 we will
present a discussion regarding this experience, which we believe will be of
interest of those who are always searching for a better way of teaching
mathematics and have used or are considering using these type of strategies in
their courses.

##  2 Examples of incorrect problems

###  2.1 Motivating the Implicit Function Theorem from a limit problem in two
variables

Limit problems in Multivariable Calculus are naturally useful to analyze and
establish conjectures. One interesting example is studying the following limit

$\lim_{(x,y)\to(0,0)}\frac{xy}{x+y}.$ (2)

These problems are often approached choosing trajectories that pass through
the origin $(0,0)$ and using the theorem that states that if the limit of a
function along one trajectory differs from the limit on another trajectory,
then the limit does not exist [11]. Hence, probably the first approach, which
is standard, used by students will be to consider straight lines through the
origin as trajectories, resulting in the limit to be zero, and then study the
limit along other trajectory such as parabolas or square root functions that
pass through the origin, also resulting in the limit to be zero. Then, sensing
that the limit exists and is zero– since different trajectories result in the
same limit– they may try polar coordinates $x=r\cos(\alpha)$,
$y=r\sin(\alpha)$, $0\leq\alpha<2\pi$, $r\geq 0$ in the attempt to prove it,
obtaining

$\lim_{r\to 0}\frac{r\sin(\alpha)\cos(\alpha)}{\sin(\alpha)+\cos(\alpha)}.$
(3)

In what follows we have a first learning opportunity by identifying a mistake
which students often make. The mistake lies in concluding that the limit is
zero since $r$ approaches zero without considering that the function

$g(\alpha)=\frac{\sin(\alpha)\cos(\alpha)}{\sin(\alpha)+\cos(\alpha)}$ (4)

has to be bounded for $0\leq\alpha<2\pi$ in order to argue like that.

To make students realize their mistake, one has to prove that the limit does
not exist by for example presenting the idea of level curves to students. This
leads to an interesting technique to solve the problem. If we consider

$\frac{xy}{x+y}=a,$ (5)

solving for $y$ we obtain the curve $\displaystyle y=\frac{ax}{x-a}$, which
represents a trajectory that for every $a$ value passes through the origin and
the limit on that trajectory is equal to $a$. Choosing $a$ different from zero
we found a trajectory that gives us a limit different from the zero limit that
we obtained along a straight line, this shows then that the limit does not
exist. Fig. 1 depicts different trajectories for different values of $a.$ Once
proven that the limit does not exists, their mistake can be discussed once
again to emphasize the importance to check boundedness of the function
$g(\alpha)$ given by equation (4).

Figure 1: Trajectories $\displaystyle y=\frac{ax}{x-a}$ for of $a=1$ and
$a=3$.

Afterwards, the students may be asked to analyze the limit

$\lim_{(x,y)\to(0,0)}\frac{x^{3}+y^{3}}{x^{2}+y^{2}}.$ (6)

using level curves. Equaling to a constant $a$, we obtain level curves of the
form

$\displaystyle\frac{x^{3}+y^{3}}{x^{2}+y^{2}}=a,$ (7)

and hence, for $a=1$ we obtain the equation

$x^{3}+y^{3}-x^{2}-y^{2}=0.$ (8)

This is an equation that supposedly represents a curve passing through $(0,0)$
and hence we could argue that the limit along that trajectory is equal to
$a=1$, and one can also easily see that the limit is zero along the straight
line $y=x$, and hence the argument would conclude that the limit does not
exist. But, observe that by utilizing polar coordinates, it can be easily
proven that the given limit is zero. So the question that arises naturally is:
What went wrong with the level curves approach? After giving the question some
thought, students could observe that the reason the approach went wrong may be
that the relation in Eqn. (8) can not define a function $y=f(x)$ nor $x=f(y)$
near the origin and passing through it, hence could not represent a
trajectory. Indeed, see in Fig. 2 the graph of the curve given by Eqn. (8).
The origin satisfies the equation but no other point in its neighborhood does.

Figure 2: Graph given by the equation $x^{3}+y^{3}-x^{2}-y^{2}=0.$

From this incorrect problem the following question arises naturally: when does
a relation between to variables represent one as a function of the other in
the neighborhood of a point that verifies it? Thus, we have reached through
incorrect limit solutions with two variables an understanding of the real
importance of the Implicit Function Theorem (see the details of the Theorem in
[8]).

###  2.2 A problem that questions numerical methods

We present in this subsection an example that illustrates interesting
questions that arise from an initial value problem, for which the initial
condition escapes the interval of definition of the solution. In particular we
are going to study how a numerical solution method may fail to give expected
answer. Consider the following incorrect problem that was proposed by Simmons
in [10]:

Given the initial value problem

$y^{\prime}=y^{2}+1,\;\;y(0)=0.$ (9)

Use the Euler Method to estimate numerically $y(2)$ with steps $h=$0.4,
$h=$0.2 and $h=$0.1 starting from the origin. Discuss your results.

The problem given is an initial value problem of the form

$\frac{dy}{dx}=f(x,y),\;\;y(x_{0})=y_{0}.$ (10)

In order to find $y_{k}=y(x_{k})$, the Euler method– as also do other
numerical methods to solve differential equations– considers a sequence of
points $\\{x_{k}\\}$ starting from the initial condition $x_{0}$ and separated
by a step size $h$. It then uses the function $f(x,y)$ in Eqn. (10) that
represents the slope of the tangent line to the solution curve at any given
point to find the next image in the form $y_{k+1}=y_{k}+f(x_{k},y_{k})h$ (see
Fig. 3 ). Starting from $(x_{0},y_{0})$ several iterations then return the
desired $y_{k}=y(x_{k})$ value that approximates the image of the solution
curve at $x_{k}$ [12].

Figure 3: The figure shows three iterations to find the value of $y_{3}$ that
approximates the image $y(x_{3})$ of the solution curve of a first order
differential equation.

Table 1 shows some results for $y(2)$ when performing some iterations of
Euler’s Method for the example from Eqn. (9).

Table 1: The table shows in the last rows the approximate value of $y(2)$ obtained after 10 and 5 iterations of the Euler Method when using the step size $h=0.2$ and $h=0.4$ respectively.  $h=0.2$ | $h=0.4$
---|---
$n$ | $x_{n}$ | $y_{n}$ | $n$ | $x_{n}$ | $y_{n}$
0 | 0 | 0 | 0 | 0 |
1 | $0.2$ | $0.2$ | 1 | $0.4$ | $0.4$
2 | $0.4$ | $0.408$ | 2 | $0.8$ | $0.864$
3 | $0.6$ | $0.6413$ | 3 | $1.2$ | $1.5626$
4 | $0.8$ | $0.9235$ | 4 | $1.6$ | $2.9393$
5 | $1.0$ | $1.2941$ | 5 | $2$ | $6.795$
6 | $1.2$ | $1.8291$ |  |  |
7 | $1.4$ | $2.6982$ |  |  |
8 | $1.6$ | $4.3542$ |  |  |
9 | $1.8$ | $8.3461$ |  |  |
10 | $2$ | $22.4778$ |  |  |

The values observed in Table 1 gives a starting point to think that something
strange is happening in this example when using this method. We could ask
ourselves: Why is there such a high variability in the approximate values of
$y(2)$ when choosing different step sizes? The first approach to start
understanding this, is to consider whether the initial value problem has a
unique solution, since the variability could be due to the method finding
different solutions, and hence $y(2)$ could be different for each solution.
Analyzing the hypothesis of the Theorem of Existence and Uniqueness for first
order initial value problems [12], one concludes easily that the initial value
problem from Eqn. 9 has a unique solution. Therefore the variation is likely
due to some other cause.

A second reflection could be to look at the differential equation in Eqn. (9)
as an autonomous equation and examine the dynamics of its family of solutions.
Since for the given equation $y^{\prime}=y^{2}+1$ there are no real constant
solutions, the solutions could perfectly tend to infinity when approaching
$x=2$. Hence students could want to justify the high variability of the
solution $y$ at $x=2$ stating that the solution goes to infinity at $x=2$.

A third reflection could be to try to find an exact solution of the initial
value problem, which when solving the differential equation for instance as a
separable equation and using the initial condition gives us $y=\tan(x).$ The
student can note that the interval of definition of this solution is the
interval $\displaystyle]-\frac{\pi}{2},\frac{\pi}{2}[$ and that the solution
does not exist beyond $\pi/2\approx$ 1.57079. Thus, there is no point in
asking for $y(2)$ because $x=2$ is not within the interval of definition of
the solution. This reflection can lead to students realizing that the theorem
of existence and uniqueness is a local theorem, and therefore only guarantees
the existence of a solution in a neighborhood of the initial condition– in
this example in the neighborhood of $(0,0)$– and that the interval of
definition of the solution of a differential equation depends on the given
initial condition.

After the above reflections, a question that students should be confronted
with and will need to answer is: why when using the Euler method do we not
realize that it does not make sense to ask for $y(2)$? If students are
sufficiently persistent, they should reach to the conclusion this is the case
since the value of each step size $h$ is a rational number and hence starting
from the origin, no matter how many steps of length $h$ we make, the point on
the $x$-coordinate will never coincide with the irrational number $\pi/2$.
Hence, the tangent line taken at the last partition point before $\pi/2$ will
cross $\pi/2$ without noticing that it has already escaped the interval of
definition of the solution as can be seen in Fig. 4 for the step sizes $h=0.2$
and $h=0.4$.

Figure 4: The figure shows the approximate solution curve for the initial
value problem given in Eqn. (9). The solid line shows the curve using the
Euler Method with step size $h=0.2$ and the dashed line using the step size
$h=0.4$.

###  2.3 A problem that makes us reflect on how temperature of an object
decreases.

In this subsection we present a problem that requires to model– using
differential equations– how the temperature of an object changes with time, as
well as to use calculus and algebra skills to analyze the model and find a
solution. This problem leads to an interesting reflection regarding the
differential equation model used that we will discuss further in Sec. 3. Let
us consider the following incorrect problem:

We have a pool with water at a constant temperature of $T_{M}.$ We place a
thermometer into the pool, which initially shows $40^{\circ}C.$ After half a
minute it marks $36^{\circ}C$ and after one minute $30^{\circ}C$. What is the
temperature of the water in the pool?

Newton’s Law of Cooling states that the temperature variation of an object in
time $t$ is proportional to the difference between the temperature of the
object and the temperature of the medium [12]. In this case the temperature of
the medium is the temperature of the pool’s water, which is assumed to have
constant temperature $T_{M}$. In other words, if $T(t)$ represents the
temperature of the thermometer at time $t,$ then

$\frac{dT}{dt}=k(T-T_{M}).$ (11)

By solving the differential equation for instance as a separable equation we
obtain

$T(t)=T_{M}+(40-T_{M})e^{kt}.$ (12)

Replacing the data points provided in the problem, we obtain the following
system

$\left.\begin{array}[]{ccc}T_{M}+(40-T_{M})e^{k/2}&=&36\\\
T_{M}+(40-T_{M})e^{k}&=&30\end{array}\right\\}.$ (13)

If we solve for the exponential, we obtain

$e^{k/2}=\frac{36-T_{M}}{40-T_{M}}\;\;\mbox{and}\;\;e^{k}=\frac{30-T_{M}}{40-T_{M}}.$
(14)

This leads to the following equation which must be satisfied by $T_{M}$:

$(36-T_{M})^{2}=(30-T_{M})(40-T_{M}).$ (15)

After manipulation, the term $T_{M}^{2}$ is canceled, which results in
$T_{M}=48^{\circ}C.$ In other words, by placing a thermometer showing an
initial temperature of $40^{\circ}C$ in a pool at $48^{\circ}C$, the
thermometer lowers the water’s temperature, which is something that goes
against common sense. Hence one realizes that something went wrong and the
problem is an incorrect problem. This motivates a discussion and a deeper
understanding of the modeling aspects of the problem when we ask students:
Where is the mistake in the reasoning and/or solution method of this exercise?

##  3 Exposing students to an incorrect problem in Differential Equations

The problem that we presented in Subsec. 2.3 arose from an exam proposal for a
basic course of Differential Equations for engineering students and was
intended to be a relatively basic application of Newton’s Law of Cooling.
After revising the proposal, the professors lecturing the course realized that
this is an incorrect problem and how many interesting questions it provokes
that we could ask our students to think about. Hence we decided to design a
homework assignment to 168 undergraduate engineering students enrolled in the
course Differential Equations. In the assignment, we firstly asked for the
solution of a “correct” problem and secondly asked to analyze and discuss an
incorrect problem that led to contradictions. Students worked on it for two
weeks after which they had to turn in their assignment. Then, the assignments
were graded and students’ performance was analyzed. In the following Subsec.
3.1 we describe the assignment that was given to the students and in Subsec.
3.2 we discuss students’ performance in both, the “correct” and the incorrect
part of the problem that was assigned.

###  3.1 Students’ assignment: the “correct” problem and the incorrect problem

Students were confronted with two problems in their homework assignment. The
first one is a “correct” problem that can be solved directly since the problem
is well posed and does not provide any contradictions in the solution process.
Students in this course are generally used to these types of “correct”
problems. The problem is the following:

A thermometer showing $40^{\circ}$C is placed inside a pool filled with water
that has a constant temperature of $T_{M}$. An observer notices that the
thermometer shows $34^{\circ}$C after $1/2$ minute and $30^{\circ}$C after 1
minute.

  1. a)

Determine the differential equation which models the dynamic of the
temperature shown by the thermometer while it is in the pool.

  2. b)

Using the equation from the previous item and the given data points: what is
the temperature $T_{M}$ of the pool’s water?

Using the model and solution method of Subsec. 2.3, without further
difficulties one obtains that $T_{M}=22^{\circ}C$, which makes perfect sense
and solves the problem without further analysis.

The second problem that we assigned to students is the incorrect problem,
which we also presented in Subsec. 2.3, it states as follows:

A thermometer showing $40^{\circ}$C is placed inside a pool filled with water
that has a constant temperature of $T_{M}$. An observer notices that the
thermometer shows $36^{\circ}$C after $1/2$ minute and $30^{\circ}$C after 1
minute. Discuss and answer the following questions:

  1. a)

What is the temperature $T_{M}$, of the pool’s water?

  2. b)

Think about how exactly the temperature of the thermometer within the pool
changes with time. Discuss the contradiction that arises given the value of
$T_{M}$ obtained in a).

  3. c)

Plot the solution curve of the problem for the temperature $T_{M}$ obtained in
a).

  4. d)

What happens to the value of $T_{M}$ when changing $T(1/2)=36^{\circ}$C to
$T(1/2)=35^{\circ}$C ?

  5. e)

What can you say about the value of $k$ so that the differential equation
models correctly a real situation such as the one observed in the pool?

  6. f)

For what range $T(1/2)$ values will the differential equation model the real
situation correctly, given the presented data?

As discussed in Subsec. 2.3, we denote this problem incorrect problem since
when solving item $a)$ we obtain for the temperature of the water in the pool
$T_{M}=48^{\circ}C$, which is a result that does not make sense in the given
context, as we asked our students to discuss in item $b)$. Next, in Subsec.
3.2 we discuss the answers to items $b)-f)$ and how students performed in each
of the items.

###  3.2 Answers to the incorrect problem and students’ performance

We collected and analyzed the answers of the 168 undergraduate engineering
students across 7 course sections that were enrolled in the Differential
Equations course during one semester. Table 2 summarizes the results on
student performance, showing the percentage students that provided a correct
solution for each of the items of the assignment described in Subsec. 3.1.

Table 2: The table shows the percentage of correct answers of the students’ assignment for the “correct” problem and for the different items $a)-c)$, $d)$, $e)$ and $f)$ of the incorrect problem.  Type of | Correct | Incorrect
---|---|---
Problem |  |  |  |  |
|  $a)$, $b)$ | $a)-c)$ | $d)$ | $e)$ | $f)$
$\%$ of correct | 91$\%$ | 62$\%$ | 10$\%$ | 21$\%$ | 0$\%$
answers |  |  |  |  |

The results of analyzing students’ performance in solving the “correct”
problem show that most of the students were able to answer that problem. In
fact, 91$\%$ of the students answered this question correctly. This shows that
students did not have any difficulties in solving a problem that is well
posed: they new the model that needed to be used and had the necessary skills
to solve the differential equation and the algebraic skills to obtain the
value for $T_{M}$ using the data points given in the problem.

On the contrary, the results when analyzing the performance of the same
students in solving the incorrect problem show the lack of deep understanding
students have in what really matters when modeling a cooling process: Item
$a)$ of the incorrect problem could be solved directly using the typical
solution method as in Subsec. 2.3, obtaining $T_{M}=48$. As mentioned before,
this result for $T_{M}$ does not make sense in the given context. We asked
students in item $b)$ to discuss this contradiction. Observe that according to
the data points given in the problem, a cooling process should be happening
(the temperature shown by the thermometer decreases) and hence the temperature
of the pool’s water has to be at most $30^{\circ}$ to allow it. The
mathematical skills to solve items $a)-c)$ were similar to the ones for the
“correct” problem, the difference now was that the result did not make sense.
The number of students answering these items of the incorrect problem
correctly decreased to 62 $\%$ compared to the “correct” problem that was
answered by 91$\%$ of the students correctly (see Table 2).

When analyzing the answer for item $d)$ we observed that a small number of
students realized that since the dynamics of the temperature is modeled by a
decreasing exponential (see Equation (12)), which is a convex function, and
any image on the graph of this function between the points $(0,40)$ and
$(1,30)$ will be smaller than the corresponding image on the graph of the line
unifying these two points (see Fig. 5). Hence, since the point $(1/2,35)$ is
colinear with the points $(0,40)$ and $(1,30)$, the exponential curve can not
be fitted to contain all three points and hence the value of $T_{M}$ can not
be obtained in particular with the solution method from Subsec. 2.3. Students
had difficulty analyzing why the solution method failed, in fact, only 10 $\%$
of the students justified correctly why $T_{M}$ could not be found.

Figure 5: The figure shows an exponential solution curve for the equation
$\frac{dT}{dt}=k(T-T_{M})$, $k<0$, for a model with data points $T(0)=40$,
$T(1/2)<35$ and $T(1)=30$. It also depicts the secant line between the points
$(0,40)$ and $(1,30)$ and illustrates that those two points are colineal with
the point $(1/2,35)$.

In item $e)$ we expected students to think about under which conditions the
model actually describes the cooling process that we observe without obtaining
a contradiction, and specifically asked for the value of the constant $k$.
Observe that in a cooling process the temperature of the medium is less than
the temperature of the object at any moment in time, and hence in Eqn. (11) we
have that $T-T_{M}$ is positive and hence the constant $k$ must be negative in
order for the temperature to be decreasing ($dT/dt<0$) and for the model to
describe well the cooling process. This does not happen here, since if we
solve for $k$ from Eqn. (14) when replacing by $T_{M}=48^{\circ}C$, we obtain
$k=0.81$, and hence the solution method produces a contradiction with the
model. Only a small group of 21 $\%$ of the students was able to explain the
reason as to why the constant should be negative and that the solution method
produces a contradiction because it produces a positive $k$ value.

There were no students who analyzed correctly item $f)$. Observe that if we
replace the data points $T(1)=30$ and $T(1/2)=c$ in the solution process
described in Subsec. 2.3 and then solve for $T_{M}$ in Eqn. (15) we obtain
$T_{M}$ as a function of $T(1/2)=c$ in the following way

$T_{M}=\frac{1200-c^{2}}{70-2c}.$ (16)

Observe that $\displaystyle\lim_{c\rightarrow 35^{-}}T_{M}=-\infty$ and thence
the real situation gets modeled correctly only for $c$ values for which
$T_{M}$ attains temperature values larger then the absolute minimum
($-273.15^{\circ}C$). In other words, what cannot happen is that
$T_{M}<-273.15$ and using the expression in Eqn. (16) we obtain that this
happens if $T(1/2)=c$ is in the range $34.9594<c<35$. So, for the model to
describe a real situation, the data point $T(1/2)=c$ has to be in the range
$30<c\leq 34.9594$. It can be seen from Fig, 6 what happens, the figure
depicts solution curves $T(t)=T_{M}+(40-T_{M})e^{kt}$, $k<0$, for different
data points $T(1/2)$.

Figure 6: solid for $T_{M}=29.99$, $30.31$, point dashed $T_{M}=29$,
$T(1/2)=32.32$, dashed for $T_{M}=20$, $T(1/2)=34.14$, point for $T_{M}=-10$
$T(1/2)=34.72$, light grey the secant line

Observe that $\displaystyle\lim_{t\rightarrow\infty}T(t)=T_{M}$ and observe
from the figure that the closer $T(1/2)$ is to $35$, the lesser the limit
$T_{M}$ is. Hence we conclude that it is important to notice that the value of
the data point $T(1/2)$ has a threshold below $35^{\circ}C$– which here is
$34.9594^{\circ}C$– such that above this threshold the model does not make
sense. Questions like whether there are materials that do not behave
exponentially as temperature decreases did not appear in the student’s
analyses and would be interesting to explore further.

##  4 Discussion

In this work we showed examples of incorrect problems and displayed possible
ways these could be used to motivate critical mathematical thinking in
students. In particular, in Sec. 2 we described incorrect problems that
illustrated how important hypothesis are when modeling a problem or using a
technique and how important it is to understand the restrictions of methods
(numerical methods). We exposed a group of students to an incorrect problem in
Differential Equations which gave us important insight on how students react
to a problem, which gives rise to a contradiction in their mathematical
solution method or in their interpretation in a real life context.

We observed several important aspects when confronting our Differential
Equations students with an incorrect problem: First, in order use incorrect
problems as a teaching opportunity, we think it is very important to first
engage students with these types of incorrect problems in a way they reach a
level of natural curiosity regarding the questions posed by the problem. This
is very important, since nobody makes discoveries without first being
interested in the topic. Second, we think that incorrect problems that are
properly posed and guided can lead to a better analysis and understanding of
the hypotheses and methods used to solve a problem. For many students reaching
this analytic level is complicated due to the weaknesses and mathematical
qualifications prior to the course, which need to be consolidated in order to
tackle these topics. For instance, those who have not heard of convexity in
Subsec. 3.2 are clearly at a disadvantage for properly performing a stronger
analysis of the causes of the contradictions in the model. Third, we must be
self-critical with regards to having accustomed our students to respond rather
direct “correct” questions, which is why some fail when analyzing situations
that escape the established boundaries.

In history, people questioning their learning, rebelling against the
established models are the ones that have been capable of great scientific
revolutions, as shown by Einstein or Galileo, who questioned their academic
realities and discovered new representations of our universe (see [6, 3, 1, 4]
for great literature on the life of individuals that changed the vision of the
universe). Thus, we believe it to be necessary– after the initial knowledge of
a topic is learned– to present students also with questions that require “out
of the box” thinking, additional analysis and even contradictions, allowing
notable students in our course to discover new viewpoints and transforming
themselves, viewing their knowledge from a different perspective. It is
important to develop reflections on mathematical topics that infuse life to
our science and show it as novel, defying and attractive to all who we invite
to enter the wonderful world of discovery and mathematics.

## Acknowledgements

We thank our colleagues from the Mathematics Department of the Universidad
Adolfo Ibáñes that taught Differential Equations that semester for giving us
important feedback on this work. We also thank Clemens Vogt Geisse for his
valuable contribution in translating this manuscript from its Spanish version.

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