Skip to main content
Mathematics LibreTexts

4.2: Other Base Systems

  • Page ID
    132881
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Previously, we used cups, pints, etc. to model the grouping that forms the place values in a base two system. It is helpful to have a more abstract way to model this that can be applied to any base. For this, we have multibase blocks. Most people have seen the blocks that are used for a base ten system because they are used in many classrooms. When the base is known, we can use that when referring to the different types. Below, we will look at base three blocks, which are the multibase blocks used for working in base three.

    The ones place is also called the units place. This is why the fundamental type of block in multibase blocks is the unit block. We simply refer to it as a unit. Just like how for the number three we must jump to the threes place in base three, when we group three units together, we create a new kind of block: a long. When we group three longs together, we get a new shape to represent the nines place: a flat. Different people use different terminology for the next type of multibase block. When three flats are grouped together, we get a new shape to represent the twenty-sevens place. Some people call this a "cube", others a "big cube", and still others a "block". For now, we will use "cube". After that, instead of creating new names, we will just put multiple names together. Three cubes will form a "cube-long" (or "long cube" or "long block"). Three cube-longs will form a "cube-flat" (or "flat cube" or "flat block"). At some point (possibly one we have already passed), this becomes tedious. But that's a good time to transition from using manipulatives to using the symbols only. Below is an amateur drawing of what a unit, long, flat and cube or block really look like in base three.

    base three unit

    Screen Shot 2021-04-18 at 12.05.03 PM.png

    base three long

    Screen Shot 2021-04-18 at 12.05.12 PM.png

    base three flat

    Screen Shot 2021-04-18 at 12.05.23 PM.png

    base three cube or block

    Screen Shot 2021-04-18 at 12.05.33 PM.png

     

    Notice that in base three, a unit is made up of 1 small cube, a long is made up of 3 units (small cubes), a flat is made up of 9 units (small cubes), a cube or block is made up of 27 units (small cubes), a cube-long (or long cube or long block) is made up of 81 units (small cubes), a cube-flat (or flat cube or flat block) is made up of 243 units (small cubes), a "cube-cube" (or "block block") is made up of 729 units (small cubes), and so on. These are the place values in base three numerals.

    Think about how you would form the base two blocks. What would a long look like? A flat? The rest? After thinking about this and possibly drawing pictures, continue reading.

    Below is an amateur drawing of a unit, long, flat and block in base two. Note that the pictures are not quite accurately scaled.

    base two unit

    Screen Shot 2021-04-18 at 12.31.42 PM.png

    base two long

    Screen Shot 2021-04-18 at 12.32.03 PM.png

    base two flat

    Screen Shot 2021-04-18 at 12.32.12 PM.png

    base two cube or block

    Screen Shot 2021-04-18 at 12.32.19 PM.png

    In base two, a long is made up of 2 units, a flat is made up of 4 units, a cube or block is made up of 8 units, a cube-long (or long cube or long block) is made up of 16 units, a cube-flat (or flat cube or flat block) is made up of 32 units, a cube-cube (or block block) is made up of 64 units, and so on.

    On your own, try to draw a picture of a base four unit, long, flat and cube (or block).

    Example \(\PageIndex{1}\)

    Fill in the blanks.

    In base four, it takes ____ units to make a long, it takes ____ longs to make a flat, it takes ____ flats to make a cube, it takes ____ cube to make a cube-long, and it takes ____ cube-longs to make a cube-flat.

    In base four, a long is made up of ______ units, a flat is made up of ______ units, a cube is made up of ______ units, and a cube-long is made up of ______ units.

    In base five, it takes ______ flats to make a cube.

    In base six, a flat is made up of ______ units.

    Solution

    The blanks should have the following numbers in them: 4, 4, 4, 4, 4, 4, 16, 64, 256, 5, and 36.

    Previously you practiced converting from other bases to base ten; now we'll focus on converting from base ten to other bases. There is more than one way to accomplish this task, but we will focus on only one method with an aim to understand what is actually happening.

    Example \(\PageIndex{1}\)

    Represent 42 in base two.

    Solution

    To model 42 with base two blocks, we have some options. We could use 42 units. But this would be unnecessarily cluttered. Let's imagine exchanging them to have fewer individual pieces.

    Imagine exchanging as many units as possible to form longs. Since each long is made of 2 units, we could convert all 42 units into 21 longs. Check to make sure you agree with this. So, we now have 21 longs and 0 units.

    Now imagine exchanging as many longs as possible to form flats. Since each flat is made of 2 longs, we could convert 20 longs into 10 flats and have one long left over. Do you agree with this? Now we have 10 flats, 1 long, and 0 units.

    Now imagine exchanging as many flats as possible to form cubes. Since each cube is made of 2 flats, we could convert all 10 flats into 5 cubes. Do you agree with this? Now we have 5 cubes, 0 flats, 1 long, and 0 units.

    Before we keep going, let's see why we know we can keep going. Since there are 5 cubes and we only need 2 to move up to the next shape size (because we are in base two), we keep going!

    Now imagine exchanging as many cubes as possible to form cube-longs. Since each cube-long is made of 2 cubes, we could convert 4 cubes into 2 cube-longs with one cube left over. Do you agree with this? Now we have 2 cube-longs, 1 cubes, 0 flats, 1 long, and 0 units.

    Now imagine exchanging as many cube-longs as possible to form cube-flats. Since each cube-flat is made of 2 cube-longs, we could convert both cube-longs into 1 cube-flat. Do you agree with this? Now we have 1 cube-flat, 0 cube-longs, 1 cube, 0 flats, 1 long, and 0 units.

    We are finished. Make sure you understand why we know this!

    This means \(42_{\text{ten}} = 101010_{\text{two}}\).

    Try to redo the example above but with base three. After you do this, read the next example.

    Example \(\PageIndex{1}\)

    Represent 42 in base three.

    Solution

    Let's start with 42 units again. Now we imagine we are using (or we actually use) base three blocks.

    • Exchange as many units as possible to get longs. Since each long is 3 units, all 42 units can be exchanged for 14 longs. We have 14 longs and 0 units.
    • Exchange as many longs as possible to get flats. Since each flat is 3 longs, 12 longs can be exchanged for 4 flats with 2 longs left over. We have 12 flats, 2 longs, and 0 units.
    • Exchange as many flats as possible to get cubes. Since each cube is 3 flats, 3 flats can be exchanged for 1 cube with 1 flat left over. We have 1 cube, 1 flat, 2 longs, and 0 units.

    We stop now (make sure you understand why!) and have \(42_{\text{ten}} = 1120_{\text{three}}\).

    Try again but with base four. If you did it correctly, you should have \(42_{\text{ten}} = 222_{\text{four}}\).

    Now, let's look at that problem (converting to base four) but discuss it with groups of values rather than blocks.

    Example \(\PageIndex{1}\)

    Represent 42 in base four.

    Solution
    • We start with 42 ones.
    • From those 42 ones, create as many groups of 4 ones as possible. Since 42 ones = 40 ones + 2 ones and 40 ones is 10 groups of 4 ones , we can think of 42 as 10 fours and 2 ones.
    • From the 10 fours, create as many groups of 4 fours as possible. Since 10 fours = 8 fours + 2 fours and 8 fours is 2 groups of 4 fours, we can think of 10 fours as 2 groups of 4 fours and 2 fours left over. Since 4 fours is sixteen, we have changes the 10 fours to 2 sixteens and 2 fours left over. Put together with the previous work, we have 2 sixteens, 2 fours, and 2 ones.

    We stop now because we only have 2 sixteens (not enough to form a group of 4). Thus, \(42_{\text{ten}} = 222_{\text{four}}\).

    If we only needed to convert relatively small numbers and we always had multibase blocks available in any base we needed to convert to, we could continue to convert to various bases by using the blocks and doing exchanges. But this is not always the case. Consider if you were asked to convert 999 to base four. This would be pretty time-consuming and tedious. So, it is important to see how it works without the manipulatives.

    The previous example probably got a little confusing because of all the numbers. We can make this a little easier by keeping things organized in a chart. Let's look at an example.

    Example \(\PageIndex{1}\)

    Represent 82 in base six.

    Solution

    Let's make a chart that corresponds to place values in base six. Since the place values (from left to right) in base six are ones, sixes, thirty-sixes, two hundred sixteens, etc. we only need three places (since 216 > 82). We will label the columns of a table with the base ten values of the places. We start with 82 in the ones place.

    36 6 1
        82

    Now, we find how many groups of 6 are in 82. Since \(82 \div 6 = 13 \frac{4}{6}\), we have 13 groups of 6 and 4 left over. In other words, we have 13 sixes and 4 ones. This means we write 13 in the 6 column, cross out the 82 and replace it with a 4 in the 1 column.

    36 6 1
      13 82 4

    Now, we find how many groups of 6 are in 13. Since \(13 \div 6 = 2 \frac{1}{6}\), we have 2 groups of 6 and 1 left over. In other words, we have 2 thirty-sixes (since 6 groups of 6 equals 36) and 1 six. This means we write 2 in the 36 column, cross out the 13 and replace it with a 1 in the 6 column.

    36 6 1
    2 13 1 82 4

    We confirm that we don't need another digit by noticing that 2 < 6. So, we are finished, and \(82_{\text{ten}} = 214_{\text{six}}\).

    Try to represent 477 in base five. After doing this, read the next example.

    Example \(\PageIndex{1}\)

    Represent 477 in base five.

    Solution

    Since 477 is greater than \(5^3 = 125\) but less than \(5^4 = 625\), we will need four places. The chart we create looks like this.

    125 25 5 1
    3 19 4 95 0 477 2

    So, \(477_{\text{ten}} = 3402_{\text{five}}\).

    Try redoing the previous example but converting to base twelve instead of five. Notice you need one fewer place. Did you get \(477_{\text{ten}} = 339{\text{twelve}}\)? If not, see if you can find your error.

    Try again but converting to base three. Now you'll need more places, 6 of them! Did you get \(477_{\text{ten}} = 122200{\text{three}}\)?

    Now, let's end with a little extension our exploration a bit. Ask yourself the following questions. Share your responses with a classmate, a tutor, or your teacher and discuss them.

    1. In base ten, what is the largest eight digit numeral?
    2. In base ten, what is the smallest ten digit numeral?
    3. In base three, what is the largest seven digit numeral?
    4. In base three, what is the smallest seven digit numeral?
    5. In base three, what number comes after \(1202010_{\text{three}}\)?
    6. In base three, what number comes after \(2100212_{\text{three}}\)?
    7. In base three, what number comes before \(1202221_{\text{three}}\)?
    8. In base three, what number comes before \(2110100_{\text{three}}\)?
    9. What is wrong with the numeral \(1022301_{\text{three}}\)?
    10. Can you count to \(113_{\text{four}}\) in base four without ever leaving base four?
    11. Can you count to \(10100_{\text{two}}\) in base two (binary) withouth ever leavning base two?

    This page titled 4.2: Other Base Systems is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Julie Harland via source content that was edited to the style and standards of the LibreTexts platform.