Marquette University

Department of Mathematics, Statistics and Computer Science

Wim Ruitenburg's Fall 2010 MATH 1300-101

Last updated: September 2010
Comments and suggestions: Email   wimr@mscs.mu.edu

---

Why Mathematics? Living with Uncertainty

A large part of our conscious lives revolves around believing things. We may have to.

• Some things we believe because we like to believe them. This is comforting, but we run a fair chance of being mistaken.
• Some things we believe because we have it on good authority, to the best of our understanding. Our ancestors usually believed that the earth was flat. Now most people believe that the earth is round.
• Some things we believe by the authority of experiment. When we release a pen or a book or a pebble, it falls to the ground. Try again and the object falls again. And again. And again. Many people believe that all objects, when released, fall to the ground. This essentially is a belief based on scientific experiment.

Are there things that we can know with certainty? Maybe not, but are there things that we can establish as true with maximum assurance and, if so, how? This is where mathematics comes into play.

Why proofs? Mathematics without proof

• Mathematical science is commonly seen as providing the highest level of certainty. However, mathematical science without proof is science, not mathematics.
• To illustrate the weakness of mathematics without proof, we introduce a statement for which we use common sense scientific reasoning by which we may come to believe it true.
  Recall that a whole number bigger than 1 is called prime if we can not write it as a product of smaller whole numbers. So 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and so on, are primes. But 4 is not prime because 4 = 2*2; and 6 is not because 6 = 2*3; and 8 is not because 8 = 2*2*2; and 9 is not because 9 = 3*3; and so on.
• Consider the function
  • g(n) = 2 *n + 3
  For all whole numbers n the function f returns another whole number. For example
  • g(0) = 3
  • g(1) = 5
  • g(2) = 7
  These five numbers are all prime. Is it possible that g(n) is prime, for all values of n? It would be careless to believe so based on just a few values. It is common sense to try a few more cases:
  • g(3) = 9
  Aha. This one case suffices to show that g(n) is not always prime, because g(3) = 9 = 3 * 3.
• Consider the function
  • f(n) = n*n+n+41
  For all whole numbers n the function f returns another whole number. For example
  • f(0) = 0+0+41 = 41
  • f(1) = 1+1+41 = 43
  • f(2) = 4+2+41 = 47
  • f(3) = 9+3+41 = 53
  • f(4) = 16+4+41 = 61
  Remarkably, these five numbers are all prime (you can check by trying all possible ways to write them as products). Is it possible that f(n) is prime, for all values of n? It would be careless to believe so based on just a few values. It is common sense to try a few more cases:
  • f(5) = 25+5+41 = 71
  • f(6) = 36+6+41 = 83
  • f(7) = 49+7+41 = 97
  • f(8) = 64+8+41 = 113
  • f(9) = 81+9+41 = 131
  These five are also prime! Now many of us will consider it likely that f(n) is prime, for all n. This is a sensible guess. Just to be on the safe side, we check a few more cases. We leave it to the reader to perform the tedious check that f(10), f(11), f(12), f(13), f(14), f(15), f(16), f(17), f(18), and f(19) are indeed prime. This is enough evidence for sensible people to believe that f(n) is prime for all n.
  • Sensible belief: f(n) is prime, for all n.
  Even if we believe something like this, it is still of value to continue checking. In fact, a further tedious check shows that f(20), f(21), f(22), f(23), f(24), f(25), f(26), f(27), f(28), f(29), f(30), f(31), f(32), f(33), f(34), f(35), f(36), f(37), f(38), and f(39) are all prime. It is common sense for many of us to claim:
  • Strong belief: f(n) is prime, for all n.
  But f(40) = 40^2 + 40 + 41 = 40 * (40 + 1) + 41 = 40 * 41 + 1 * 41 = 41^2 is not prime. As a mathematically thinking student in class observed: f(41) is not prime for the obvious reason that it is divisible by 41. Remarkably, f(n) is usually not prime.
• Conclusion: Extensive experimentation, however persuasive, is not the same as knowing something with maximum certainty.

Mathematics with proof

• Let us try another case. We easily check the following sums:
  • 1 = 1 = 1^2, that is, equals 1 squared
  • 1 + 3 = 4 = 2^2, that is, equals 2 squared
  • 1 + 3 + 5 = 9 = 3^2, that is, equals 3 squared
  • 1 + 3 + 5 + 7 = 16 = 4^2
  • 1 + 3 + 5 + 7 + 9 = 25 = 5^2
  • 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6^2
  Now it is sensible to believe that the sum of the first n odd numbers equals n^2, that is, equals n squared. But we may be wrong. In such cases a mathematician may make a conjecture.
  • Conjecture: For all n, the sum of the first n odd numbers equals n^2.
  When using formulas, one can also write
  • Conjecture: For all n, the sum 1 + 3 + 5 + ... + (2n-5) + (2n-3) + (2n-1) equals n^2.
  The notation   + ... +   means that if n is big enough, then we are supposed to also include the numbers in between, like   7 + 9 + 11 + 13 +   and so on. There happen to be multiple ways to prove the conjecture.
  • Proof 1 Proof by induction. The idea is that if (1) you know how to step on the bottom rung of a ladder, and (2) you know how to step from one rung to the next one up, then you can climb to any level on the ladder that you may wish.
    • The conjecture holds when n equals 1. We tested that case above.
    • Suppose that the conjecture has been proved for a certain value, say p. So we suppose that 1 + 3 + 5 + ... + (2p-3) + (2p-1) equals p^2. Then 1 + 3 + 5 + ... + (2p-3) + (2p-1) + (2p+1) equals p^2 + 2p + 1, which simplifies to (p+1)(p+1) or, in other notation, (p+1)^2. Thus the conjecture also holds for p+1. We call this the induction step.
    • So, by mathematical induction, the conjecture holds for all n, that is, the conjecture is proven true.
  We don't have to give another proof of the conjecture. We are done. Still, we give some more proofs just to show that there may be several ways to prove one conjecture. So here is another proof.
  • Proof 2 Proof by algebraic manipulation. Write S(n) as short for 1 + 3 + 5 + ... + (2n-3) + (2n-1). So also:
    • S(n) = 1 + 3 + 5 + ... + (2n-3) + (2n-1)
    • S(n) = (2n-1) + (2n-3) + ... + 5 + 3 + 1
    Note that 1 + (2n-1) equal 2n. And 3 + (2n-3) equals 2n. And 5 + (2n-5) equals 2n. And so on until the last pairs: (2n-3) + 3 = 2n and (2n-1) + 1 = 2n. So when we add the two lines above, we get
    • S(n) + S(n) = 2n + 2n + 2n + ... + 2n + 2n, where there are n occurrences of 2n on the right hand side. So 2S(n) = 2(n^2). Thus S(n) = n^2.
  • Proof 3 Finally, here is a picture proof of the same conjecture. We can build squares in layers, like an onion. In the picture, start with a smallest square formed by the single box This gives a 1 by 1 small square. To extend this to a 2 by 2 square, we must add a layer of 3 boxes Next, the 3 by 3 square. The added layer consists of 5 boxes And so on. So, in the picture below, the 5 by 5 square has 5^2 = 25 fields, built as a sum of layers with 1, 3, 5, 7, and 9 fields:

---