WARPING NORMAL DISTANCE ON THE NUMBER LINE
We usually say the number 5, for instance, is a distance five from 0 on the number line because 5 is five unit lengths away from the zero. (We usually use absolute value notation for this distance: ∣5∣=5.)
And 3.7 is a distance 3.7 from 0, ∣3.7∣=3.7, because three-and-seven-tenths of a unit fit between 0 and 3.7 on the number line. And so on.
This is a very additive way of thinking about distance: adding five 1s gets you from 0 to 5; adding 3.7 1s get you from 0 to 3.7; and so on. We can say that the distance of a point a on the number line, in this thinking, is the number of 1s that go additively into a.
But much of mathematics is not only concerned with the additive properties of numbers, but also the multiplicative properties of numbers. For example, many people are interested in the prime factorizations of numbers (for example, 1000=2353 and 105=3⋅5⋅7). There are so many unanswered questions about the prime numbers and prime factorizations still in mathematics today. These questions are, in general, very hard!
Is there are way to bring the geometry of the number line into play to possibly help with multiplicative questions? Is there a way to think about the number line itself as perhaps structured multiplicatively rather than additively?
To think about this, rather than focus on all possible factor of numbers, let’s focus on one possible factors of numbers. And to keep matters relevant to our base-ten arithmetic thinking, let’s focus on the number 10.
In our additive thinking for distance on the number line we use the unit of 1 and ask how many ones (additively) go into each number for its distance from 0. We now want to use the unit of 10 and ask how many times ten goes multiplicatively into each number.
What could that mean?
In the world of integers the number 0 is the most divisible number of all: it can be divided by any integer any number of times and still give an integer result (namely 0) each and every time. Focusing on our chosen factor of ten, we can divide 0 by ten once, or twice, or thirty-seven times, and still have an integer.
The number 40 is a little bit “zero-like” in this sense in that we can divide it by ten once and still have an integer. The number 1700 is more zero-like as it can be divided by ten twice and still give an integer result. A googol is very much more zero-like: it can be divided by ten one hundred times and still stay an integer.
The integer 5 is not very zero-like at all: one can’t divide it by ten even once and stay an integer.
In this setting the more times ten “goes into” into a number multiplicatively, the more zero-like it is. So in this sense, a googol is much closer to zero than 5 is.
So let’s develop a distance formula that regards numbers with large powers of ten as factors as closer to zero than numbers with less counts of powers of ten as factors. There are a number of ways one might think to do this, but let’s try to mimic the additive properties of the number line we are familiar with.
Normally we would say that 850 is further from zero than 85 is, and, in fact, we might even say 850 is ten times further from zero as 85 is. In our multiplicative thinking, 850 is now closer to zero than 85 is and it would be natural to have it as ten times closer.
The following formula seems a natural way to have this happen.
If a can be divided by ten a maximum of k times and remain an integer, then set ∣a∣ten=10k1.
For example, then, ∣850∣ten=1011=0.1 and ∣8500∣ten=1021=0.01 and ∣8500000∣ten=1051=0.00001. Also, since ∣85∣ten=1001=1. we see, indeed, that 850 is ten times closer to zero than 85 is.
We can also measure the distance between any two numbers in this multiplicative way. For example, the distance between 3 and 33 is ∣33−3∣ten=∣30∣ten=1011=0.1.
With this new way to measure distance, we see that
1, 10, 100, 1000, …
is a sequence of numbers getting closer and closer to zero. We have ∣1∣ten=1 and ∣10∣ten=0.1 and ∣100∣ten=0.01 and ∣1000∣ten=0.001, and so on, indeed approaching a distance of zero from 0.
In terms of values in a 1←10 machine, we see that boxes far to the left in the machine, representing high powers of ten, are representing values very close to zero. (Before, in our additive thinking, boxes to the far right for decimals represented values very closer and closer to zero.)
Mathematicians call this way of viewing distances between the non-negative integers ten-adic arithmetic. (The suffix adic means “a counting of operations” and here we are counting factors of ten.) It is fun to think how to extend this notion of distance to fractions too, and then to all real numbers.
THE NUMBER …9999
Let’s look now at the sequence of numbers 9 and 99 and 999 and so on marching off to the right on the number line. Could they possibly be marching closer and closer to the value −1?
Yes, if by “closer” we mean this new multiplicative way to think of distance.
The numbers 9, 99, 999, 9999, … are indeed approaching the value 0−1=−1.
Comment: We can now justify the (very) long addition computation given below.
We first compute 9+1=10, and then we add 90 to this to obtain 100, and then we add 900 to obtain 1000, and so on. The further along we go with the computation the closer our results are to the number zero.
You can intuitively see this in the 1←10 machine: when you add one more dot to this loaded machine and perform the explosions, one clears away dots, pushing what remains further and further to the left where boxes have less and less significant value.
Computation of this next (very) long multiplication is justified in a similar way.
We first compute 3×7=21, and then add to this 3×60 to get 201, and then add to this 3×600 to get 2001, and so on. Now the numbers 20, 200, 2000, … are getting closer and closer to zero, so the numbers 21=20+1, 201=200+1, 2001=2000+1, … are getting closer and closer to 1. The further along we go with this computation, the closer our answers are to the number 1.
Again, we can intuitively see this reasoning at play by tripling all the values in this loaded 1←10 machine.
And did you discover that ⋯∣3∣3∣3∣3∣4 behaves like the fraction 32?
Question: Doubling ⋯∣3∣3∣3∣3∣4 gives ⋯∣6∣6∣6∣6∣8 which is one more than ⋯∣6∣6∣6∣6∣7, which is 31. Is this consistent?
Challenge: Show that in a 2←3 machine that ⋯∣1∣1∣1∣1∣2 is negative one! Show that ⋯∣0∣1∣0∣1∣0∣1∣0∣2 when multiplied by 5 gives 1, and so represents 51. (What measure of distance might we be using on the number line this time for these “numbers” to make sense?)
CONSTRUCTING NEGATIVE INTEGERS
In our base-ten thinking with our multiplicative notion of distance on the number line, we set
∣a∣ten=10k1 where k is the largest count of times a can be divided by ten and remain an integer.
And we have made sense of ⋯9999 as a meaningful number with value −1.
So what’s −2 in this unusual system of arithmetic?
Let’s think in terms of a 1←10 machine. Since −1=⋯9∣9∣9∣9 , and −2 is double −1, we should have
With explosions we get
And one can check that this long addition does give zero.
We can see now how to readily construct any negative integer. For example, we can see that adding 47 to ⋯999953 will give zero and so this latter quantity must be −47, and that adding 3000 to ⋯9997000 gives zero and so this quantity must be −3000.
Challenge: What is −2 in a 2←3 machine? What is −5?
We saw that ⋯66667 is the fraction 31: multiply this quantity by three and you get 1.
The 1←10 machine provides a natural way to compute such fractions. For example, let’s find the ten-adic representation of 74 . That is, let’s find a number x such that 7×x=4. Start by writing
as for a 1←10 machine. Then
We want 7a ,after explosions, to leave a 4. So we need a multiple of 7 four greater than a multiple of 10. We see that 7a=14 is good. So let’s set a=2.
Now we want 7b+1 to be a multiple of 10 so that all dots in that box explode to leave zero behind. This suggests b=7.
Now we need 7c+5 a multiple of 10. Choose c=5.
Now choose d=8.
And then e=2.
And now I am doing the same work as I did for a value b, making 7b+1 a multiple of 10. We are in a cycle and so x=74 is represented as
Challenge: This process felt reminiscent of the task of writing 74 as a decimal in ordinary arithmetic using a 1←10 machine with decimals. We argued there too that the decimal represent had to fall into a cycle.
Can you argue that the fraction 132 will also have a repeating ten-adic expansion?
Challenge: What is the ten-adic expansion of −74 ?
Write −74 as −1∣−4∣−2∣−8∣−5∣−7∣−2 and add some dots and antidot pairs to make all the terms positive.
Let’s try to compute the ten-adic representation of the fraction 21. Here we seek a number
This means we have a number a so that, after explosions, 2a leaves a single dot. That is, we need 2a to be one more than a multiple of ten. This is not possible!
Challenge: Contemplate the ten-adic expansions for 51 and 103 and 352.
In general, which fractions qp seem to be problematic?
Challenge: Develop a general theory that if qp is a reduced fraction with q sharing no factor in common with ten (other than 1), then it is for certain possible to express qp as a ten-adic number ⋯hgfedcba. Show further that its expression is sure to fall into a repeating cycle.
BROADENING OUR DEFINITION A TAD
It seems we have defined a ten-adic value to be an expression of the form ⋯edcba with each digit one of the standard digits 0 through 9, allowing for non-zero digits to appear infinitely far to the left.
In this system we have the ordinary positive integers,
eg 5 is ⋯00005,
the negative numbers
eg −5 is ⋯99995
and some fractions
eg 31 is ⋯66667.
But not all fractions. It turns out that the troublesome fractions are the ones qp which, when written in reduced form, have a denominator a multiple of 2 or 5 or both.
We can obviate this problem if we allow a ten-adic number to extend finitely far into the decimal places on the right. That is, set a ten-adic expression to be one of the form ⋯edcba.xy…z with each digit one of the standard digits 0 through 9, allowing for non-zero digits to appear infinitely far to the left of the decimal point, and only finitely far to its right. (After all, we do the analogous thing in ordinary arithmetic by writing 33.3333⋯, for example, for thirty-three and a third.)
Now we have
21=0.5 is ⋯0000.5
10023=0.23 is ⋯0000.23.
We can also handle 352 by thinking of this as 7×52=7×102×2=104/7. Since 74 is 1428572 we must have 352=142857.2.
Challenge: Show that 61=⋯3333.5 and hence find the ten-adic expression for 125=61+41.
What is the ten-adic expression for 121?
Challenge: Explain why every fraction is now sure to have a ten-adic representation.
Challenge: Show that 5323=⋯53535323 is the number −331000 in ten-adic arithmetic. (Hint: Multiply the quantity by 100 and subtract.)
One can use the technique of this question to show that every ten-adic number that eventually falls into a cycle going leftwards is a rational number.
Challenge: In ordinary arithmetic, the quantity 0.abcabcabc⋯=0.abc is the fraction 999abc. We see this by setting x=0.abcabcabc⋯ and noticing that 1000x=abc.abcabcabc⋯. Subtracting then yields 999x=abc.
Show that the same algebra applied to the ten-adic number ⋯abcabcabc=abc shows that it this number has value −999abc.
In fact, prove the following general result. Suppose b1, b2, …, bk are single digits.
If 0.b1b2…bk is the fraction qp in ordinary arithmetic,
then b1b2…bk is the fraction −qp in ten-adic arithmetic, and vice versa.
Challenge: Explore a theory of “3/2-adic” representations of fractions using a 2←3 machine.