Marquette University

Department of Mathematics, Statistics and Computer Science

Wim Ruitenburg's Spring 2008 MATH025.1001

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Proofs in mathematics?

A large part of our conscious lives revolves around believing things. Some things we believe because we like to believe them. That may be comforting, but we ought to know that there is a fair chance that we are mistaken. Other things we believe because we have it on good authority, to the best of our understanding. For example, we believe that the earth is round. Are there statements that we can establish as true with maximum assurance and, if so, how? This is where Mathematics comes into play.

Mathematics without proof

To illustrate the extreme rigor of mathematics, we introduce a statement which by use of common sense reasoning we may come to believe true. Later we raise the level of rigor towards that of proving the truth of statements.

• 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
  • 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, way ahead of the crowd: 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:

Mathematics is hard

• Here is an example of something that was proved, but that is very difficult. Simple experimenting shows that we can write all small numbers as sums of four squares, sometimes in more than one way. For example,
  • 0 = 0+0+0+0
  • 1 = 1+0+0+0
  • 2 = 1+1+0+0
  • 3 = 1+1+1+0
  • 4 = 1+1+1+1 = 4+0+0+0
  • 5 = 4+1+0+0
  • 6 = 4+1+1+0
  • 7 = 4+1+1+1
  • 8 = 4+4+0+0
  • 9 = 9+0+0+0 = 4+4+1+0
  • blablabla
  • 107 = 81+25+1+0 = 81+16+9+1 = 64+25+9+9
  • blablabla
  By checking many cases we may believe, but will not know, that all numbers are sums of four squares. In the eighteenth century the Frenchman Lagrange proved that all numbers are sums of four squares. Shortly thereafter, the Swiss Euler gave a nice proof which is the better known one. The proof is by mathematical induction. The proofs essentially show for all positive numbers n that if one already knows that all numbers less than n are sums of four squares, then n itself is also a sum of four squares. We already saw that 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are sums of four squares. So by their proof, 10 is also a sum of four squares. Now we know that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are sums of four squares. So, again applying their proof, 11 is also a sum of four squares. Now we know that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11 are sums of four squares. So, again applying their proof, 12 is also a sum of four squares. And so on. Thus, by mathematical induction, all numbers are sums of four squares.

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Last updated: January 2008
Comments & suggestions: wimr@mscs.mu.edu