Base 16: Hexadecimal Numbers

Hexadecimal

Hexadecimal is regarded as base-$16.$ Each place value in the number is some power of $16.$ The hexadecimal values are: $$0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9,\,A,\,B,\,C,\,D,\,E,\,F$$ The letters $A$ through $F$ represent $10$ through $15,$ respectively.

From Decimal to Hexadecimal

Get ready, this method can be a pain to use. Just like with ternary, we want to focus on the remainder.

Let's work through an example to see how it's done. Let's convert $42$ to hexadecimal.

The first step is to divide our number by $16,$ taking the remainder.

We get a remainder of $10.$ Remember that hexadecimal uses the letter $A$ to represent this. So the "ones place" of our hexadecimal number is $A$.

Next, we take the non remainder portion, $2$, and divide it again by $16.$

We get $0$ as our non remainder part so we know we're done. The remainder is $2$ so this will be the "tens place" of our hexadecimal number. $$42_{10}=2A_{16}$$

From Hexadecimal to Decimal

Just like with binary and ternary, to get to decimal you want to take the sum of each digit multiplied by the powers of the base you're in. Since we're in base-$16,$ we multiply by powers of $16.$

Let's convert $1C2$ to a decimal number. Starting with $2,$ multiply by $16$ to the zero power, then $C$ gets multiplied by $16$ to the first power, and $1$ gets multiplied by $16$ to the second power. Remember that $C$ represents the number $12.$ $$\begin{align} &1C2\\ &=(1\cdot16^2)+(C\cdot16^1)+(2\cdot16^0)\\ &=(1\cdot16^2)+(\color{red}{12}\cdot16^1)+(2\cdot16^0)\\ &=(1\cdot256)+(12\cdot16)+(2\cdot1)\\ &=256+192+2\\ &=450 \end{align}$$

So, $$1C2_{16}=450_{10}$$

Practice

Try some on your own!

Solution

Because I'm lazy, I'm not going to make a diagram for the long division. Regardless, the first step is dividing $162$ by $16.$ $$162\div16=10\,r2$$ Since we get a remainder of $2,$ the "ones place" of our hexadecimal number is $2.$

Next, divide the non remainder portion by $16.$ $$10\div16=0\,r10$$ Since we got a $0$ we end here. The remainder, $10,$ corresponds to the letter $A$. This will be the "tens place" of our hexadecimal number.

So, $$162_{10}=A2_{16}$$

Solution

$$\begin{align} &C4\\ &=(C\cdot16^1)+(4\cdot16^0)\\ &=(12\cdot16^1)+(4\cdot16^0)\\ &=(12\cdot16)+(4\cdot1)\\ &=192+4\\ &=196 \end{align}$$

How many numbers can be represented in Hexadecimal?

The standard way of writing numbers, base $10,$ is where we use the digits $0, 1, 2, 3, 4, 5, 6, 7, 8,$ and $9.$ However, using base $16$ gives us more digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E,$ and $F.$ Thus we'd think that if we limit the number of digits we can use, we'd be able to make more numbers in base $16$ than in base $10.$ This intuition is actually helpful, as the claim I made is true.

If we only have $3$ digits, we can make less numbers in base $10$ than in base $16,$ because we have more options for each digit. This is true no matter how many digits we allow ourselves to use. It lets us get more information out of fewer digits; that is, it takes less writing or typing to represent bigger numbers.

Now we can work to answer the question of how many different numbers we can make in a given base if we limit the number of digits. We'll answer this question for base $10,$ and then for base $16.$ We start with base $10$ since it's more familiar to most people. For that we turn to the following diagram to guide us:

Suppose the number of digits we are allowing is $d.$ In the above picture we pick three digits so $d = 3.$ Then to count how many numbers we can make, we count how many options there are for each digit, and multiply them together. To pick the first digit, we can use $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9,$ giving $10$ possible options. We allow $0$ to be the first digit because then we technically have a number with less than $d$ digits, which is OK; we just can't use more than $d$ digits.

In the same way, we can pick one of these same digits for the second digit, and then again for the third. This means we have $10$ choices for each digit, and then multiply these numbers together.

$$10 \times 10 \times 10 = 1000$$

If we use no more than $3$ digits, we can make $1000$ different numbers in base $10.$ You might realize these are the first $1000$ whole numbers: $0, 1, 2, ..., 999!$

The answer is $10^5 = 10 \times 10 \times 10 \times 10 \times 10 = 100000,$ where we have $10$ options for each digit: $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$

Now let's start over and have the same discussion for hexadecimal, base $16.$ The only difference in solving the problem is that we have $16$ options for each digit: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E,$ and $F.$ Again, we'll focus on the case where $d = 3,$ with having three digits to choose.

There are $16$ ways to choose each digit, and there are three digits. We multiply the number of choices for each digit together, giving

$$16 \times 16 \times 16 = 4096$$

options for constructing base $16$ numbers with three digits or less. The two-digit numbers will begin with one $0,$ and the one-digit numbers will begin with two $0$s, and then a third digit.

The calculation for this question is

$$16 \times 16 \times 16 \times 16 \times 16 = 16^5 = 1048576.$$

The digits in hexadecimal (or base $16$) start with $0,1,2,3,4,5,6,7,8,9$ (just like in base $10$). The remaining base-$16$ digits are $A,B,C,D,E,F,$ corresponding in order to the remaining base-$10$ numbers less than $16$ (namely $10,11,12,13,14,15$).

Hexadecimal (aka base-$16$) numbers have $2$ properties:

1. Each place numeral is in the set $\{0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f\},$ with $a-f$ representing the decimal values $10-15.$
2. Each place value is equivalent to ($\text{number} \times 16^{place}$).

To get the decimal equivalent add the place values.

E.g. The decimal equivalent of $9af2$ is:

$= (9 \times 16^{3}) + (10 \times 16^{2}) + (15 \times 16^{1}) + (2 \times 16^{0})$

$= (9 \times 4096) + (10 \times 256) + (15 \times 16) + (2 \times 1)$

$= 36864 + 2560 + 240 + 2$

$= 39,666$

(Video) Hexadecimal

by Carl Herold

Here's a video by Carl Herold showing you how to convert from hexadecimal to base-$10$ and base-$2$.

$\large{\text{Summary}}$

Here's a video by Carl Herold showing you how to convert from hexadecimal to base-$10$ and base-$2$. This isn't a terribly complicated process but it does require more calculations than the previous sections. What the video doesn't explain is how to convert from base-$10$ to base-$16.$

It's pretty simple so I'll show an example. Take a number you want to convert. Let's say it's $145$. First divide the number by $16.$ $$\frac{145}{16}=9.0625$$ Now take the remainder ($0.0625$) and multiply it by $16,$ rounding up if necessary. This will become the $1$'s place. $$16\times0.0625=1$$ So the $1$'s place is actually $1.$ Now take the number before the remainder $(9).$ Since it's less than $16$ we know that it's the digit for the $10$s place and we can stop. So the base-$16$ version of $145$ is $\boxed{91}.$