Number Bases

Lessons

Senior 3 · Mathematics

Number Bases

5 lessons

Lesson 3 of 5

Adding and subtracting in other bases

25 min

The same column method, a different trigger

Inside every phone and computer, numbers are added in base two, billions of times a second, by the same column method you learned in primary school. The only change from base ten is the moment you carry: in base $b$ you carry as soon as a column reaches $b$ , not 10. Subtraction changes in the same way: a borrowed 1 is worth $b$ , not 10.

Definition

Carry: when the digits in a column add up to the base $b$ or more, write (sum $- b$ ) in that column and add 1 to the next column on the left.

Notation: the carried 1 is written small, above the next column.

Example: in base five, $4 + 3 = 7 = 1 \times 5 + 2$ : write 2 and carry 1.

Definition

Borrow: when a top digit is smaller than the digit below it, take 1 from the next column on the left. In base $b$ that 1 is worth $b$ units of the column you are working on.

Notation: cross out the digit you borrowed from and write it one smaller.

Example: in base five, for $2 - 3$ , borrow: $2 + 5 = 7$ , and $7 - 3 = 4$ .

Important

Why the base is the trigger. In base $b$ , each place is worth $b$ times the place on its right. So $b$ units of one column can always be exchanged for 1 unit of the next column (a carry), and 1 unit of a column can be exchanged for $b$ units of the column on its right (a borrow). Neither exchange changes the value of the number.

A carry exchanges 5 ones for 1 five; a borrow exchanges 1 five for 5 ones. Neither exchange changes the value of the number.

Addition facts in base five

In base five, any sum of 5 or more is written with two digits. Learn to use this table, as you once learned the base-ten addition facts.

$+$	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	10
2	2	3	4	10	11
3	3	4	10	11	12
4	4	10	11	12	13

For example, $4 + 3 = 12_{5}$ (one five and two ones) and $3 + 2 = 10_{5}$ .

Worked example 1: adding in base five

Work out $342_{5} + 143_{5}$ .

	$5^{3}$	$5^{2}$	$5^{1}$	$5^{0}$
carry	1	1	1
		3	4	2
$+$		1	4	3
$=$	1	0	4	0

Ones: $2 + 3 = 5 = 10_{5}$ . Write 0, carry 1.
Fives: $4 + 4 + 1 = 9 = 14_{5}$ . Write 4, carry 1.
Twenty-fives: $3 + 1 + 1 = 5 = 10_{5}$ . Write 0, carry 1 into a new column.

So $342_{5} + 143_{5} = 1040_{5}$ . Check in base ten: $97 + 48 = 145$ , and $1040_{5} = 125 + 20 = 145$ . ✓

Worked example 2: adding in base two

Work out $10111_{2} + 1101_{2}$ .

	$2^{5}$	$2^{4}$	$2^{3}$	$2^{2}$	$2^{1}$	$2^{0}$
carry	1	1	1	1	1
		1	0	1	1	1
$+$			1	1	0	1
$=$	1	0	0	1	0	0

In base two the only facts you need are $1 + 1 = 10_{2}$ and $1 + 1 + 1 = 11_{2}$ .

$2^{0}$ : $1 + 1 = 10_{2}$ . Write 0, carry 1.
$2^{1}$ : $1 + 0 + 1 = 10_{2}$ . Write 0, carry 1.
$2^{2}$ : $1 + 1 + 1 = 11_{2}$ . Write 1, carry 1.
$2^{3}$ : $0 + 1 + 1 = 10_{2}$ . Write 0, carry 1.
$2^{4}$ : $1 + 0 + 1 = 10_{2}$ . Write 0, carry 1 into a new column.

So the sum is $100100_{2}$ . Check: $23 + 13 = 36 = 32 + 4$ . ✓

Worked example 3: adding in base eight and in base sixteen

(a) $5674_{8} + 2357_{8}$

	$8^{4}$	$8^{3}$	$8^{2}$	$8^{1}$	$8^{0}$
carry	1	1	1	1
		5	6	7	4
$+$		2	3	5	7
$=$	1	0	2	5	3

Ones: $4 + 7 = 11 = 1 \times 8 + 3$ : write 3, carry 1. Eights: $7 + 5 + 1 = 13 = 8 + 5$ : write 5, carry 1. Next: $6 + 3 + 1 = 10 = 8 + 2$ : write 2, carry 1. Next: $5 + 2 + 1 = 8$ : write 0, carry 1. Answer $10253_{8}$ . Check: $3004 + 1263 = 4267 = 4096 + 128 + 40 + 3$ . ✓

(b) $9\mathrm{C}7_{16} + 4\mathrm{A}9_{16}$

	$16^{2}$	$16^{1}$	$16^{0}$
carry	1	1
	9	C	7
$+$	4	A	9
$=$	E	7	0

Ones: $7 + 9 = 16$ : write 0, carry 1. Sixteens: $\mathrm{C} + \mathrm{A} + 1 = 12 + 10 + 1 = 23 = 16 + 7$ : write 7, carry 1. Next: $9 + 4 + 1 = 14 = \mathrm{E}$ : write E. Answer $\mathrm{E}70_{16}$ . Check: $2503 + 1193 = 3696 = 14 \times 256 + 7 \times 16$ . ✓

Worked example 4: subtracting in base five, with a chain of borrows

Work out $4012_{5} - 1243_{5}$ .

	$5^{3}$	$5^{2}$	$5^{1}$	$5^{0}$
top after borrowing (in base ten)	3	4	5	7
	4	0	1	2
$-$	1	2	4	3
$=$	2	2	1	4

Ones: $2 \lt 3$ . Borrow 1 five from the fives column: $2 + 5 = 7$ and $7 - 3 = 4$ . The fives digit drops from 1 to 0.
Fives: $0 \lt 4$ , and the twenty-fives digit is also 0. Borrow from the 125s column: 4 becomes 3 and the twenty-fives column gets 5. It lends 1 to the fives column, keeping 4, so the fives column has $0 + 5 = 5$ , and $5 - 4 = 1$ .
Twenty-fives: $4 - 2 = 2$ . 125s: $3 - 1 = 2$ .

So $4012_{5} - 1243_{5} = 2214_{5}$ . Check: $507 - 198 = 309 = 2 \times 125 + 2 \times 25 + 5 + 4$ . ✓

Worked example 5: subtracting in base two and in base sixteen

(a) $110010_{2} - 10111_{2}$

	$2^{5}$	$2^{4}$	$2^{3}$	$2^{2}$	$2^{1}$	$2^{0}$
top after borrowing (in base ten)	0	2	1	1	2	2
	1	1	0	0	1	0
$-$		1	0	1	1	1
$=$		1	1	0	1	1

Each borrow in base two brings 2 into the column: for example $0 - 1$ becomes $2 - 1 = 1$ . Answer $11011_{2}$ . Check: $50 - 23 = 27 = 16 + 8 + 2 + 1$ . ✓

(b) $\mathrm{B}04_{16} - 3\mathrm{E}7_{16}$

	$16^{2}$	$16^{1}$	$16^{0}$
top after borrowing (in base ten)	10	15	20
	B	0	4
$-$	3	E	7
$=$	7	1	D

Ones: $4 \lt 7$ ; the sixteens digit is 0, so borrow from B: it becomes A, the sixteens column gets 16 and lends 1, keeping 15. Ones: $4 + 16 = 20$ , $20 - 7 = 13 = \mathrm{D}$ . Sixteens: $15 - 14 = 1$ . Last column: $\mathrm{A} - 3 = 10 - 3 = 7$ . Answer $71\mathrm{D}_{16}$ . Check: $2820 - 999 = 1821 = 7 \times 256 + 16 + 13$ . ✓

Warning

Do not think in base ten by accident. In base eight, $7 + 5 = 12$ in base ten, which is written $14_{8}$ : write 4 and carry 1, not "write 2, carry 1". And a borrow in base five adds 5, not 10. When in doubt, work the column out in base ten, then change the result into the base.

Tip

Check a subtraction by adding. The answer plus the number you subtracted must give the number you started with, and that check uses the same base: $2214_{5} + 1243_{5} = 4012_{5}$ .

Jen Laughlin: Subtracting in Bases (4 min). Watch how each borrow adds the base, not 10, to the column that needs to be made bigger.

✦

Practice

Show the carries and borrows. Check each answer in base ten. The last three are harder.

$34_{5} + 23_{5}$
$1011_{2} + 111_{2}$
$657_{8} + 435_{8}$
$8\mathrm{D}_{16} + 5\mathrm{F}_{16}$
$402_{5} - 143_{5}$
$10100_{2} - 1011_{2}$
$6030_{8} - 2547_{8}$
Find the digit $x$ if $3 x 4_{5} + 132_{5} = 1031_{5}$ .

Show answersHide answers

$112_{5}$ ( $19 + 13 = 32$ )
$10010_{2}$ ( $11 + 7 = 18$ )
$1314_{8}$ ( $431 + 285 = 716$ )
$EC_{16}$ ( $141 + 95 = 236$ )
$204_{5}$ ( $102 - 48 = 54$ )
$1001_{2}$ ( $20 - 11 = 9$ )
$3261_{8}$ ( $3096 - 1383 = 1713$ )
$1031_{5} - 132_{5} = 344_{5}$ , so $x = 4$ . Check: $99 + 42 = 141$ .

Sources and credits

4.3: Addition and Subtraction in Base Systems, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The carry and borrow tables, the base-five addition facts and the worked examples. Written by Ganzaa.
Video: Subtracting in Bases, Jen Laughlin, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1The same column method, a different trigger

2Addition facts in base five

The same column method, a different trigger

Addition facts in base five