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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.

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## 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 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

3

D

3

F

3

7

D

7

D

B

21

C

25

16

B

Unreasonable Effectiveness

of Mathematics

### 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

• 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 = w2or

(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 – t2u = 2stv = 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 each

individual factor (s, t and s2 + t2).

• This means that

s = x12t = y12s2 + t2 = w12

and hence

x14+ y14= w12

• Finally, note that

x1< xy1 < yw1< w

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