02 824 378
Unusualus
Unusualus
Unusualus

Station  10C
Some Unusual Mathematics for Unusual Numbers 

It is possible to develop an arithmetic system of numbers for which a quantity like 9999\ldots9999 actually is meaningful (and by what we proved in the previous section has value 1-1).

Here’s one such system, one that requires we change our sense of distance between numbers on the number line. It is a system that will allow us to say, for instance, that the numbers 99, 9999, 999999, … are indeed marching closer and closer to 1-1 on the number line despite what our geometric training says!

I10S10B - Image01

WARPING NORMAL DISTANCE ON THE NUMBER LINE

We usually say the number 55, for instance, is a distance five from 00 on the number line because 55 is five unit lengths away from the zero. (We usually use absolute value notation for this distance: 5=5|5|=5.)

I10S10C - Image02

And 3.73.7 is a distance 3.73.7 from 00, 3.7=3.7|3.7|=3.7, because three-and-seven-tenths of a unit fit between 00 and 3.73.7 on the number line. And so on.

This is a very additive way of thinking about distance: adding five 11s gets you from 00 to 55; adding 3.73.7 11s get you from 00 to 3.73.7; and so on. We can say that the distance of a point aa on the number line, in this thinking, is the number of 11s that go additively into aa.

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=23531000=2{3}5{3} and 105=357105=3\cdot 5 \cdot 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 1010.

In our additive thinking for distance on the number line we use the unit of 11 and ask how many ones (additively) go into each number for its distance from 00. We now want to use the unit of 1010 and ask how many times ten goes multiplicatively into each number.

What could that mean?

In the world of integers the number 00 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 00) each and every time. Focusing on our chosen factor of ten, we can divide 00 by ten once, or twice, or thirty-seven times, and still have an integer.

The number 4040 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 17001700 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 55 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 55 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 850850 is further from zero than 8585 is, and, in fact, we might even say 850850 is ten times further from zero as 8585 is. In our multiplicative thinking, 850850 is now closer to zero than 8585 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 aa can be divided by ten a maximum of kk times and remain an integer, then set aten=110k|a|_{ten}=\dfrac{1}{10^{k}}.

For example, then, 850ten=1101=0.1|850|_{ten}=\dfrac{1}{10^{1}}=0.1 and 8500ten=1102=0.01|8500|_{ten}=\dfrac{1}{10^{2}}=0.01 and 8500000ten=1105=0.00001|8500000|_{ten}=\dfrac{1}{10^{5}}=0.00001. Also, since 85ten=1100=1|85|_{ten}=\dfrac{1}{10^{0}}=1. we see, indeed, that 850850 is ten times closer to zero than 8585 is.

We can also measure the distance between any two numbers in this multiplicative way. For example, the distance between 33 and 3333 is 333ten=30ten=1101=0.1|33-3|{ten}=|30|{ten}=\dfrac{1}{10^{1}}=0.1.

With this new way to measure distance, we see that

11, 1010, 100100, 10001000, \ldots

is a sequence of numbers getting closer and closer to zero. We have 1ten=1|1|_{ten}=1 and 10ten=0.1|10|_{ten}=0.1 and 100ten=0.01|100|_{ten}=0.01 and 1000ten=0.001|1000|_{ten}=0.001, and so on, indeed approaching a distance of zero from 00.

In terms of values in a 1101 \leftarrow 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.)

I10S10C - Image03

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\ldots9999

Let’s look now at the sequence of numbers 99 and 9999 and 999999 and so on marching off to the right on the number line. Could they possibly be marching closer and closer to the value 1-1?

I10S10B - Image01

Yes, if by “closer” we mean this new multiplicative way to think of distance.

We have

I10S10C - Image05

The numbers 99, 9999, 999999, 99999999, … are indeed approaching the value 01=10-1=-1.

Comment: We can now justify the (very) long addition computation given below.
I10S10B - Image02

We first compute 9+1=109+1=10, and then we add 9090 to this to obtain 100100, and then we add 900900 to obtain 10001000, 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 1101 \leftarrow 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.

I10S10B - Image03

Computation of this next (very) long multiplication is justified in a similar way.

I10S10B - Image04

We first compute 3×7=213 \times 7 = 21, and then add to this 3×603 \times 60 to get 201201, and then add to this 3×6003 \times 600 to get 20012001, and so on. Now the numbers 2020, 200200, 20002000, … are getting closer and closer to zero, so the numbers 21=20+121 = 20+1, 201=200+1201 = 200+1, 2001=2000+12001=2000+1, … are getting closer and closer to 11. The further along we go with this computation, the closer our answers are to the number 11.

Again, we can intuitively see this reasoning at play by tripling all the values in this loaded 1101 \leftarrow 10 machine.

I10S10B - Image05

And did you discover that 33334\cdots|3|3|3|3|4 behaves like the fraction 23\dfrac{2}{3}?

Question: Doubling 33334\cdots|3|3|3|3|4 gives 66668\cdots|6|6|6|6|8 which is one more than 66667\cdots|6|6|6|6|7, which is 13\dfrac{1}{3}. Is this consistent?

Challenge: Show that in a 232 \leftarrow 3 machine that 11112\cdots|1|1|1|1|2 is negative one! Show that 01010102\cdots|0|1|0|1|0|1|0|2 when multiplied by 55 gives 11, and so represents 15\dfrac{1}{5}. (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

aten=110k|a|_{ten}=\dfrac{1}{10^{k}} where kk is the largest count of times aa can be divided by ten and remain an integer.

And we have made sense of 9999\cdots9999 as a meaningful number with value 1-1.

So what’s 2-2 in this unusual system of arithmetic?

Let’s think in terms of a 1101 \leftarrow 10 machine. Since 1=9999-1=\cdots 9|9|9|9 , and 2-2 is double 1-1, we should have

2=18181818-2=\cdots 18|18|18|18.

With explosions we get

2=18181818-2=\cdots 18|18|18|18

=1818198=\cdots 18|18|19|8

=181998=\cdots 18|19|9|8

=19998=\cdots 19|9|9|8

=9998=\cdots 9|9|9|8.

And one can check that this long addition does give zero.

I10S10C - Image10

We can see now how to readily construct any negative integer. For example, we can see that adding 4747 to 999953\cdots999953 will give zero and so this latter quantity must be 47-47, and that adding 30003000 to 9997000\cdots9997000 gives zero and so this quantity must be 3000-3000.

Challenge: What is 2-2 in a 232 \leftarrow 3 machine? What is 5-5?

CONSTRUCTING FRACTIONS

We saw that 66667\cdots66667 is the fraction 13\dfrac{1}{3}: multiply this quantity by three and you get 11.

The 1101 \leftarrow 10 machine provides a natural way to compute such fractions. For example, let’s find the ten-adic representation of 47\dfrac{4}{7} . That is, let’s find a number xx such that 7×x=47 \times x = 4. Start by writing

x=hgfedcbax=\cdots h|g|f|e|d|c|b|a

as for a 1101 \leftarrow 10 machine. Then

7x=7h7g7f7e7d7c7b7a7x=\cdots 7h|7g|7f|7e|7d|7c|7b|7a.

We want 7a7a ,after explosions, to leave a 44. So we need a multiple of 77 four greater than a multiple of 1010. We see that 7a=147a=14 is good. So let’s set a=2a=2.

x=hgfedcb2x=\cdots h|g|f|e|d|c|b|2

7x=7h7g7f7e7d7c7b147x=\cdots 7h|7g|7f|7e|7d|7c|7b|14

=7h7g7f7e7d7c7b+14=\cdots 7h|7g|7f|7e|7d|7c|7b+1|4

Now we want 7b+17b+1 to be a multiple of 1010 so that all dots in that box explode to leave zero behind. This suggests b=7b=7.

x=hgfedc72x=\cdots h|g|f|e|d|c|7|2

7x=7h7g7f7e7d7c5047x=\cdots 7h|7g|7f|7e|7d|7c|50|4

=7h7g7f7e7d7c+504=\cdots 7h|7g|7f|7e|7d|7c+5|0|4

Now we need 7c+57c+5 a multiple of 1010. Choose c=5c=5.

x=hgfed572x=\cdots h|g|f|e|d|5|7|2

7x=7h7g7f7e7d40047x=\cdots 7h|7g|7f|7e|7d|40|0|4

=7h7g7f7e7d+4004=\cdots 7h|7g|7f|7e|7d+4|0|0|4

Now choose d=8d=8.

x=hgfe8572x=\cdots h|g|f|e|8|5|7|2

7x=7h7g7f7e600047x=\cdots 7h|7g|7f|7e|60|0|0|4

=7h7g7f7e+60004=\cdots 7h|7g|7f|7e+6|0|0|0|4

And then e=2e=2.

x=hgf28572x=\cdots h|g|f|2|8|5|7|2

7x=7h7g7f2000047x=\cdots 7h|7g|7f|20|0|0|0|4

=7h7g7f+200004=\cdots 7h|7g|7f+2|0|0|0|0|4

And f=4f=4.

x=hg428572x=\cdots h|g|4|2|8|5|7|2

7x=7h7g30000047x=\cdots 7h|7g|30|0|0|0|0|4

=7h7g+3000004=\cdots 7h|7g+3|0|0|0|0|0|4

And g=1g=1.

x=h1428572x=\cdots h|1|4|2|8|5|7|2

7x=7h100000047x=\cdots 7h|10|0|0|0|0|0|4

=7h+10000004=\cdots 7h+1|0|0|0|0|0|0|4

And now I am doing the same work as I did for a value bb, making 7b+17b+1 a multiple of 1010. We are in a cycle and so x=47x=\dfrac{4}{7} is represented as

1428571428571428572=1428572\cdots142857 : 142857 : 142857 : 2 = \overline{142857} : 2.

Challenge: This process felt reminiscent of the task of writing 47\dfrac{4}{7} as a decimal in ordinary arithmetic using a 1101 \leftarrow 10 machine with decimals. We argued there too that the decimal represent had to fall into a cycle.

Can you argue that the fraction 213\dfrac{2}{13} will also have a repeating ten-adic expansion?

Challenge: What is the ten-adic expansion of 47-\dfrac{4}{7} ?
One approach:

Write 47-\dfrac{4}{7} as 1428572\overline{-1|-4|-2|-8|-5|-7} :|- 2 and add some dots and antidot pairs to make all the terms positive.

1428572\overline{-1|-4|-2|-8|-5|-7} :|- 2

=1428572+(8+8)=\overline{-1|-4|-2|-8|-5|-7} :|- 2+\left(-8+8\right)

=14285718=\overline{-1|-4|-2|-8|-5|-7-1} :|8

==\cdots

A GLITCH

Let’s try to compute the ten-adic representation of the fraction 12\dfrac{1}{2}. Here we seek a number

x=hgfedcbax=\cdots h|g|f|e|d|c|b|a

so that

2x=2h2g2f2e2d2c2b2a2x=\cdots 2h|2g|2f|2e|2d|2c|2b|2a

equals 11.

This means we have a number aa so that, after explosions, 2a2a leaves a single dot. That is, we need 2a2a to be one more than a multiple of ten. This is not possible!

Challenge: Contemplate the ten-adic expansions for 15\dfrac{1}{5} and 310\dfrac{3}{10} and 235\dfrac{2}{35}.
In general, which fractions pq\dfrac{p}{q} seem to be problematic?

Challenge: Develop a general theory that if pq\dfrac{p}{q} is a reduced fraction with qq sharing no factor in common with ten (other than 11), then it is for certain possible to express pq\dfrac{p}{q} as a ten-adic number hgfedcba\cdots 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\cdots edcba with each digit one of the standard digits 00 through 99, allowing for non-zero digits to appear infinitely far to the left.

In this system we have the ordinary positive integers,

eg 55 is 00005\cdots00005,

the negative numbers

eg 5-5 is 99995\cdots99995

and some fractions

eg 13\dfrac{1}{3} is 66667\cdots66667.

But not all fractions. It turns out that the troublesome fractions are the ones pq\dfrac{p}{q} which, when written in reduced form, have a denominator a multiple of 22 or 55 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.xyz\cdots edcba.xy\ldots z with each digit one of the standard digits 00 through 99, 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 33.333333.3333\cdots, for example, for thirty-three and a third.)

Now we have

12=0.5\dfrac{1}{2}=0.5 is 0000.5\cdots0000.5

and

23100=0.23\dfrac{23}{100}=0.23 is 0000.23\cdots0000.23.

We can also handle 235\dfrac{2}{35} by thinking of this as 27×5=2×27×10=4/710\dfrac{2}{7 \times 5}=\dfrac{2\times 2}{7 \times 10}=\dfrac{4/7}{10}. Since 47\dfrac{4}{7} is 1428572\overline{142857} : 2 we must have 235=142857.2\dfrac{2}{35}=\overline{142857}.2.

Challenge: Show that 16=3333.5\dfrac{1}{6}=\cdots3333.5 and hence find the ten-adic expression for 512=16+14\dfrac{5}{12}=\dfrac{1}{6}+\dfrac{1}{4}.
What is the ten-adic expression for 112\dfrac{1}{12}?

Challenge: Explain why every fraction is now sure to have a ten-adic representation.

Challenge: Show that 5323=53535323\overline{53}23=\cdots53535323 is the number 100033-\dfrac{1000}{33} in ten-adic arithmetic. (Hint: Multiply the quantity by 100100 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.abc0.abcabcabc\cdots=0.\overline{abc} is the fraction abc999\dfrac{abc}{999}. We see this by setting x=0.abcabcabcx=0.abcabcabc\cdots and noticing that 1000x=abc.abcabcabc1000x=abc.abcabcabc\cdots. Subtracting then yields 999x=abc999x=abc.

Show that the same algebra applied to the ten-adic number abcabcabc=abc\cdots abcabcabc=\overline{abc} shows that it this number has value abc999-\dfrac{abc}{999}.

In fact, prove the following general result. Suppose b1b_{1}, b2b_{2}, …, bkb_{k} are single digits.

If 0.b1b2bk0.\overline{b_{1} b_{2}\ldots b_{k}} is the fraction pq\dfrac{p}{q} in ordinary arithmetic,

then b1b2bk\overline{b_{1} b_{2} \ldots b_{k}} is the fraction pq-\dfrac{p}{q} in ten-adic arithmetic, and vice versa.

Challenge: Explore a theory of “3/2-adic” representations of fractions using a 232 \leftarrow 3 machine.

Unusualus
Vision byPowered by
Contact GMP:
  • The Global Math Project on Twitter
  • The Global Math Project on Facebook
  • Contact The Global Math Project
Contact BM:
  • Buzzmath on Twitter
  • Buzzmath on Facebook
  • Visit Buzzmath's Website
  • Read Buzzmath's blog!