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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

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Mathematical problems inquiry in mathematics l.jpg

Mathematical Problems & Inquiry in Mathematics

AME Tenth Anniversary Meeting

May 29 2004

A/P Peter Pang

Department of Mathematics and

University Scholars Programme, NUS


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Four Important Concepts

Specificity

Generality

Specialization

Generalization


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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?


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  • 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?


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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


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  • 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.


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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)?


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  • 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)


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Tension between Specificity and Generality

  • Generality is often accompanied by loss of context (i.e., abstractness)


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D

3

D

3

F

3

7

D

7

D

B

21

C

25

16

B


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Unreasonable Effectiveness

of Mathematics


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Comparison with other disciplines

  • Literature

  • Science

  • Social science


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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?


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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.


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  • 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


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  • 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.


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  • 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


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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.


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  • 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


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  • 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).


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  • 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.


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Conclusions


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