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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**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 • 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 = 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 – 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 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.

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