1.4: Addition and Subtraction in Base Systems

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A person holding a magnifying glass is observing binary code. — Figure $\PageIndex{1}$: All information in computers is represented by 0's and 1's, including quantity, which means computers use Base 2 for arithmetic. (credit: modification of work “Magnifying glass and binary code” by Marco Verch Professional Photographer/Flickr, CC BY 2.0)

Learning Objectives

Add and subtract in bases $2$ to base $16.$

Once we decide on a system for counting, we need to establish rules for combining the numbers we’re using. This begins with the rules for addition and subtraction. We are familiar with base 10 arithmetic, such as or . How does that change if we use a different base instead? A larger base? A smaller one? In particular, computers use base $2$ for all number representations. When your calculator adds or subtracts, multiplies or divides, it uses base $2.$ This is because the circuitry recognizes only two things, high current and low current, which means the system is uses only has two symbols. Which is what base $2$ is.

In this section, we use addition and subtraction in bases other than $10$ by referencing the processes of base $10,$ but applied to a new base system.

Addition in Bases Other Than Base 10

Now that we understand what it means for numbers to be expressed in a base other than $10,$ we can look at arithmetic using other bases, starting with addition. When you think back to when you first learned addition, it is very likely you learned the addition table. Once you knew the addition table, you moved on to the addition of numbers with more than one digit. The same process holds for addition in other bases. We begin with an addition table and then move on to adding numbers with two or more digits.

Let's create the base $6$ addition table.

Look in the row that starts with “$5$” and select the number in the column that starts with “$1$”. At the intersection of the selected row and column, you will find the number $10.$
Is $5_6 + 1_6=6_6?$ Since $6$ is not one of the digits in base $6.$ So, we need to convert $6$ to base $6.$

$\begin{array}{ll}
6 \div 6= & 1\quad \mathrm{R} \quad 0 \\
\end{array}$

We'll say that $6_{10}=10_{6}$

Is $5 _6+ 2_6 = 7_6$? Since $7$ is not one of the digits in base $6$. So, we need to convert $7$ to base $6$.

$\begin{array}{ll}
7\div 6 = & 1 \quad \mathrm{R} \quad 1 \\
\end{array}$

We'll say that $7_{10}=11_{6}$

Similarly, $5 _6+ 3_6 = 12_6$

And so it goes. Using that process, stepping one more along the list, we can create the addition table for base $6$.

Careful

When adding two single-digit numbers in the given base, if the sum is greater than or equal to the base, we need to convert the sum to the given base by dividing it by the base.

Base 6 Chart .png — Figure $\PageIndex{1}$: Base $6$ chart

Your Turn $\PageIndex{1}$: Adding Single Digit Number in Different Bases

Example $\PageIndex{2}$: Adding Base Six Numbers

Calculate $251_{6}+ 133_{6}$

Answer

\[
\begin{array}{r}
\phantom{+}251_{\text{six}} \\
+ \phantom{0}133_{\text{six}} \\
\hline
\phantom{+}424_{\text{six}}
\end{array}
\]

Let’s do for the place of one's first. According to the base $6$ addition table see above, $ 1_6 + 3_6 = 4_6.$

Now, we do the “tens” place (it’s the sixes place). According to the base $6$ addition table, we have $5_6 + 3_6 = 12_6.$

So, like in base $10,$ we use the $2$ and carry the $1.$ Now the “hundreds” place (really, thirty-sixes place). There, we have $1\text{(carried)} + 2_6 + 1_6 = 4_6.$

Therefore \[251_{6}+ 133_{6} = 424_{6}\nonumber\]

Note: When you add sixes places (middle column), do not write $8_6.$ Be Careful! We are adding in base $6.$

In base $7$, the number that follows $6$ is $10$ (since we’ve run out of symbols!). So, $67 + 17 = 107$. In another word $6_7+1_7=7$, but $7$ is not base $7$ number so we need to convert $7$ to base $7$ dividing by $7$ (See previous section).Once that is established,$ 67 + 27$ will be two numbers past $6$, which is (11\) in base $7$ (That is $11_7$). Continuing, we can fill in the rows as we would in base $10$, but be aware that we are working in base $7$.

$Base 7 Chart .png$

Figure $\PageIndex{2}$: Base $7$ chart

Your Turn $\PageIndex{2}$: Base $7$ Addition Table

Example $\PageIndex{3}$: Adding Base Seven Numbers

Create the addition table for base $7.$ and add $536_{7}+ 433_{7}.$

Answer

\[
\begin{array}{r}
\phantom{+}536_{\text{seven}} \\
+ \phantom{0}433_{\text{seven}} \\
\hline
\phantom{+}1302_{\text{seven}}
\end{array}
\]

Let’s do for the place of one's first. According to the base $7$ addition table, $6_7 + 3_7 = 12_{7}.$ We will carry the $1$. Now, we do the “tens” place (it’s the sevens place). According to the base $7$ addition table, we have $ 1(\text{carried}) + 3_7 + 3_7 = 7 = 10_{7}$. We use the $0$ and carry the $1.$ Now the “hundreds” place (really, forty-nines place). There, we have $1(\text{carried}) + 5_7 + 4_7 = 10 = 13_{7}.$

So, \[536_{7}+ 433_{7} = 1302_{7}\nonumber\]

Your Turn $\PageIndex{3}$: Add in Different Bases

As seen previously, when performing addition in another base, set up the problem exactly as you would for addition in base $10.$ At each step, check the addition table for the base. As in base $10$ addition, move right to the left, adding down the columns using the rules in the addition table. When necessary and just as in base $10,$ be sure to carry the $1.$

Adding and Subtracting in a Base Higher Than Ten

As we know from the previous section, when the base is higher than $10,$ we usually use letters to represent digits beyond $9.$ For example:

Base $11$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}$, where $10=\text{A}$

Base $12$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}, \text{B}$, where $10=\text{A}, 11= \text{B}$

Base $13$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}, \text{B}, \text{C}$, where $10=\text{A}, 11= \text{B}, 12= \text{C}$

Base $14$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}, \text{B}, \text{C}, \text{D}$, where $10=\text{A}, 11= \text{B}, 12 =\text{C}, 13=\text{D}$

Base $15$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}, \text{B}, \text{C}, \text{D}, \text{E}$, where $10=\text{A}, 11= \text{B}, 12=\text{C}, 13= \text{D},14=\text{E}$

Base $16$ : Digits are $0,1,2,3,4,5,6,7,8,9,\text{A}, \text{B},..., \text{F}$, where $10=\text{A}, 11= \text{B}, 12=\text{C}, 13= \text{D},14=\text{E}, 15= \text{F}$

Let’s make an addition table for base $12,$ just as an example:

Example $\PageIndex{4}$: Adding Base Twelve Numbers

Create the addition table for base $12$ and calculate $3A7_{12}+ 9BA_{12}.$

Answer

When the sum is up to $9,$ we are good. If sum is $10,$ we use A for $10$ as a digit in base $12.$ When sum is $11$ we use B for $11$ as digit in base $12.$ What if sum is $12,$ for example, $6 + 6= 12,$ we do not have digits that represent $12$ in base $10.$ (We only have up to $11$, which is B)

In other words, since $12$ is not one of the digits in base $12.$ So we need to convert $12$ to base $12$ by using division.

$\begin{array}{ll}
12 \div 12= & 1\quad \mathrm{R} \quad 0 \\
\end{array}$

We'll say that $12=10_{12}$

Again, what is $9 + 7 $ in base $12?$ It is $16,$ not one of the digits in base $12.$ So we need to convert $16$ to base $12$ by division.

$\begin{array}{ll}
16\div 12 = & 1 \quad \mathrm{R} \quad 4 \\
\end{array}$

We'll say that $16=14_{12}.$

Finally, what is $B+B$? $ B (11) +B (11)=22$, and $22$ is more than base $12.$ So we need to convert $22$ to base $12$ by division.

$\begin{array}{ll}
22\div 12 = & 1 \quad \mathrm{R} \quad 10 (A) \\
\end{array}$

We'll say that $22=1A_{12}$

Base 12 Addition Table
+	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	A ($10$)	B ($11$)
$0$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	A	B
$1$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	A	B	$10$
$2$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	A	B	$10$	$11$
$3$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	A	B	$10$	$11$	$12$
$4$	$4$	$5$	$6$	$7$	$8$	$9$	A	B	$10$	$11$	$12$	$13$
$5$	$5$	$6$	$7$	$8$	$9$	A	B	$10$	$11$	$12$	$13$	$14$
$6$	$6$	$7$	$8$	$9$	A	B	$10$	$11$	$12$	$13$	$14$	$15$
$7$	$7$	$8$	$9$	A	B	$10$	$11$	$12$	$13$	$14$	$15$	$16$
$8$	$8$	$9$	A	B	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$
$9$	$9$	A	B	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$
A ($10$)	A	B	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$	$19$
B ($11$)	B	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$	$19$	1A

\[\begin{array}{r}
\phantom{+}3A7_{\text{12}} \\
+ \phantom{0}9BA_{\text{12}} \\
\hline
\phantom{+}11A5_{\text{12}}
\end{array}\nonumber\]

Let’s do the for the place of ones According to the base $12$ addition table, $7 + A = 15_{12}.$ We will carry the $1.$ Now, we do the “tens” place (it’s really a twelves place). According to the base $12$ addition table, we have $ 1 + A + B = 22 = 1A_{12}.$ So, like in base $10$, we use the $A$ and carry the $1.$ Now the “hundreds” place (really, $144$ s place). There, we have $1 + 3 + 9 = 13 = 11_{12}.$

So, \[3A7_{12}+ 9BA_{12} =11A5_{12}\nonumber\]

Subtraction in Bases Other Than Base $10$

Subtraction in bases other than base $10$ follows the same processes as base $10$ subtraction, but, as with addition, using the addition table for the base.

Example $\PageIndex{5}$: Subtracting in Base Lower than Ten

Calculate $52_{6} - 34_{6}.$
Calculate $563_{7} - 164_{7}.$

Answer 1

For the place of ones (right column), when we borrow a $1$ from the $5$, we borrow one group of $6$. After borrowing, $2$ becomes $2 + 6 =8,$ so $8-4 =4.$ Now top $5$ on the left column becomes new $4$ and $4 − 3 = 2.$

\[\begin{aligned}
\phantom{+}52_{\text{6}} \\
- \phantom{0}34_{\text{6}} \\
\hline
\phantom{+}14_{\text{6}}
\end{aligned}\nonumber\]

Answer 2

For the place of ones (right column), when we borrow a $1$ from the $6$, we borrow one group of $7$. After borrowing, $3$ becomes $3 + 7 =10$, so $10 - 4 =6.$ In the middle column, top $6$ becomes new $5.$ Again, we borrow $1,$ from $5$ (the left column's top number), we borrow one group of $7,$ new $5$ becomes $5+7=12,$ so $12-6 =6.$ In the left column (third column from the right), $5$ becomes new $4,$ and $4 − 1 = 3.$

\[\begin{array}{r}
\phantom{+}563_{\text{7}} \\
- \phantom{0}164_{\text{7}} \\
\hline
\phantom{+}366_{\text{7}}
\end{array}\nonumber\]

Your Turn $\PageIndex{5}$: Subtraction in Different Bases

Example $\PageIndex{6}$: Subtracting in Base $12$

Calculate $A17_{12}- 4B3_{12}$

Answer

\[\begin{array}{r}
\phantom{+}A17_{\text{12}} \\
- \phantom{0}4B3_{\text{12}} \\
\hline
\phantom{+}524_{\text{12}}
\end{array}\nonumber
\]

For one place, $7 - 3 = 2,$ all good. For middle column, we borrow $1,$ from $A =10,$ we borrow one group of $12,$ after borrowing $1$ becomes $1+12=13,$ so $13-B(11) =2.$ Now in third column from the right, $A=10,$ becomes new $9,$ and $9 − 4= 5.$

Your Turn $\PageIndex{6}$: Subtraction in Base Twelve

Errors When Adding and Subtracting in Bases Other Than Base 10

When computing in bases other than $10$, errors often involve applying base $10$ rules or symbols to an arithmetic problem in a base other than base $10$. The first type of error is using a symbol that is not in the symbol set for the base. For instance, if a $9$ shows up when working in base $7$, you know an error has happened because $9$ is not a legal symbol in base $7$.

Example $\PageIndex{7}$: Identifying an Illegal Symbol in Arithmetic in a Base Other Than Base $10$

Explain the error in the following calculation:

Answer: Since the problem is in base $6,$ the symbol set available is $0, 1, 2, 3, 4,$ and $5.$ The $9$ in the answer is clearly not a legal symbol for base $6.$ Returning to the base $6$ addition table, we see that . Correcting the error, we know the sum is .

Remember: The second type of error is using a base $10$ rule when the numbers are not in base $10.$ For instance, if you are working in base $13,$ then 9₁₃ + 9₁₃ is not 18₁₃, even though $18$ is the correct answer in base $10.$

Your Trun $\PageIndex{8}$: Identifying an Arithmetic Error in a Base Other Than Base $10$

Explain the error in the following calculation and correct the error:

\begin{array}{r}
\phantom{+}89_{\text{12}} \\
+ \phantom{0}76_{\text{12}} \\
\hline
\phantom{+}165_{\text{12}}
\end{array}

Answer

This would be the correct answer if this problem were a base $10$ problem. However, in base $12,$ $9 + 6$ is not $15,$ but is instead $13.$ To correct this error, use the addition table for base $12.$ If properly used, the correct answer would be , as seen below:

\begin{array}{r}
\phantom{+}89_{\text{12}} \\
+ \phantom{0}76_{\text{12}} \\
\hline
\phantom{+}143_{\text{12}}
\end{array}

This page titled 1.4: Addition and Subtraction in Base Systems was last modified on Sun, 26 Apr 2026 11:04:20 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Cheban Acharya via source content that was edited to the style and standards of the LibreTexts platform.

+	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A (\(10\))	B (\(11\))
\(0\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B
\(1\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)
\(2\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)
\(3\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)
\(4\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)
\(5\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)
\(6\)	\(6\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)
\(7\)	\(7\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)	\(16\)
\(8\)	\(8\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)	\(16\)	\(17\)
\(9\)	\(9\)	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)	\(16\)	\(17\)	\(18\)
A (\(10\))	A	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)	\(16\)	\(17\)	\(18\)	\(19\)
B (\(11\))	B	\(10\)	\(11\)	\(12\)	\(13\)	\(14\)	\(15\)	\(16\)	\(17\)	\(18\)	\(19\)	1A

Contents

Text

Learning Objectives

Addition in Bases Other Than Base 10

Careful

Your Turn \(\PageIndex{1}\): Adding Single Digit Number in Different Bases

Example \(\PageIndex{2}\): Adding Base Six Numbers

Your Turn \(\PageIndex{2}\): Base \(7\) Addition Table

Example \(\PageIndex{3}\): Adding Base Seven Numbers

Your Turn \(\PageIndex{3}\): Add in Different Bases

Adding and Subtracting in a Base Higher Than Ten

Example \(\PageIndex{4}\): Adding Base Twelve Numbers

Subtraction in Bases Other Than Base \(10\)

Example \(\PageIndex{5}\): Subtracting in Base Lower than Ten

Your Turn \(\PageIndex{5}\): Subtraction in Different Bases

Example \(\PageIndex{6}\): Subtracting in Base \(12\)

Your Turn \(\PageIndex{6}\): Subtraction in Base Twelve

Errors When Adding and Subtracting in Bases Other Than Base 10

Example \(\PageIndex{7}\): Identifying an Illegal Symbol in Arithmetic in a Base Other Than Base \(10\)

Your Trun \(\PageIndex{8}\): Identifying an Arithmetic Error in a Base Other Than Base \(10\)