Exploding DotsUnusual Numbers
In the previous section, we looked at a number with infinitely many 9s to the right of the decimal point:
0.999999…
Now, let’s see what happens if we add infinitely many 9s to the left of the decimal point:
…999999
If we assume that this is a meaningful number (and not, for example, just “infinity”), we can try to use the same algebraic argument as before, to work out its value:
- Let’s start by giving the number a name, say Afor Allistaire:A= …999999
- Now multiply it by 10. This gives us10A= …999990
- Notice that Aand 10Aonly differ in their final digit. Therefore, if we subtract the equation in step 1 from the equation in step 2, we get9A= –9
- Finally, if we divide both sides by 9, we getA=
In other words, we have just shown that …999999 = −1. Apparently, if we pulled out an infinite calculator and computed the sum of 9 + 90 + 900 + 9000 + …, the result would be −1!
Do you believe that?
Unusual Arithmetic
Even though …9999999 is clearly not a “normal” number, let’s assume for now that it exists, and that it follows the basic laws of arithmetic. If that is the case, we’d expect …9999999 + 1 =
Let’s use a
Looks like this actually worked! If we add 1 to …9999999, the result is 0.
But remember: all we have shown is that IF we choose to believe that …999999 is a meaningful number that follows our usual laws of arithmetic, THEN it must have value –1. Most people simply say that it isn’t a number and stop there – and that is a perfectly valid view.
This begs the question: is there an unusual system of arithmetic for which …999999 is a meaningful number?
Challenge
Let’s make matters worse! Consider the number with infinitely many 9s both to the left and to the right of the decimal point: …9999.9999…. Try to use the same algebraic argument to show that this equals zero.
Somehow this makes sense, because …9999.9999… = …9999 + 0.9999… = −1 + 1 = 0.
Warping the Number Line
In the previous chapter, we saw that 0.999999… = 1. This seems somewhat plausible, because the sequence of approximations 0.9, 0.99, 0.999, 0.9999, and so on, get closer and closer to 1.
In this example, the exact opposite happens: the numbers 9, 99, 999, 9999, and so on, are marching further and further away from –1. That’s why it is so abstruse to think that …999999 could possibly equal –1.
It turns out, however, that it is possible to develop a new arithmetic system in which numbers like …999999 are meaningful. To do that, we just have to change how we measure “distance” between numbers on the number line.
Usually, distance is defined using addition and subtraction. For example, the distance between 2 and 6 is
Instead, we can define a “different kind” of distance using multiplication and division.
In the world of integers, 0 is the most divisible number of all. It can be divided any number of times by any integer, and still give an integer result (namely 0). If we focus on our number base of 10, we can see that 0 can be divided by 10 once, or twice, or thirty-seven times, or a million times.
- The number 40 is a little bit “zero-like”, in this sense in that we can divide it by ten and still have an integer.
- The number 1700 is more zero-like: it can be divided
by 10, and still give an integer result. - The number 230,000 is even more zero-like. It can be divided
times by 10, and still stay an integer. - The number 5, on the other hand, is not very zero-like. We can’t divide it by ten even once, and have it stay an integer.
We can now develop a distance formula, based on how often 10 “goes into” into a number multiplicatively. If we can divide a number a by ten a maximum of k times while remaining an integer, let’s write
For example,
We can also measure the distance between any two different numbers. For example, the distance between 3 and 33 is
With this new way to measure distance, 1, 10, 100, 1000, … is a sequence of numbers getting closer and closer to
Mathematicians call this way of viewing distances between the non-negative integers
Negative Numbers and Fractions
We’ve already seen that our new, ten-adic system supports negative integers: …999999 = –1. We can do something similar for other negative numbers. How much do you have to add to …999998, to get it to explode?
In other words, …999998 =
•
Constructing ten-adic fractions is a bit more difficult. Let’s see what happens if we multiply …6666667 by 3:
Since …6666667 × 3 =
Challenge
Can you work out which ten-adic number behaves like
What about other fractions like
It turns out that there are a few fractions that cannot be expressed in our ten-adic number system: all fractions that, in their reduced form, have a
A Serious Flaw
We’ve now seen that every integer an fraction has a ten-adic equivalent, and that we can add, subtract and multiply ten-adic numbers, just like we would normal integers. Unfortunately there is one serious flaw: we cannot divide by all ten-adic numbers.
To see why that’s the case, we need to look at the powers of 2 and 5:
…
…
Notice how many of the powers of 5 end in
M = …33554432
N = …1953125
If we try to multiply powers of 2 and 5, we get a sequence of products that get closer and closer to zero (in our 10-adic sense):
2 | × | 5 | = | 10 |
4 | × | 25 | = | 100 |
8 | × | 125 | = | |
16 | × | 625 | = |
The same happens if we try to multiply M and N:
… | 3 | 1 | 2 | 5 | |
× | … | 4 | 4 | 3 | 2 |
… | 6 | 2 | 4 | 10 | |
… | 3 | 6 | 15 | ||
… | 8 | 20 | |||
+ | … | 20 | |||
= | … | 37 | 28 | 19 | 10 |
= | … | 0 | 0 | 0 | 0 |
In other words, we have found two non-zero numbers M and N so that M × N = 0.
This means that in ten-adic arithmetic, it is impossible to divide by M or N. (If it were possible, we could divide the equation M × N = 0 by N, and get M = 0. That is a contradiction.)