Mathematical Problems

& Inquiry in Mathematics

AME Tenth Anniversary Meeting May 29 2004

A/P Peter PangDepartment of Mathematics and

University Scholars Programme, NUS

Four Important Concepts

SpecificityGenerality

SpecializationGeneralization

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 A is drinking beer Customer B is drinking coke Customer C is 25 years old Customer 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 F 3 73

C 25 16B

D

21 B

D 73 D

Comparison with other disciplines

Literature Science

Social science

A less trivial example x2 + y2 = z2 has an infinite number of positive

integer solutions x = u2 – v2

y = 2uv z = u2 + v2

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

(xq)p + (yq)p = (zq)p

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 z = e/f

(a/b)p + (c/d)p = (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 + t2

Recalling that y = 2uv, we have

y2 = 2(2st)(s2 + t2) and hence

(y/2)2 = st(s2 + t2)

Note that s, t and s2 + t2 are relatively prime.

As their product is a perfect square, so must eachindividual factor (s, t and s2 + t2).

This means that

s = x12 t = y1

2 s2 + t2 = w12

and hence

x14 + y1

4 = w12

Finally, note that

x1 < x y1 < yw1 < w

This leads to infinite descent, which is not possible as we are dealing with positive integers.

Conclusions