mathematical problems & inquiry in mathematics ame tenth anniversary meeting may 29 2004 a/p...
TRANSCRIPT
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