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Prime Research on Education (PRE)
ISSN: 2251-1253. Vol. 4(4), pp. 702-707, June 30th, 2014
www.primejournal.org/PRE
© Prime Journals
Communication
Theorem Proving: An Algorithmic Pedagogical Strategy
Marion G. Ben-Jacob, Ph.D.
Mercy College, Department of Mathematics and Computer Sciences. 555 Broadway, Dobbs Ferry, New York 10522.
E-mail: [email protected]. Tel.: 914-674-7524; Fax: 914-674-7528
Accepted 17th June, 2014
Many students of mathematics find theorem proving difficult. To facilitate the art and science of mathematical
theory, we present an algorithmic approach to proving theorems that is easily integrated into the use of
common technological tools, for example, Blackboard and Iclickers. Examples from different areas of
mathematics are demonstrated in detail.
Key words: mathematics, theorem proving; pedagogical strategy; technology
INTRODUCTION
Advances in recent technologies in communication and
computing affect Pedagogy and learning. The influence
of technology on the learning environment of this
millennium cannot be over emphasized. There are
courses of study that are completely online, hybrid,
blended and even those onsite rely to some degree on
technology, be it via research assignments or online
tutorials. Online learning platforms, for example,
Blackboard, allow for asynchronous learning supported
via the posting of lecture notes, a discussion area, and a
synchronous chat room. One is able to post
announcements, interact via email and make use of tools
for the facilitation of grading and tracking of student posts
(Blackboard 2009). Clickers are another popular
technology tool that has been integrated into the
pedagogical environment. Using clickers students can
select their answers to multiple choice questions. These
answers can be identified with each student individually
or just collected as group responses. Clickers can be
used for surveys, multiple choice tests, and recently, for
formative assessment of student learning (Iclicker 2008).
With the integration of technology into pedagogy we have
to insure that the same material is covered in all types of
learning environments.
There is a strong need for mathematics majors to be
facile with theorem proving. It is an integral part of the
discipline that reinforces the beauty of the field in addition
to the inherent logic and critical thinking. Many
undergraduate major programs today focus on the
applied aspects of the subject matter and students find
proving of the theory rather difficult. They tend to gloss
over the material never fully understanding it, let alone
being able to produce a proof on their own. We contend
that with the proper material students can be become
adept
at
theorem
proving,
enhancing
their
comprehension of mathematics.
This paper presents an algorithmic approach to the
pedagogy. We will introduce the concepts of hypotheses
and conclusion, necessary and sufficient conditions and
how to identify the aforementioned when attempting to
prove a theorem. We will use examples from different
mathematics fields and present the material in an
interactive fashion. As a matter of fact, we will develop
slides to reinforce the interactive aspect of our approach
and so that the material can easily be incorporated into
the use of clickers or Blackboard if the course is online,
hybrid, or integrates the use of Blackboard in any way.
Our presentation here will be in terms of slides.
Pedagogy
Slide 1
What are hypotheses?
Hypotheses are the given information. They are the
assumptions that are assumed true. They are the starting
point for proving the conclusion of the theorem.
When given a hypothesis, it is common to ask, what the
hypothesis means, usually in terms of its mathematical
definition.
Slide 2
What is the conclusion?
The conclusion is the logical implication of the given
hypotheses. It is the truth or goal we want to establish by
means of the accepted assumptions, that is, the
703 Prim. Res. Edu.
hypotheses. Before beginning an actual proof, it
oftentimes helps to consider the relationship between the
given information and the conclusion in terms of
definitions.
Slide 3
Consider the following theorem from first year calculus:
Theorem: If f is differentiable at a point a in its domain,
then f is continuous at the point a.
Slide 4
In this case, it is not hard to identify the hypothesis as
there is only one given condition, namely that f is
differentiable at a point a in its domain. Notice that this
theorem says nothing about the points that are outside
the function's domain. We point this out because
mathematics is a very precise discipline.
Slide 5
Which form of the hypothesis is the best one to use for
the proof?
f'(a) exists
f'(a) exists for a point a belonging to the domain of f
f'(a) = lim f(a+h) -f(a)
h0 h
d) f'(a) = lim f(a+h)- f(a) for a point a belonging to the
domain of f
h0 h
Slide 6
The correct answer is d because it is the most specific
(as opposed to c) and it involves the definition of f’(a) (as
opposed to a and b).
Slide 7
Let's examine the conclusion. Based on our analysis of
the hypothesis which penultimate form of the conclusion
do you think is the best one to work with? Keep in mind
there should be some relationship between the form of
the hypothesis and the form of the conclusion.
a) f is continuous
b) lim f(a+h) - f(a) = 0
h0
c) f is continuous at a point a in its domain
d) lim f(a+h) - f(a) = 0 for a point in its domain
h0
Slide 8
Once again the correct response is d because it is
specific and involves the definition.
Slide 9
Let us continue with the proof itself. We want to take the
given information and logically arrive at the conclusion.
Given: f'(a) = lim f(a+h) - f(a)
h0 h
We want to arrive at an equation involving just the
numerator of the expression.
Slide 10
What should our next step be?
a) Multiply both sides of the equation by h
b) Divide both sides of the equation by h
c) Add h to both sides of the equation
d) Subtract h from both sides of the equation
Slide 11
The correct answer is a. If we multiply both sides of the
equation by h, we arrive at
lim f(a+h) -f(a) = lim f(a+h) -f(a) . h = f'(a).0 =0
h0 h0 h
This is the conclusion we were looking for.
Slide 12
We need to be able to justify every step we take in the
proof.
Why are we able to multiply both sides of the equation by
h?
- h is it just a letter so it does not make a difference
- h is like 0 so it is ok
- h is going to 0 but not zero and so as in algebra if we
multiply both sides of an equation by the same non-zero
quantity, - the results are still equal.
Slide 13
The correct answer is c. We note that h is not 0, merely
approaching it so we are allowed to multiply both sides of
the equations by h.
Slide 14
Another mathematical fact we used in this proof is that
the limit of products is the product of limits. This was
used to determine that
lim f(a+h) -f(a) .h = f'(a). 0 = 0
h0 h
Slide 15
We now have determined that lim f(a+h)- f(a)=0:
h0
a) The proof is concluded.
b) We need to add f(a) to both sides of the equation
without justification.
c) We need to add f(a) to both sides of the equation and
state we can do this because the limit of a sum = the sum
of the limits.
Slide 16
The correct answer is c because we need to make things
explicit in a proof with justification.
Slide 17
We want out final statement to match the definition of
continuity, that is, lim f(x) = f(a).
xa
Marion 704
In order to do this we need to make a substitution. Which
of the following is the correct one?
a) Let x be a+h then h= x-a and our result can be written
as lim f(x) = f(a)
x-a0
b) Let x be a-h then h=x-a and our result can be written
as lim f(x) = f(a)
x-a0
c) Let x be a-h then h= x-a and our result can be written
as lim f(x) =f(a)
x-a0
Example 1
Slide 1
Let's try and prove another theorem from calculus,
namely that to compute the derivative of the product of
two functions, we need to hold the first one constant and
multiply by the derivative of the second then, add the
second function multiplied by the derivative of the first.
More explicitly
Given f(x) and g(x) defined and differentiable on an
interval I,
[f(x).g(x)]' = f(x).g'(x) + g(x).f(x)'
Slide 18
The correct answer is a. it is the only substitution that will
lead to the necessary conclusion.
Slide 2
Which form of the hypotheses should we use?
f' and g' exist
f'(x) and g'(x) exist for all x in the interval I
f'(x) = lim f(x+h)-f(x) and
h0 h
g'(x) = lim g(x+h)- g(x) exist for all x in the interval I
h0 h
Slide 19
Take a moment and write the entire proof beginning with
the statement and justify each line along the way.
Slide 20
The entire proof should look something like the following:
Theorem: If f is a differentiable function at a point a on an
interval, I, in its domain, then f is continuous at the point.
Given that f is differentiable. This means that
f'(a) = lim f(a+h) -f(a) exists.
h0 h
h is not zero. It is going to zero. We can multiply both
sides of the equation by h and
take the limit of both sides as h0, to obtain lim .
f'(a). h = lim f(a+h) - f(a) .h
h0 h0 h
If you take the same limit on both sides of an equation,
the equation remains true.
Now, the limit of the product is the product of the limits,
lim f'(a) = f'(a) and the h's on the h0 right side cancel
each other. Remember this is ok because h is not 0. It is
approaching 0.
We get 0 = lim f(a+h) - f(a), or adding f(a) to both sides
ho
we get f(a) = lim f(a+h). Now
h0
make the substitution h= x-a so h0 => x-a->0 or x a
to obtain f(a) = lim f(x).
xa
QED
Slide 19
At the end of many proofs we see the letters QED. This
stands for Quod Erat Demonstratum, Latin for "that which
was to be demonstrated."
Also note that we have not proven anything about the
converse of the theorem, that is, if a function is
continuous, it is differentiable. As a matter of fact this is
not true in general. To prove something false, one needs
to provide only one counterexample.
Slide 3
The correct answer is c. It goes back to the definition and
that is almost always necessary for a mathematics proof.
Slide 4
Which form of the conclusion will we arrive at, at the
penultimate step of the proof?
a) the derivative of the product of two functions exist
b) [f(x).g(x)]' = f(x).g(x)' + g(x).f(x)'
c) [f(x).g(x)]' = f(x). lim g(x+h) - g(x) + g(x). lim f(x+h) - f(x
h0 h
h0 h
Slide 5
It is c, the option with the most detail. We will use the
form of the hypotheses with the most detail and so we will
deduce a comparable form. We note that once we arrive
at choice c, the stated conclusion follows as c is
equivalent to
[f(x).g(x)]' = f(x).g(x)' + g(x).f(x)'
Slide 6
Is the following the correct definition of [f(x).g(x)]’ ?
[f(x).g(x)]' = lim f(x+h).g(x+h) - f(x).g(x)
h0 h
a) True
b) False
Slide 7
It is true. Yet when we look at slide 4, we see that we
need to manipulate the right hand side of the equation to
match the right hand side on slide 6. We are going to add
and subtract the same quantity on the right hand side of
option c, slide 4. (If you are concerned about how we
know to do this, it is experience and now that you have
seen this technique once, you will consider using it when
you prove another theorem when necessary)
705 Prim. Res. Edu.
Slide 8
Our "working equation" becomes
[f(x).g(x)]' = lim f(x+h).g(x+h) - f(x).g(x) –
h0 h
lim f(x).g(x+h) + lim f(x).g(x+h)
h0 h
h0 h
Slide 9
Rearranging our terms we get
[f(x).g(x)]' = lim f(x).g(x+h) - lim f(x).g(x) +
h0 h
h0 h
lim f(x+h).g(x+h) - lim f(x).g(x+h)
h0 h
h0 h
= lim f(x).g(x+h) - f(x).g(x) + lim f(x+h).g(x+h) - f(x).g(x+h)
h0 h
h0 h
Slide 10
In order to justify the rearrangement of our terms which
reason is the one we must use?
a) The limit of the sum is the sum of the limits
b) The limit of the product is the product of the limit
c) Both of the above
d) Neither a) nor b)
Slide 11
‘a’ is the correct answer because all we did was add and
subtract the terms involving limits.
Slide 12
If we factor out f(x) from our first two terms and g(x+h)
from the latter two terms we get
[f(x).g(x)]' = f(x). lim g(x+h)-g(x) + lim g(x+h).
h0 h
h0 h
lim f(x+h) -f(x)
h0 h
Slide 13
What is the justification for the last result?
a) The limit of the sum is the sum of the limits
b) The limit of the product is the product of the limits
c) Both of the above
d) Neither of the above
Slide 14
The correct answer is b as we are dealing with products.
We note that lim f(x) = f(x)
h0
as there is no h involved in the argument of this term.
Slide 15
We have arrived at our penultimate step, that is,
[f(x).g(x)]' = f(x). lim g(x+h) -g(x) + g(x). lim f(x+h) -f(x)
h0 h
h0 h
using the fact that
lim g(x+h) = g(x)
h0
Slide 16
We rewrite the results in the form the theorem was
originally stated (and for the "sticklers" add QED at the
end).
[f(x).g(x)]' = f(x).g(x)' + g(x).f(x)'
Slide 17
You try to rewrite the entire proof now, justifying each
step or logical inference that you make.
Slide 18
Theorem: Given that f(x) and g(x) are defined and
differentiable on an interval I, then the derivative of the
product exists and
[f(x).g(x)]' = f(x).g(x)' + g(x).f(x)'
Proof: We are given that f(x)' = lim f(x+h) - f(x) and g'(x)=
lim g(x+h)-g(x) exist
h0 h
h 0 h
We want to show that [f(x).g(x)]' = f(x).g'(x) + g(x).f(x)'
It will suffice to show that [f(x).g(x)]' = f(x). lim g(x+h)-g(x)
+ g(x).lim f(x+h) -f(x)
h0 h
h0 h
by the definition of the derivative.
Consider the definition of the derivative of the product
[f(x).g(x)]' = lim f(x+h).g(x+h)- f(x).g(x)
h0 h
Add and subtract lim f(x).g(x+h) to the right side of the
equation and rearrange the
h0 h
order of the terms to get
[f(x).g(x)]' = lim f(x) .g(x+h) -g(x) + lim g(x+h). f(x+h)-f(x)
h0 h h0 h
We can do this because adding and subtracting the same
value to one side of an
equation is adding zero and does not change the value.
Also, addition is commutative
and so we can change the order of the terms.
Now, using the facts that the limit of the product of terms
is the product of the limits and
that lim f(x) = f(x) and that the lim g(x+h) = g(x) we get
h0
h0
[f(x).g(x)]' = f(x).lim g(x+h)-g(x) + g(x). lim f(x+h) -f(x)
h0 h
h0 h
QED
Example 2
Slide 1
We take this example from abstract algebra to 'change
the flavor' of the mathematics a bit. We note, the strategy
is the same. List the hypotheses in the form of their
definitions and logically arrive or deduce the conclusion.
We remark that we will need to use all the hypotheses or
most of the hypotheses if they are part of a definition;
otherwise the assumptions in the proof would have been
Marion 706
relaxed (reduced).
Slide 2
Theorem: If G is a group with binary operation *, then the
left and right cancellation laws hold in G, that is, a*b = a*c
implies b= b, and b*a = c*a implies b=c for all a,b,c in G.
Slide 3
The characteristics of a group will be our hypotheses.
Consider the following characteristics:
A group (G, *) is a non-empty set, G, with a binary
operation *such that the following hold:
i) Closure: If a and b belong to G then a*b is an element
in G as well.
ii) Associativity: If a, b, and c belong to G, then (a*b)*c =
a*(b*c).
iii) Identity: There exists an element e in G such that for
every element a in G we have a*e = e*a =a.
iv) Inverses: For every a in G, there exists a-1 such that
a*a’ = a’*a = e.
Slide 4
Which of the aforementioned four conditions are part of
the definition of a group (G, *)?
a) i, ii, i
b) ii, iii, iv
c) i, ii, iii, iv
d) i, ii ,iii, iv and one more that is not listed
Slide 5
The correct answer is c. If it does not look familiar to you,
please refer to your textbook to confirm.
Slide 6
If a and b belong to G, a*b belongs to G. Why is this
true?
a) Closure property
b) Associativity property
c) Identity property
d) Inverse property
Slide 7
The correct answer is a. This is exactly what closure
means.
Slide 8
Suppose that a*b = a*c. Then there exists a’, the inverse
of a such that;
a’*(a*b) = a’*(a*c).
This is true by:
a) Closure property
b) Associativity property
c) Identity property
d) Inverse property
Slide 9
The answer is d.
Slide 10
The last step leads us to (a’*a)*b = (a’*a)*c because of
the:
a) Closure property
b) Associativity property
c) Identity property
d) Inverse property
Slide 11
The correct answer is c.
Slide 12
But, (a’*a) = e because:
a) Closure property
b) Associativity property
c) Identity property
d) Inverse property
Slide 13
The correct answer is c.
Slide 14
We conclude b = c.
QED
Slide 15
1) Starting with b*a = c*a one can deduce that b= c upon
multiplication on the right by a’ using the same axioms.
2) Note that we used all the hypotheses in the proof.
Slide 16
Try to reproduce the entire proof now, with justification for
each step.
Slide 17
Theorem: If G is a group with binary operation*, then the
left (and right) cancellation laws hold in G, that is, a*b =
a*c (and b*a = c*a) implies b= c for all a, b, c in G.
Proof: For any two elements a,b in G, the product of the
elements is in G as well because of closure. Thus, we
have a*b and a*c are in G. We are given that a*b=a*c.
Every element in G has an inverse; in particular, let a’ be
the inverse of a.
We get a’*(a*b) = a’*(a*c) by closure. Using the
associative law we get
(a’*a)*b = (a’*a)*c. Using the identity property we find that
a’*a= e and so we get
e*b= e*c which implies b=c, using the identity property
again.
QED
CONCLUSION
Our contention is that by breaking down a mathematical
proof into small pieces or bits of information, and
reinforcing the step-by-step approach, we will help
707 Prim. Res. Edu.
students learn to prove theorems. Our examples in this
paper are just that, examples. The first two are from first
semester calculus, a course taken by every math major
and other students who want to further their study of
mathematics. The third example comes from a
mathematics elective, algebraic structures or abstract
algebra, to illustrate that our approach to the pedagogy
holds for higher level courses as well.
REFERENCES
Blackboard. (2009). Blackboard Inc. (Accessed 10
September 2013). <http://www.blackboard.com/>
Iclicker. (2008).iclicker. (Accessed 11 September 2013).
<http://www.iclicker.com/dnn/>