Transcript Research in Mathematics - National Institute of Education

Mathematical Problems
& Inquiry in Mathematics
AME Tenth Anniversary Meeting
May 29 2004
A/P Peter Pang
Department of Mathematics and
University Scholars Programme, NUS
Four Important Concepts
Specificity
Generality
Specialization
Generalization
D
F
3
7

Each card has a number on one side and a letter
on the other.

Claim: “If a card has ‘D’ on one side, then it has
a ‘3’ on the other.”

Which cards do you need to turn over to find out
if this is true?

You are a bouncer in a bar. You must make sure
that there are no under-age (below 21) drinkers.

There are 4 customers (A—D) in the bar. You
know what 2 of them are drinking and you know
the age of the other 2.





Customer
Customer
Customer
Customer
A is drinking beer
B is drinking coke
C is 25 years old
D is 16 years old
Which of the 4 customers do you need to check to
do your job?
What is a maths problem?

One (possibly the most?) important aspect of
inquiry in mathematics is to find a problem

One important quality of a maths problem has to
do with the notions of specificity and generality

Consider the following problems:

If x = 2 and y = 4, show that x + y = 6

If x is even and y is even, show that x + y is
even






Define “even”
A number is even is it is two times a natural
number; equivalently, an even number is divisible
by 2, i.e., when divided by 2, the remainder is 0.
x is even if x = 2n for some natural number n
Let x = 2n and y = 2m where n and m are
natural numbers
x + y = 2n + 2m = 2(n+m)
As n and m are natural numbers, so is n + m.
This shows that x + y is even.
Can this be generalized?





If x is a multiple of 3 and y is a multiple of 3,
then so is x + y.
If x and y are multiples of p, then so is x + y.
If x is odd and y is odd, is x + y odd?
The number x is odd if, when divided by 2, the
remainder is 1. We denote this by
x = 1 (mod 2).
If x = 1 (mod 2) and y = 1 (mod 2), is x + y = 1
(mod 2)?



If x = 1 (mod 2) and y = 1 (mod 2), then
x + y = 1 + 1 (mod 2)
= 2 (mod 2)
= 0 (mod 2)
This means that x + y is even.
If x = p (mod r) and y = q (mod r), where p, q, r
are natural numbers, then
x + y = p + q (mod r)
Tension between
Specificity and Generality

Generality is often accompanied by loss of
context (i.e., abstractness)
D
3
F
3
D
7
21
B
C
25
16
B
Comparison with other disciplines
Literature
 Science
Social science


A less trivial example

x2 + y2 = z2 has an infinite number of positive
integer solutions
2
2
 x = u – v
 y = 2uv
2
2
 z = u + v

This result is believed to be due to Pythagoras

What about powers higher than 2?
Fermat’s Last Theorem

xn + yn = zn has no integer solution when n > 2

Observation #1 (specialize to prime powers):

It suffices to look at powers n that are prime
Suppose there is a solution (x, y, z) for
n = p x q, where p is prime. Then

xpq + ypq = zpq
q p
q p
q p
(x ) + (y ) = (z )

Thus, (xq, yq, zq) would be an integer solution for
the power p.

Very important note:

If you have a solution for the power pq, then you
have a solution for the power p (and q)

However, if you have a solution for the power p, it
does not mean that you have a solution for the
power pq

(xq)p + (yq)p = (zq)p

Observation #2 (specialize to “primitive
solutions”)

It suffices to look at solutions that are pairwise
relatively prime, i.e., between any two of the
three numbers x, y and z there are no common
factors other than 1

For example, suppose x and y have a common
factor of 2. Then, as
xn + yn = zn,
z will also have a factor of 2. Thus I can divide
the equation through by the common factor 2.

Observation #3 (generalize to rational solutions)

Instead of asking for integer solutions, the
problem can be equivalently stated by asking for
rational solutions
x = a/b
y = c/d
p
p
(a/b) + (c/d) = (e/f)

z = e/f
p
Put the three fractions under a common
denominator g
(a’/g)p + (c’/g)p = (e’/g)p
a’p + c’p = e’p
A special case : n = 4

To show that x4 + y4 = z4 has no integer
solution

Strategy: Proof by contradiction

Suppose there were a solution. Will show that
this supposition will lead to a logical
contradiction, i.e., something will go wrong.

As a result, the supposition cannot be correct,
and hence its opposite is correct.

The solution for n = 4 uses the very interesting
idea of “infinite descent”


Suppose there were a solution (x, y, z),
i.e., x4 + y4 = z4
Write z2 = w. Then x4 + y4 = w2 or
(x2)2 + (y2)2 = w2

By Pythagoras
x2 = u2 – v2, y2 = 2uv, w = u2 + v2

From this, we get
x2 + u2 = v2

Again by Pythagoras,
x = s2 – t2

u = 2st
v = s2 + t 2
Recalling that y = 2uv, we have
y2 = 2(2st)(s2 + t2)

and hence

Note that s, t and s2 + t2 are relatively prime.

(y/2)2 = st(s2 + t2)
As their product is a perfect square, so must each
individual factor (s, t and s2 + t2).

This means that
s = x12
t = y12
s2 + t2 = w12
and hence
x14 + y14 = w12

Finally, note that
x1 < x

y1 < y
w1 < w
This leads to infinite descent, which is not
possible as we are dealing with positive integers.
Conclusions