ABSTRACT
The proof of Pythagoras’s theorem given in Box 3.1 uses a number of operations that were discussed in Chapter 2, namely,
1. am * an = a (m+n) 2. am/an =am * a-n = a (m-n) 3. (am)n = a (m*n)
The proof also introduces the use of brackets so that
(a + b) * (c + d) = a * (c + d) + b * (c + d)
or
(a + b) * (c + d) = (a + b) * c + (a + b) * d
This in both cases is equal to
(a * c) + (a * d) + (b *c) + (b * d)
It also uses indices so that
(x + y)1 * (x + y)1 = (x + y)(1+1)
= (x + y)2 = x2 + y2 + 2xy
It also makes the point that we can often dispense with the “multiply” and “divide” symbols by writing the above as
(a + b)(c + d) = ac + ad + bc + bd
Using indices, we can also write
( ) ( )
+ +
a b
c d = (a + b) * (c + d)–1
using the –1 index to mean “divide by” or “one over.” Pythagoras deals with the sum of two squares. An equation, in which the highest
power of any term is 2, is known as a quadratic. A particular quadratic form that is
BOX 3.1 ONE PROOF OF PYTHAGORAS’S THEOREM
Consider a simple right-angled triangle with sides of length x, y, and z (the hypotenuse) as in Figure 3.1a. Copy the triangle three times and rotate it each time to form Figure 3.1b. This builds a square on the hypotenuse or long side.
