Place values as powers of the base, the symbols each system uses, and converting a number written in binary, octal, duodecimal or hexadecimal into a decimal number.

The decimal system is not the only one possible - it is simply the one we are used to. Every system rests on the same principle: a place value is a power of the base. The base and the set of symbols change, everything else stays. This lesson shows four such systems and one conversion method that works in all of them.

Number systems - the decimal number system

	2	7	6	5,	3	2	3
place values	$10^{3}$	$10^{2}$	$10^{1}$	$10^{0}$	$10^{-1}$	$10^{-2}$	$10^{-3}$
	$1000$	$100$	$10$	$1$	$\tfrac{1}{10}$	$\tfrac{1}{100}$	$\tfrac{1}{1000}$

The place values are powers of 10, hence the decimal number system.

The binary number system (base 2)

	1	0	1	1,	1	0	1
place values	$2^{3}$	$2^{2}$	$2^{1}$	$2^{0}$	$2^{-1}$	$2^{-2}$	$2^{-3}$
	$8$	$4$	$2$	$1$	$\tfrac{1}{2}$	$\tfrac{1}{4}$	$\tfrac{1}{8}$

The place values are powers of 2.

Converting a binary number to a decimal number

\text{a)}\quad 1011.101_{2} = 11.625_{10}

\begin{aligned}&1 \cdot 8 + 0 \cdot 4 + 1 \cdot 2 + 1 \cdot 1 + 1 \cdot \tfrac{1}{2} + 0 \cdot \tfrac{1}{4} + 1 \cdot \tfrac{1}{8} \\&= 11 + \tfrac{1}{2} + \tfrac{1}{8} = 11\tfrac{5}{8} = 11.625\end{aligned}

\text{b)}\quad 1101.011_{2} = 13.375_{10}

\begin{aligned}&1 \cdot 8 + 1 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 + 0 \cdot \tfrac{1}{2} + 1 \cdot \tfrac{1}{4} + 1 \cdot \tfrac{1}{8} \\&= 13 + \tfrac{1}{4} + \tfrac{1}{8} = 13\tfrac{3}{8} = 13.375\end{aligned}

The small subscripts 2 and 10 indicate the bases of the two number systems.

The octal number system (base 8)

The octal number system uses the following symbols: 0, 1, 2, 3, 4, 5, 6, 7.

	3	5	7,	3	2	1
place values	$8^{2}$	$8^{1}$	$8^{0}$	$8^{-1}$	$8^{-2}$	$8^{-3}$
	$64$	$8$	$1$	$\tfrac{1}{8}$	$\tfrac{1}{64}$	$\tfrac{1}{512}$

The place values are powers of 8.

Converting an octal number to a decimal number

\text{a)}\quad 357.321_{8} = 239.408_{10}

\begin{aligned}&3 \cdot 64 + 5 \cdot 8 + 7 \cdot 1 + 3 \cdot \tfrac{1}{8} + 2 \cdot \tfrac{1}{64} + 1 \cdot \tfrac{1}{512} \\&= 192 + 40 + 7 + \tfrac{3}{8} + \tfrac{1}{32} + \tfrac{1}{512} \\&= 239\tfrac{209}{512} = 239.408\end{aligned}

correct to three decimal places

\text{b)}\quad 263.452_{8} = 179.582_{10}

\begin{aligned}&2 \cdot 64 + 6 \cdot 8 + 3 \cdot 1 + 4 \cdot \tfrac{1}{8} + 5 \cdot \tfrac{1}{64} + 2 \cdot \tfrac{1}{512} \\&= 128 + 48 + 3 + \tfrac{1}{2} + \tfrac{5}{64} + \tfrac{1}{256} \\&= 179\tfrac{149}{256} = 179.582\end{aligned}

correct to three decimal places

The duodecimal number system (base 12)

The duodecimal number system uses the following symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X, Λ.

	2	X	5,	1	3	6
place values	$12^{2}$	$12^{1}$	$12^{0}$	$12^{-1}$	$12^{-2}$	$12^{-3}$
	$144$	$12$	$1$	$\tfrac{1}{12}$	$\tfrac{1}{144}$	$\tfrac{1}{1728}$

The place values are powers of 12.

Converting a duodecimal number to a decimal number

\text{a)}\quad 2X5.136_{12} = 413.108_{10}

\begin{aligned}&2 \cdot 144 + 10 \cdot 12 + 5 \cdot 1 + 1 \cdot \tfrac{1}{12} + 3 \cdot \tfrac{1}{144} + 6 \cdot \tfrac{1}{1728} \\&= 288 + 120 + 5 + \tfrac{1}{12} + \tfrac{3}{144} + \tfrac{1}{288} \\&= 413\tfrac{31}{288} = 413.108\end{aligned}

correct to three decimal places

\text{b)}\quad 3\Lambda 4.265_{12} = 568.211_{10}

\begin{aligned}&3 \cdot 144 + 11 \cdot 12 + 4 \cdot 1 + 2 \cdot \tfrac{1}{12} + 6 \cdot \tfrac{1}{144} + 5 \cdot \tfrac{1}{1728} \\&= 432 + 132 + 4 + \tfrac{1}{6} + \tfrac{1}{24} + \tfrac{5}{1728} \\&= 568\tfrac{365}{1728} = 568.211\end{aligned}

correct to three decimal places

The hexadecimal number system

The hexadecimal number system uses the following symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.

	2	A	7,	3	E	2
place values	$16^{2}$	$16^{1}$	$16^{0}$	$16^{-1}$	$16^{-2}$	$16^{-3}$
	$256$	$16$	$1$	$\tfrac{1}{16}$	$\tfrac{1}{256}$	$\tfrac{1}{4096}$

The place values are powers of 16.

Converting a hexadecimal number to a decimal number

\text{a)}\quad 2A7.3E2_{16} = 679.243_{10}

\begin{aligned}&2 \cdot 256 + 10 \cdot 16 + 7 \cdot 1 + 3 \cdot \tfrac{1}{16} + 14 \cdot \tfrac{1}{256} + 2 \cdot \tfrac{1}{4096} \\&= 512 + 160 + 7 + \tfrac{3}{16} + \tfrac{7}{128} + \tfrac{1}{2048} \\&= 679\tfrac{497}{2048} = 679.243\end{aligned}

correct to three decimal places

\text{b)}\quad 3C4.21F_{16} = 964.133_{10}

\begin{aligned}&3 \cdot 256 + 12 \cdot 16 + 4 \cdot 1 + 2 \cdot \tfrac{1}{16} + 1 \cdot \tfrac{1}{256} + 15 \cdot \tfrac{1}{4096} \\&= 768 + 192 + 4 + \tfrac{1}{8} + \tfrac{1}{256} + \tfrac{15}{4096} \\&= 964\tfrac{543}{4096} = 964.133\end{aligned}

correct to three decimal places

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Frequently asked questions

How do you convert a binary number to a decimal number?

Multiply each digit by its place value, that is by the appropriate power of two, and add the results. For 1011.101 that gives 8 + 2 + 1 + 1/2 + 1/8, which is 11.625.

What symbols does the hexadecimal system use?

The digits 0 to 9 and the letters A to F, where A stands for ten and F for fifteen. That way each position holds sixteen different values using a single character.

What does the small subscript next to a number mean?

The base of the system the number is written in. The notation 1011₂ = 11₁₀ reads: the number 1011 in base two is 11 in base ten.

Why does the duodecimal system need extra symbols?

Because a single position has to hold twelve values, and there are only ten digits from 0 to 9. The two missing ones are written with the separate symbols X and Λ, standing for ten and eleven.

A lesson from the PhiBoard knowledge base: phiboard.com/en/knowledge-base/from-zero-to-engineer/number-systems. On the lesson page you will also find games and practice exercises - free and without an account.

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Converting numbers from other systems to decimal - an alternative method