Archive for November, 2009

Why it’s good to be confused by Algebra’s Negative Numbers

Friday, November 27th, 2009

If you, or someone you know, is having difficulty “getting” negative numbers, be happy.  They’re a very strange concept and an understanding, as distinct from a rote application of rules, is going to be hard to come by because they’re one of the earliest mathematical leaps from the concrete to the abstract.

Negative numbers are a great concept, but that’s what they are, a concept.  Negative quantities do not exist in our physical experience.  I can point to two apples or two dollars, but I can’t point to minus two apples or minus two dollars.

Math can be a source of wonder, but too often we’re in too much of a rush and it’s like that with negative numbers. We present them to students as if they’re obvious and difficulty with them in the sign of a shallow mind, when actually the opposite is true.  If they don’t seem weird, you’re probably not really thinking about them.

Here’s what some great mathematicians thought about them:

“Pascal regarded the subtraction of 4 from 0 as utter nonsense.” (Kline, Morris, Mathematical Thought form Ancient to Modern Times, Oxford University Press, New York, 1972, p 252)

Antoine Arnauld, a theologian and mathematician, invoked ratios and “questioned that  -1:1 = 1:-1 because, he said, -1 is less than +1; hence,  How could a smaller be to a greater as a greater is to a smaller?” (Kline, Morris, Mathematical Thought form Ancient to Modern Times, Oxford University Press, New York, 1972, p 252)

The person generally credited with the idea of the number line, John Wallis, “rejected as absurd the … idea of a negative number as being less than nothing, but accepted the view that it is something greater than infinity — a viewpoint shared by Swiss mathematician Leonhard Euler…” (http://en.wikipedia.org/wiki/John_Wallis)

“As recently as the 18th century, it was common practice to ignore any negative results returned by equations on the assumption that they were meaningless, just as René Descartes did with negative solutions…” (http://en.wikipedia.org/wiki/Number)

“In 1759, Francis Maseres, an English mathematician, wrote that negative numbers ‘darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple’. Because of their dark and mysterious nature, Maseres concluded that negative numbers did not exist…” (http://www.bbc.co.uk/radio4/history/inourtime/inourtime_20060309.shtml)

“In the 15th century, Nicolas Chuquet, a Frenchman, used negative numbers as exponents and referred to them as ‘absurd numbers’” http://en.wikipedia.org/wiki/Negative_and_non-negative_numbers

So, if there’s something in you that thinks negative numbers are strange, you’re in great company!

A method for using a negative base in a logarithm

Tuesday, November 10th, 2009

You can download a spreadsheet to accompany this explanation into which you can enter your own values (lines 8 =  10) here

Common wisdom says you can’t use a negative base in a logarithm if you’re working with real numbers

i.e. Y=log-b(X) is not valid, where –b is a negative number

However, consider this

For every integer a, there is some negative number –p, such that

1) (-p)(2a+1) = -b : because 2a+1 is an odd number and a negative number to an odd power is a negative number.

P(2a + 1) = b : (the relationship between positive p and positive b)

P = b1/(2a+1) (spreadsheet line 14 solves for –p)

There is some z, such that

Pz = X  (z solved for on spreadsheet line 16)

If  c = int(z/2) (c is on line 18 of the spreadsheet) then

X = p(z-2c) * p2c

Because the exponent of p2c is even, we can rewrite it as (-p)2c

2) Let W = (-p)2c , (spreadsheet line 22) therefore

X = p(z-2c) * W

we know that (-p)2c is positive because the exponent 2c is an even integer

Taking log-b of expression 1) yields

(2a+1) log-b(-p) = log-b(-b) : since the log of a number to its base is 1, we have

3) Log-b(-p) = 1/(2a+1)

Taking log-b of expression 2 yields

Log-b(W) = log-b[(-p)2c] = 2c * log-b(-p)

Substituting 1/(2a+1) for log-b(-p) from equation 3 yields

Log-b(W) = 2c/(2a+1)

As a approaches infinity, p approaches one, p2 approaches 1, W approaches X and log-b(W) approaches log-b(X), so

Log-b(X) = 1/(2a+1) * 2c

4) Log-b(X) = 1/(2a+1) * [2 * int( logb(X) * (2a+1)/2)]

You can see this formula implemented step by step in the spreadsheet starting on line 27

The answer can be checked (starting on spreadsheet line 37) by the following process

Log-2(X) = 1/(2a+1) * 2c is

X =((-b) 1/(2a + 1)) 2c

First, take the (2a + 1) root of –b which is the same as minus the (2a+1) root of b.

Then take that number to the 2c power.