???
Well done!
Archie

Send us Feedback

Please let us know if you have any feedback and suggestions, or if you find any errors and bugs in our content.

Reset Progress

Are you sure that you want to reset your progress, response and chat data for all sections in this course? This action cannot be undone.

Glossary

Select one of the keywords on the left…

Exploding DotsUnusual Numbers

Reveal All Steps

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:

1. Let’s start by giving the number a name, say
A
for Allistaire:
A
= …999999
2. Now multiply it by 10. This gives us
10
A
= …999990
3. Notice that
A
and 10
A
only differ in their final digit. Therefore, if we subtract the equation in step 1 from the equation in step 2, we get
9
A
= –9
4. Finally, if we divide both sides by 9, we get
A
=

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 110 machine to see if that is actually the case. Click somewhere in the 1s cell to add 1:

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 , because 2+4=6.

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 twicethree timesfour times 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 and, it 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

aten=110k

For example, 850ten=1101=0.1, and 8500ten=1102=0.01, and 850000ten= .

We can also measure the distance between any two different numbers. For example, the distance between 3 and 33 is 333ten=30ten=1101=0.1.

With this new way to measure distance, 1, 10, 100, 1000, … is a sequence of numbers getting closer and closer to zero1–1infinity. Similarly, 9, 99, 999, 9999, … is getting closer and closer to , just like we saw above.

Mathematicians call this way of viewing distances between the non-negative integers ten-adic arithmetic. The suffix “adic” means “a counting of operations”. Here we are counting factors of ten.

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 = . We can similarly calculate that …999997 = , …999953 = , …999700 = –300 and so on. Every negative integer has a ten-adic equivalent.

Constructing ten-adic fractions is a bit more difficult. Let’s see what happens if we multiply …6666667 by 3:

Since …6666667 × 3 = , we know that …6666667 = 13.

Challenge

Can you work out which ten-adic number behaves like 23?

What about other fractions like 47 or 213?

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 denominator that is a multiple of 2 or 5 (or both). You can fix this by allowing ten-adic numbers to have a finite number of decimal places. Now, every rational number as a ten-adic equivalent.

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:

21 = 2
22 = 4
23 = 8
24 = 16
25 =
26 =
27 = 128

51 = 5
52 = 25
53 =
54 =
55 = 3,125
56 = 15,625
57 = 78,125

Notice how many of the powers of 5 end in other, smaller powers of 5. The same is also true for powers of 2. It turns out that we can create two infinite, 10-adic numbers, that always end in powers of 2 or 5 respectively:

21 = 2
25 = 32
225 = 33554432
M = …33554432

51 = 5
52 = 25
53 = 125
N = …1953125

If we try to multiple 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.)