### History of Mathematics Assignment 3

From chapter 2, exercises 8, 9, 10, 11, and 13.

#### Exercise 8.

Show that the nth triangular number is represented algebraically as Tn = n(n+1)/2, and therefore that an oblong number is double a triangular number.

There are a number of ways to do this. The Pythagoreans probably would have done it in the opposite order. That is, they would show that an oblong number n(n+1) could be seen as a pair of equal triangular numbers. Pictorially, the oblong number 30, 5 times 6, is a pair of triangular numbers 15 + 15.

```     o\o o o o o
\
o o\o o o o
\
o o o\o o o
\
o o o o\o o
\
o o o o o\o
```
A modern method would use mathematical induction (occasionally used since the 10th century, but not stated as a general principal until much later). In mathematical induction, you assume that the statement is true for one value of n, then prove it's true for the next value n+1. In this case, you would assume that the n-th triangular number Tn equals n(n+1)/2, and prove that the next triangular number Tn+1 equals (n+1)(n+2)/2. The difference between Tn+1 and Tn is just the row that has to be added, namely, n+1. But a simple algebraic manipulation shows that the difference between (n+1)(n+2)/2 and n(n+1)/2 also equals n+1. Hence, if Tn equals n(n+1)/2, then Tn+1 equals (n+1)(n+2)/2. There is one more step to mathematical induction, namely a base case. In this case, that would be to show that T1 equals 1(1+1)/2, which is easily verified, as most base cases are for mathematical induction.

#### Exercise 9.

Show algebraically that any square number is the sum of two consecutive triangular numbers.

It's easier to do it pictorially, as the Pythagoreans would have:

```     o\o o o o
\
o o\o o o
\
o o o\o o
\
o o o o\o
\
o o o o o
```
but that's not the question.

The algebraic demonstration is pretty easy, too. You just need to compute Tn plus Tn+1. You get n(n+1)/2 + (n+1)(n+2)/2 which simplifies to n2 + 2n + 1, which you recognize as (n+1)2. Since every square is of that form, you're done.

#### Exercise 10.

Show using dots that eight times any triangular number plus 1 makes a square. Conversely, show that any odd square diminished by 1 becomes eight times a triangular number. Show these results algebraically as well.

Pictorially first. One picture does the statement and the converse.

```     o\o o o o o|o o o o o
\        |        /
o o\o o o o|o o o o/o
\      |      /
o o o\o o o|o o o/o o
\    |    /
o o o o\o o|o o/o o o
\  |  /
o o o o o\o|o/o o o o
---------+-+/
o o o o o|o|o o o o o
/+-+---------
o o o o/o|o\o o o o o
/  |  \
o o o/o o|o o\o o o o
/    |    \
o o/o o o|o o o\o o o
/      |      \
o/o o o o|o o o o\o o
/        |        \
o o o o o|o o o o o\o
```
One equation will do it, too.
8Tn + 1 = 8n(n+1)/2 + 1 = 4n2 + 4n + 1 = (2n+1)2.

#### Exercise 11.

Show that in a Pythagorean triple, if one of the terms is odd, then two of them must be odd and one even.

(Note that the Pythagoreans didn't know all the Pythagorean triples, as did the Babylonians. Euclid, much later, did.) Suppose you've got three numbers, x, y, and z, and x2 + y2 = z2. We need to eliminate two cases as impossible: (1) all three terms are odd, (2) one term is odd and two are even.

Here's one argument you could give. Recall that a number is even if and only if its square is even, and it's odd if and only if its square is odd. In case (1) where all three terms are odd, so are their squares, and we get a sum of two odd numbers equal to an odd number, which is impossible. In case (2) one term is odd and two even, so we get either an even sum of an even number and an odd number, which is impossible, or an odd sum of two even numbers, which is also impossible. Those two cases eliminated, that leaves only the result: two of the terms are odd and one even.

#### Exercise 13.

Show that if a right triangle has one leg of length 1 and a hypotenuse of length 2, then the second leg is incommensurable with the first leg. (In modern terms, this is equivalent to showing that the square root of 3 is irrational.) Use an argument similar to the proposed Pythagorean argument that the diagonal of a square is incommensurable with the side (that is, that square root of 2 is irrational).

There are several proofs by contradiction. Suppose that the square root of 3 equals m/n where m and n have no common factors. Then 3n2 = m2.

Proof 1. Since 3 is odd, you can quickly show both m and n are odd. Write m as 2i+1, and write n as 2j+1. Then 3(2i+1)2 = (2j+1)2. Simplify and divide the equation by 2. You'll get an equation which says an even number equals an odd number, a contradiction.

Proof 2. Another proof is just like the proof that the square root of 2 is incommensurable with 1 except that it's harder to show that the square n2 of a number is divisible by three, then the number n is divisible by 3. That, in turn, uses a proof by contradiction, in other words, a proof of the contrapositive statement that if 3 doesn't divide n, then 3 doesn't divide n2.

Begin by assuming that 3 doesn't divide n. Since every third number is divisible by 3, that means that either n+1 or n+2 is divisible by 3. First, take the case that 3 divides n+1. That is, n+1 = 3k. Then n = 3k–1, so n2 = (3k–1)2 = 9k2 – 6k + 1. Since 3 divides 9k2 – 6k, it does not divide n2. Second, take the case that 3 divides n+2 and show similarly that 3 does not divide n2. Therefore, if 3 doesn't divide n, then 3 doesn't divide n2.

(Since the rest of the proof is the same as that for the square root of 2, it's not reproduced here.)

Proof 3. Alternatively, if you're willing to use the prime number theorem (which is something the ancient Greeks may have known but there's no record that they did), then you can give a shorter proof that the square root of 3 is irrational. Suppose that it equals m/n. Then 3n2 = m2. Since n2 and m2 are both squares, they both have an even number of prime factors. The number 3 has an odd number of prime factors, namely 1. Therefore, 3n2 has an odd number of prime factors while m2, which it's equal to, has an even number of prime factors, a contradiction. Therefore the square root of 3 is irrational.

Proof 4. It's more work, but also a useful result, to show that if any whole number is the square of a rational number, then that rational number is actually a whole number. That proof is omitted here.