基底为3：三进制数

Ternary

Ternary is a numeral system with base $3.$ This essentially means it only uses $3$ digits to make up its numbers: $$0,\,1,\,2$$

Our normal number system is called the decimal or base-$10.$ As you may have guessed it is because we use $10$ digits. $$0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9$$

From Decimal to Ternary

To convert from a regular decimal number to a ternary number can be a little tedious, but it only requires division. Let's work through an example to see how we do it.

Suppose we want to convert $42$ from base-$10$ to base-$3.$ One way we show a number is base-$10$ is to put a subscript after it with $10.$ $$42_{10}$$

To begin the conversion to base-$3$, we divide our number by $3.$ We don't need to worry about decimal points, just what the remainder is.

Let's divide $42$ by $3.$

We get $14$ with remainder $0.$ The remainder will be the digit in the "one's place" of our ternary number.

Next, take the number we get as our solution (minus the remainder) and divide that by $3.$ Let's divide $14$ by $3.$

We get $4$ remainder $2.$ This remainder will be the digit in the "ten's place" of our ternary number.

We again divide our result by $3.$ This will go till we get a result of $0.$ So let's divide $4$ by $3.$

We get $1$ remainder $1.$ This will be the digit in the "hundred's place" of our ternary number.

We're almost done! One more division. Divide $1$ by $3.$

We get another remainder of $1.$ This will be the digit in the "thousand's place" of our ternary number. We know to stop here because we finally get $0$ up top.

So, we got the remainders $0,\,2,\,1,\,1$. The ternary version of $42$ will be these numbers but written in reverse order. So: $$42_{10}=1120_3$$

From Ternary to Decimal

Don't worry, this way is much simpler in my opinion. Let's use the number we just found, $1120$, and convert it back to $42$.

Starting with the last digit, $0,$ we want to multiply each digit by $3$ raised to a power starting from $0$ and then add them all together. In other words, $0$ is multiplied by $3^{\color{red}{0}}$, $2$ is multiplied by $3^{\color{red}{1}}$, $1$ is multiplied by $3^{\color{red}{2}}$, and $1$ is multiplied by $3^{\color{red}{3}}$. $$\begin{align} &1120\\ &=1\cdot3^3+1\cdot3^2+2\cdot3^1+0\cdot3^0\\ &=1\cdot27+1\cdot9+2\cdot3+0\cdot1\\ &=27+9+6+0\\ &=42 \end{align}$$

Practice

Try converting each way on your own.

Solution

I won't show all of the division because I'm lazy but start by dividing $19$ by $3.$ $$19\div3=6\,r1$$ Next, divide $6$ by $3.$ $$6\div3=2\,r0$$ Finally, divide 2 by 3. $$2\div3=0\,r2$$ We know to stop here because we get a $0$ as our end result. The ternary number, from left to right, will be the last remainder up to the first remainder. $$19_{10}=201_3$$

Solution

We want to multiply each digit by $3$ raised to increasing powers. $$\begin{align} &10212\\ &=1\cdot3^4+0\cdot3^3+2\cdot3^2+1\cdot3^1+2\cdot3^0\\ &=1\cdot81+0\cdot27+2\cdot9+1\cdot3+2\cdot1\\ &=81+0+18+3+2\\ &=104 \end{align}$$ So, $$10212_3=104_{10}$$

All your life you've been counting in what is called the base $10$ number system, where we build numbers using the digits $0, 1, 2, 3, 4, 5, 6, 7, 8,$ and $9.$

But it turns out we don't have to use all these digits. In particular, if we only use the first three, then we are writing numbers using a base $3$ number system. It is important to know that we don't use the number $3$ in this system, but only $0, 1,$ and $2.$ Also, "ternary" is just a fancy word for base-$3.$ So, let's try counting in base $3$ for a little while.

$$0, 1, 2...$$

The first three numbers are easy, but what's next? We can't use $3.$ To get an answer, remember that in base $10,$ we can write every number with an invisible $0$ in front of it. That is, counting in base $10$ gives

$$00, 01, 02, 03, 04, 05, 06, 07, 08, 09...$$

We can actually do the same thing in base $3,$ so in base $3$ we start counting as

$$00, 01, 02...$$

In base $3,$ the $1$ still comes after $0.$ This makes us think we can keep counting in base $3$ as

$$00, 01, 02, 10, 11, 12, ...$$

and in fact that's exactly what we do. And we can do this as many times as we want, adding more digits as we go along:

$$00, 01, 02, 10, 11, 12, 20, 21, 22, \color{red}{100}\color{black}{...}$$

When we count we usually use the digits $0, 1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ to make numbers. Well, as it might surprise you, that's not the only way to do it. The way we normally do it is called base $10$ representation because we use $10$ different digits, but we can use more or less digits to count. One common example, sometimes used in computer operations, is the base $3$ representation, which only uses the digits $0,$ $1,$ and $2.$

We can count any number in base $3$ that we could in base $10,$ and we usually use subscripts to indicate which base we're in. A base-$10$ number will have a $_{10}$ stuck to the end, while a base-$3$ number will have a $_{3}$ stuck to the end. The reason for this is clear upon answering the next question.

The correct answer was a $22$ and $101,$ because those were the numbers in the list that ONLY used the digits $0, 1,$ and $2.$ Thus we use the subscripts: $22_3$ means $22$ in base $3,$ while $22_{10}$ means $22$ in base $10,$ and these numbers have different values!

So now you might be asking, how do we compare these values? For that, I introduce you to the $\color{brown}{\text{Conveyor Belt of Chocolate}}:$

So how on earth does this work? Well, we have three digits to work with, so we use three different types of chocolate to represent the different digits: a white chocolate will be a $0,$ a dark chocolate will be a $1$ (the dark brown one), and a milk chocolate will be a $2$ (the light brown one).

The above picture clarifies which number belongs with which piece of chocolate. The number at the TOP of the wheel shows which digit we're actually using, so the picture above is how we represent the number $0_3.$

The next question is: how do we represent $1_3?$ Well, we can turn the conveyor belt clockwise until the number $1$ (and the dark chocolate) is at the top:

Finally, using just a single conveyor belt, we can spin the conveyor belt clockwise some more until the $2$ (and the milk chocolate) is at the top.

Here's where it gets a bit trickier. This conveyor belt only represents one digit of a base $3$ number, as we can only show the numbers $0_3, 1_3,$ and $2_3$ with this conveyor belt. How do we represent $3_{10}$ with the conveyor belts?

The answer is to use another conveyor belt. We have run out of digits in base $3$ to write $3_{10}$ with only a single base $3$ digit, so we will need two conveyor belts. Think of $2_3$ as $02_{3}:$ in base $10$ we can add a $0$ to the front of a number and not change its value, and in fact this works in any base. So let's display the number $2_3$ again with two conveyor belts.

We find the digits of our number by reading the conveyor belts from left to right, which gives us the first digit of $0$ and second digit of $2.$

Now we can represent $3_{10}$ with the conveyor belts. To increase the number represented by $1$ we use the following rule which resembles "carrying" when we add numbers together. Rotate the rightmost conveyor belt clockwise. If this leads to having a $0$ at the top, then rotate the next conveyor belt clockwise. If that conveyor belt shows a $0$ after doing so, keep moving to the left (if we run out of conveyor belts again, add another one).

This will let us keep counting in base $3.$ It also tells us how to convert numbers from base $10$ into base $3.$ Our previous picture showed us how to write $2_3$ with conveyor belts, so let's make $3$ in base $3:$ rotate the right conveyor belt to get $0,$ then rotate the left conveyor belt clockwise since the previous conveyor belt gave us $0.$

Now that we have that picture, we see that we write $3_{10}$ as $10_3$ because the left conveyor belt shows a $1,$ and the right conveyor belt shows a $0.$

The natural thing to explore is what happens to the conveyor belts if we keep increasing the number. Well, if the right conveyor belt does not give a $0$ when we rotate it clockwise, we don't touch the other ones. For example, to go from $3$ to $4$ in base $10,$ we rotate the right conveyor belt clockwise to turn the $0$ into a $1,$ and the other conveyor belt stays put.

The picture above represents $4_{10}$ in base $3,$ and from reading the top numbers we conclude $4_{10} = 11_3.$

Eventually, we're going to run out of numbers we can represent with $2$ conveyor belts, but how big can we get? Let's keep counting and see what happens.

To get $5,$ rotate the right conveyor belt clockwise to get $2,$ meaning $5_{10} = 12_3.$

To get $6,$ rotate the right conveyor belt clockwise to get $0,$ so we rotate the left conveyor belt clockwise to get $2,$ meaning $6_{10} = 20_3.$

To get $7,$ rotate the right conveyor belt clockwise to get $1,$ meaning $7_{10} = 21_3.$

To get $8,$ rotate the right conveyor belt clockwise to get $2,$ meaning $8_{10} = 22_3.$

Now, we run into a problem. If we rotate the right conveyor belt clockwise, we get $0,$ meaning we also need to rotate the left conveyor belt clockwise, meaning we get another $0.$ If counting like this gives all conveyor belts showing $0,$ we need to add another conveyor belt, and we set the new conveyor belt to the $1$ setting.

This is how we write $100_3$ with the conveyor belts, and since we added $1$ to $8$ to get this picture, we know that $9_{10} = 100_3.$

You might now be noticing a pattern. The smallest number for which we needed two conveyor belts is $3,$ and the smallest number for which we needed three conveyor belts is $9.$ These are powers of $3,$ as $3^1 = 3$ and $3^2 = 9.$ In fact this pattern keeps going no matter how big we make the numbers. In general, to find how many conveyor belts we need to represent a number in base $3,$ use the following trick: find the biggest power of $3$ that is smaller than the number you want to represent. Take the exponent and add $1$ to it. That's how many conveyor belts you need.

In the previous problem, we see that $3^3 = 27$ and $3^4 = 81,$ so $27$ is the largest power of $3$ that is smaller than $3.$ Then we add $1$ to the exponent to get $\boxed{4}.$

Hopefully that demonstration helped you understand how base $3$ and base $10$ are related, and how to turn a base $10$ number into base $3.$ Just don't eat any of the chocolates off the conveyor belts or you'll have a harder time counting.

Ternary (aka base-$3$) numbers have $2$ properties:

1. Each place numeral is in the set $\{0, 1, 2\}.$
2. Each place value is (number x $3^{place}$).

To get the base-$10$ value add up the place values.

E.g. What is base-$10$ equivalent to the base-$3$ number $2210?$

$$= (2 \times 3^{3}) + (2 \times 3^{2}) + (1 \times 3^{1}) + (0 \times 3^{0})$$ $$= (2 \times 27) + (2 \times 9) + (1 \times 3) + (0 \times 1)$$ $$= 54 + 18 + 3 +0$$ $$= 75$$

Base 3

The base $3,$ or ternary, system, uses only the digits $0, 1,$ and $2.$ For each place, instead of multiplying by the power of $10,$ you multiply by the power of $3.$

For example, $12012_3\rightarrow 1\times3^4+2\times3^3+0\times3^2+1\times3^1+2$.

Base $3$ is a largely unused system for computers, as computers mostly use base $2$ by tradition. However, some ternary computers have actually been built.

(Video) Number Systems 3: Ternary

by Jacob Schrum

Here's a good video by Jacob Schrum showing you how to convert from base-$10$ to base-$3$ and from base-$3$ to base-$10$.

$\large{\text{Summary}}$

Base-$3$ is a number system that uses three numbers, $0, 1,$ and $2.$ All you do to convert the number into base-$3$ is to continuously divide by $3$ and use the remainders each time to figure out the the new number.