Numeration Systems and Whole Number Operations
3.1 Numeration Systems
Numbers, Numerals, and Digits
What exactly is a number? As elementary as it sounds, the question merits a pause for consideration. Depending on who you ask, a number is a mathematical object used to count things. People have counted things for thousands of years, but not everyone wrote numbers the same way. Different cultures used different symbols and different grouping rules. In order to further our understanding of what numbers are and how they behave, we're going to look at different numeration systems.
Definition: Numeral and Digit
A numeral is a written symbol (or group of symbols) that represents a number.
Both $8$ and VIII are numerals representing the number that counts groups of eight things.
A digit is one basic symbol used in a number system.
We're going to work our way up to the numbering system that you know, but we're not going to define it just yet. However, I still need to reference it to explain these other examples so you need to know its name. The symbols you're used to -
$$0,1,2,3,4,5,6,7,8,9$$
are called the Hindu-Arabic numerals. So when I use this word, don't be confused. I am referencing the numeration system that you're used to.
Historical Numeration Systems
Egyptian Numerals
Egyptian numerals are additive. You count copies of each symbol and add everything. This is a link to an photo album containing images of real egyptian numerals, but for the sake of ease we'll use the unicode counterparts here.
| Hindu-Arabic | $$1 000 000$$ | $$100 000$$ | $$10 000$$ | $$1 000$$ | $$100$$ | $$10$$ | $$1$$ |
|---|---|---|---|---|---|---|---|
| Egyptian | ๐จ | ๐ | ๐ญ | ๐ผ | ๐ข | ๐ | ๐บ |
Very similar to our own counting system, the egyptians counted in groups of 10. To represent larger numbers, they just added more symbols: 1, 2, 3, 4,...9 becomes ๐บ, ๐บ๐บ, ๐บ๐บ๐บ, ๐บ๐บ๐บ๐บ, ... ๐บ๐บ๐บ๐บ๐บ๐บ๐บ๐บ๐บ. But when a group of 10 is met, we would moved to the next position in our positional based numerbering system and write 10, where the egyptians would use their symbol ๐.
Example: Egyptian-style conversion
Covert the following egyptian numeral: ๐๐ญ๐ญ๐ผ๐ผ๐ผ๐ข๐ข๐ข๐ข๐๐๐๐๐๐บ๐บ๐บ๐บ๐บ๐บ
Show Solution
Just add what each group contributes:
The singular ๐ means there is one $100000$.
The two ๐ญ means there are two $10000$'s.
There are three ๐ผ - $3\cdot 1000$, four ๐ข - $4\cdot 100$, five ๐ - $5\cdot 10$, and six ๐บ - $6\cdot 1$. All together we have
$$1(100000)+2(10000)+3(1000)+4(100)+5(10)+6(1) = 123456$$
MyOpenMath: Numeration Systems and Base Conversion
Example: Convert to Egyptian Numerals
Covert the number $21355$ to egyptian heiroglyphs.
Show Solution
Use place-value groups of Egyptian symbols.
There are two ๐ญ, one ๐ผ, three ๐ข, five ๐, and five ๐บ.
So the Egyptian numeral is:
๐ญ๐ญ๐ผ๐ข๐ข๐ข๐๐๐๐๐๐บ๐บ๐บ๐บ๐บ
Check by adding contributions:
$$ 2(10000)+1(1000)+3(100)+5(10)+5(1)=21355 $$
MyOpenMath: Numeration Systems and Base Conversion
Babylonian Numerals
Babylonian notation used a base-60 idea, this is where our notions of grouping will start to be challenged. Think of each new place as "groups of 60" instead of "groups of 10." When we count, we count up to $9$, but then we run out of symbols so we move to a new position and increment that value. $9$ goes to $10$, $1$ group of $10$ plus $0$ groups of $1$.
The Babylonians had few symbols, they used a type of cuneiform, but the numerals were repeated in groups of 60. For example, one would count all the way up to $59$, and then move to a new postion, or "roll over", and increment that position $1$ and reset the previous position to $0$. Using Hindu-Arabic numerals to mimic the base-60 counting, it would look something like this:
$$1,2,3,...,57,58,59,(1, 0),(1,1), (1,2),(1,3),...,(1,58),(1,59),(2,0),...$$
I'm using $(a,b)$ here to denote $a$ groups of $60$ and $b$ groups of $1$. $(3,15)$ would mean I have three groups of $60$ and fifteen groups of $1$, or $3(60) + 15(1) = 195$.
Even still, I'm hiding what a Babylonia numeral would really look like, because their writing was done by chiseling symbols into clay tablets. There wasn't a lot of articulation to be had with a chisel and clay, so the symbols are simple.
Using unicode glyphs, the symbol for $1$ is ๐, the symbol for $10$ is ๐, and the digit $23$ is written as ๐๐๐๐๐.
Once the count gets to $59$, we would roll over into a new position and increment there. $58,59,60,61,62$ becomes
๐๐๐๐๐๐๐๐๐๐๐๐๐
๐๐๐๐๐๐๐๐๐๐๐๐๐๐
๐
๐ ๐
๐ ๐๐
You see the problem here? Babylonians didn't have a numeral for zero and what was written was ambiguous. The context of the writing would give an idea of what the number should have been. For the sake of simplicity, I'm going to add a comma (,) between positions so it is easier to distinguish when places and to make it clear when there was a space left for a zero.
๐๐ , ๐๐
should clearly be $2(60)+11(1)=131$, and
๐๐ , , ๐๐
should clearly be $2(3600) + 0(60) + 11(1) = 7211$. A small note here, when I moved into the third position of the numeral, that represented $60$ groups of $60$ which is $60\cdot 60=3600$. The third position represent groups of $60^2=3600$. The fourth would represent $60^3=216000$ and so on.
Example: Convert Babylonian Numeral
Covert the Babylonian number ๐๐๐,๐,,๐๐๐๐ to Hindu-Arabic.
Show Solution
Looking at the numeral, there are four positions. From right to left, the position represent groups of 1's, 60's, 3600's, and 216000's.
| $$216000$$ | $$3600$$ | $$60$$ | $$1$$ |
|---|---|---|---|
| ๐๐๐ | ๐ | ๐๐๐๐ | |
| 3 | 1 | 0 | 22 |
$$3(21600)+1(3600)+0(60)+22(1)=68422$$
MyOpenMath: Numeration Systems and Base Conversion
The other direction is a bit harder, but it is an important skill to build and will become very useful to us later in this course.
Example: Hindu-Arabic to Babylonian
Covert 111742 to a Babylonian numeral.
Show Solution
Because Babylonian's counted in groups of 60, we begin by finding the largest group of 60 that the number contains. First lets, just list powers of 60.
$$60^4=12960000,60^3=21600,60^2=3600,60^1=60,60^0=1$$
If this number were larger, we could get more powers of 60, but clearly this number is smaller than $12960000$. So $60^3=21600$ is where we're starting. How many $21600$'s are in $111742$? We could start adding multiples until we overshoot the number:
$$21600(1)=21600$$ $$21600(2)=43200$$ $$21600(3)=64800$$ $$21600(4)=86400$$ $$21600(5)=108000$$ $$21600(6)=129600$$
We find that 6 is too many, but we could take 5. A faster way to do this is to use division. $$\frac{111742}{21600}\approx 5.2$$ would tell us immediately that there are $5(21600)$ to be taken from the number. So lets do exactly that, remove $5(21600)$ and see what is left.
$$111742-5(21600)=3742$$
Now we repeat. What is the largest grouping of 60 that would go into $3742$? $3600$ is the answer, and there is only 1 in there to be taken.
$$3742-1(3600)=142$$
$$142-2(60)=22$$
$$22 - 22(1)=0$$
What we have discovered is that
$$111742 = 5(21600)+1(3600)+2(60)+22(1)$$
Check it and see. For our babylonian number, there will be 4 positions. From right to left we'll need 22, 2, 1, and 5 in those positions.
| 5 | 1 | 2 | 22 |
|---|---|---|---|
| ๐๐๐๐๐ | ๐ | ๐๐ | ๐๐๐๐ |
The final conversion being
๐๐๐๐๐,๐,๐๐,๐๐๐๐
MyOpenMath: Numeration Systems and Base Conversion
Mayan Numerals
Across the pond in the Americas, Mayan's were doing their own counting. They even had a numeral for zero, ๐ ! In that way, mathematically speaking, the Mayans were more advanced.
| Numeral | Value |
|---|---|
| ๐ | 0 |
| ๐ก | 1 |
| ๐ฅ | 5 |
Mayan Numerals 1โ30
They used dots ๐ก and lines ๐ฅ for 1's and 5's, but they counted in base 20. Here is what counting from 1 to 30 would look like.
| 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| ๐ก | ๐ข | ๐ฃ | ๐ค | ๐ฅ |
| ๐ฆ | ๐ง | ๐จ | ๐ฉ | ๐ช |
| ๐ซ | ๐ฌ | ๐ญ | ๐ฎ | ๐ฏ |
| ๐ฐ | ๐ฑ | ๐ฒ | ๐ณ | ๐ก ๐ |
| ๐ก ๐ก | ๐ก ๐ข | ๐ก ๐ฃ | ๐ก ๐ค | ๐ก ๐ฅ |
| ๐ก ๐ฆ | ๐ก ๐ง | ๐ก ๐จ | ๐ก ๐ฉ | ๐ก ๐ช |
Mayan numeral positions extended vertically, with the largest positional value on top. For simplicity I will write the numbers horizontal with the largest positional value to the left to be more familiar. ๐ก ๐ซ ๐ฑ means 1 (๐ก) group of 400, 11 (๐ซ) groups of 20, and 17 (๐ฑ) groups of 1: $$1(400)+11(20)+17(1)=637$$ Notice I used $400$ for the third position because 20 groups of 20 is 400.
Example: Mayan to Hindu-Arabic numeral
Covert ๐ฃ ๐จ ๐ณ to a Hindu-Arabic numeral.
Show Solution
In base 20, the positional values are:
| $$400$$ | $$20$$ | $$1$$ |
|---|---|---|
| ๐ฃ | ๐จ | ๐ณ |
| 3 | 8 | 19 |
So we have $$3(400)+8(20)+19(1)=1379$$
MyOpenMath: Numeration Systems and Base Conversion
Again, the other direction is a bit tougher but it gets easier with practice.
Example: Hindu-Arabic to Mayan
Convert 3519 to a Mayan numeral.
Show Solution
As before, we need to find the largest group of out counting base that is contained within the number. Since Mayan's counted in base 20 we're looking at powers of 20.
$$20^3=8000,20^2=400,20^1=20,20^0=1$$
We find that 8000 is too much, so we're counting 400's first. A quick check
$$\frac{3519}{400}\approx 8.8$$
finds that there are eight 400's that can be removed.
$$3519-8(400)=319$$
That leaves us with 319 which has fifteen 20's to take.
$$319-15(20)=19$$
and finally that leaves us with 19 ones.
$$19-19(1)=0.$$
We find that
$$3519=8(400)+15(20)+19(1)$$
Converting this to our positional system
| $$400$$ | $$20$$ | $$1$$ |
|---|---|---|
| 8 | 15 | 19 |
| ๐จ | ๐ฏ | ๐ณ |
Our mayan numeral is ๐จ ๐ฏ ๐ณ
MyOpenMath: Numeration Systems and Base Conversion
Roman Numerals
While the Egyptians, Babylonians, and Mayans were most likely completely new to you, the Romans may be a bit more familiar. Their conquest of the world left a lasting impact on cultures around the world. To this day we still see them on historical buildings and other things. Notably, the most recent Super Bowl as of the typing of these notes was LXIII, or $63$.
Luckily for us, the romans mostly counted in base 10. Unluckily for us, they didn't use a positional system. They simply strung symbols together and added them up.
| Symbol | I | V | X | L | C | D | M |
|---|---|---|---|---|---|---|---|
| Value | $$1$$ | $$5$$ | $$10$$ | $$50$$ | $$100$$ | $$500$$ | $$1000$$ |
So to count 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 romans would write
I,II,III,IV,V,VI,VII,VIII,IX,X
To disect that Super Bowl numeral above, LXIII$=$L $+$ X $+$ III $$=50+10+3=63$$
Note an odd occurance here, for 4, I wrote IV instead of IIII. To help with repetition, the Romans would write a smaller valued numeral before a larger one to show that it was less. IV means one less than 5 as opposed to VI which means one more than 5. Here are the most common differences used when building a roman numeral.
| Numeral | IV | IX | XL | XC | CD | CM |
|---|---|---|---|---|---|---|
| Value | $$4$$ | $$9$$ | $$40$$ | $$90$$ | $$400$$ | $$900$$ |
This makes reading a roman numeral a little tough. In general, reading it left from right we are adding values up. That is unless we encounter a smaller value in front of a larger one, in that case we subtract the smaller from the larger and keep adding.
Example: Roman conversions
Convert each numeral to base ten.
- XLVIII
- LXXVI
- CXLVII
Show Solution
- XLVIII$= 40+8 = 48$
- LXXVI$= 50+20+6 = 76$
- CXLVII$= 100+40+7 = 147$
MyOpenMath: Numeration Systems and Base Conversion
Example: Hindu-Arabic to Roman Numerals
Convert $1994$ to a Roman numeral.
Show Solution
Let's work through this from the largest symbol down to the smallest, one chunk at a time.
We start with the thousands. $1994$ contains one thousand, so we grab one M.
$$1994 - 1000 = 994$$
Now we handle the hundreds. $994$ contains $9$ hundreds, which is $900$. Here's where we get to use one of those subtractive pairs from the table above! $900$ is not a symbol on its own, but it is one hundred less than one thousand, so we write C in front of M to get CM.
$$994 - 900 = 94$$
Moving to the tens. $94$ contains $9$ tens, which is $90$. Another subtractive pair! $90$ is ten less than one hundred, so we write X in front of C to get XC.
$$94 - 90 = 4$$
Finally the ones. We're left with $4$, and sure enough there's a subtractive pair for that too. $4$ is one less than five, so we write I in front of V to get IV.
$$4 - 4 = 0$$
We used all four subtractive pairs in the right order, so we just concatenate them:
$$1994 = 1000 + 900 + 90 + 4$$
MCMXCIV
MyOpenMath: Numeration Systems and Base Conversion
Hindu-Arabic Place Value
Now we finally make it to the system we use now, the Hindu-Arabic numeral system. This numbering system started a little later than the other systems we've seen here, but it is still very old. It started in India somewhere between the 1st and 6th century. It was adopted by Arab and Persian scholars soon after and eventually made its way to Europe via Spain and North Africa. Famously, Fibonacci, our friend form a previous section, popularized it in his book Liber Abaci or Book of Calculation.
The reason why this numbering system has had the staying power and persistance that it has, is because it takes advantage of a positional system, (try subtracing MCMXCIV and DCCXCIX by hand), a relatively simple set of glyphs, and uses base 10 to match up with our 10 fingers making it easy to use on the fly.
The numerals are Hindu-Arabic numerals are
$${0,1,2,3,4,5,6,7,8,9}$$
When counting one starts in the first position and increments up to 9,
$$1,2,3,4,5,6,7,8,9$$
and then because we've run out of symbols we move to the next position and increment while resetting the pevious position.$10$ means we have $1$ group of $10$ and $0$ groups of $1$.
$$10=1(10)+0(1)$$
This may seem trivial now, because you're very used to the system, but it merits a second look. For example, $5123$ uses 4 positions, each a power of $10$.
$$5123 = 5(1000)+1(100)+2(10)+3(1)$$
Theorem: Expanded Form
If a numeral is $d_n d_{n-1}\cdots d_1 d_0$, then its value is
$$ d_n10^n + d_{n-1}10^{n-1} + \cdots + d_1 10 + d_0 $$
Example: Expanded form of 1,234,567
Write $1{,}234{,}567$ in expanded form.
Show Solution
$$ 1\cdot10^6+2\cdot10^5+3\cdot10^4+4\cdot10^3+5\cdot10^2+6\cdot10+7 $$
That is why we read it as one million, two hundred thirty-four thousand, five hundred sixty-seven.
Counting in Other Bases
To better understand several of the algorithms and operations we'll be using later in the semester, we're going to limit out selection of digits to count in different bases. This lets us use numbers that we're familiar with but in a way that is unusual so we can experience the confusion first time learners might have with these algorithms again.
The intent here is to remove the familiarity with the systems at work so we can focus on the how instead of the what. Any calculator can add or subtract 5 digit numbers, what we care about is how its done, not what the answer is.
Definition: Valid Digits in Base $b$
The digit set is:
$$ D_b=\{0,1,2,\ldots,b-1\} $$
If $b>10$, letters are often used for extra digits.
Here are the digits of some different base systems. Examples:
- Binary (Base-2) has two digits - $$D_2=\{0,1\}$$
- Base-5 has five digits - $$D_5=\{0,1,2,3,4\}$$
- Base-12 $$D_{12}=\{0,1,2,3,4,5,6,7,8,9,T,E\}$$
- Hexadecimal (Base-16) $$D_{16}=\{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\}$$
These might seem strange at first, but they're used the same way every positional system I've described in this section works. You start by counting in the first position until you run out of numerals, then you roll over into the next position and increment there resetting the first.
Lets start by counting in base five. Up to 4 there is no problem.
$$0,1,2,3,4$$
But we've run out of numerals. So we roll over into the next position, increment it by one, and begin again.
$$0,1,2,3,4,10,11,12,13,14$$
And now we're at $14$, which in base-5 means $1(5)+4(1)=9$. This is where it starts to get a little confusing so I'm going to add a subscript to numbers when they are in different bases.
$$14_{\text{five}}=1(5)+4(1)=9$$
When a number does not have a subscript, I mean for it to be in base 10.
For another quick example using Hexadecimal (Base-16), the number $$A3F1_{\text{hex}}=10(16^3)+3(16^2)+15(16^1)+1(16^0)=41969$$
This is one reason computer scientists uses hexadecimal. You can store larger numbers in smaller spaces.
Example: Base-5 to base-10
Convert $3243_{\text{five}}$ to base ten.
Show Solution
$$ 3243_{\text{five}}=3(5^3)+2(5^2)+4(5)+3=375+50+20+3=448 $$
So $3243_{\text{five}}=448$.
MyOpenMath: Numeration Systems and Base Conversion
Backwards is always the harder direction, but we've had practice. If we wanted to convert a number from base 10 into any other base, we need only count how many groupings there are of each power of the base and remove them.
Example: Base-10 to Base-5
Convert $$448$$ to Base-5.
Show Solution
Because we're counting in groups of 5 we start by listing the powers of 5.
$$5^4=625, 5^3=125, 5^2=25,5^1=1,5^0=1$$
$625$ is too big, so we start by seeing how many $125$'s are hidden away in $448$.
$$\frac{448}{125}\approx 3.6$$
This means that there are three $125$'s to be taken.
$$448-3(125)=73$$
Then we have two $25$'s that can be taken from $73$.
$$73-2(25)=23$$
And four $5$'s to be taken from $23$.
$$23-4(5)=3$$
Finally,
$$3-3(1)=0$$
That means that
$$448=3(125)+2(25)+4(5)+3(1)=3243_{\text{five}}$$
MyOpenMath: Numeration Systems and Base Conversion
After doing this conversion several times, one starts to notice a pattern. We get to learn our first real algorithm! Here are steps one can use to convert a number from base 10 into any other base.
Theorem: Base Conversion Algorithm (from base 10 to base $b$)
- Divide by $b$.
- Record the remainder.
- Repeat with the quotient.
- Stop when quotient is 0.
- Read remainders bottom to top.
Example: Convert $448$ to base five
Show Solution
$$ 448=5(89)+3 $$ $$ 89=5(17)+4 $$ $$ 17=5(3)+2 $$ $$ 3=5(0)+3 $$
Read remainders bottom to top: $3243_{\text{five}}$.
MyOpenMath: Numeration Systems and Base Conversion
Example: Binary and base-12 conversions from the tex topics
Convert $10111_2$ and $E2T_{12}$ to base ten.
Show Solution
For binary:
$$ 10111_2=1(2^4)+0(2^3)+1(2^2)+1(2^1)+1(2^0)=16+0+4+2+1=23 $$
For base 12 ($T=10$, $E=11$):
$$ E2T_{12}=11(12^2)+2(12)+10=1584+24+10=1618 $$
So the answers are 23 and 1618.
MyOpenMath: Numeration Systems and Base Conversion
Heres a closing graph on visualizing counting in different bases with blocks.