{"id":610,"date":"2023-03-07T19:59:26","date_gmt":"2023-03-07T19:59:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=610"},"modified":"2024-10-18T20:52:29","modified_gmt":"2024-10-18T20:52:29","slug":"positional-systems-and-bases-learn-it-3","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/positional-systems-and-bases-learn-it-3\/","title":{"raw":"Positional Systems and Bases: Learn It 3","rendered":"Positional Systems and Bases: Learn It 3"},"content":{"raw":"

Converting from Base [latex]10[\/latex] to Other Bases<\/h2>\r\n

Converting from an unfamiliar base to the familiar decimal system is not that difficult once you get the hang of it. It\u2019s only a matter of identifying each place and then multiplying each digit by the appropriate power. However, going the other direction can be a little trickier. Suppose you have a base-ten number and you want to convert to base-five.<\/p>\r\n

\r\n

How to: Convert from Base-[latex]10[\/latex] to Base-[latex]b[\/latex]<\/strong><\/p>\r\n

    \r\n\t
  1. Find the highest power of the base-[latex]b[\/latex] that will divide into the given number at least once and then divide.<\/li>\r\n\t
  2. Write down the whole number part, then use the remainder from the division in the next step.<\/li>\r\n\t
  3. Repeat step two, dividing by the next highest power of the base-[latex]b[\/latex], writing down the whole number part (including [latex]0[\/latex]), and using the remainder in the next step.<\/li>\r\n\t
  4. Continue until the remainder is smaller than the base. This last remainder will be in the \u201cones\u201d place.<\/li>\r\n\t
  5. Collect all your whole number parts to get your number in base-[latex]b[\/latex] notation.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
    \r\n

    Remember, the remainder is not the decimal place when dividing a number. A remainder is the amount left over after dividing a number by another number. In division, the remainder is the part that's left after dividing as evenly as possible. <\/p>\r\n

    For example, when you divide [latex]7[\/latex] by [latex]3[\/latex], you get the number [latex]2.33[\/latex]. The remainder is not [latex]33[\/latex]; the quotient is [latex]2[\/latex], and the remainder is [latex]1[\/latex], because [latex]
    \r\n7=3\u00d72+1[\/latex]. Think of the remainder as what's \"remaining\" after you've done the division!<\/p>\r\n<\/section>\r\n

    Let\u2019s start with a worked example before you try it on your own.<\/p>\r\n

    Convert the base-ten number [latex]348[\/latex] to base-five.\r\n\r\nThe powers of five are:\r\n\r\n

    [latex]5^{0} = 1[\/latex]
    \r\n[latex]5^{1} = 5[\/latex]
    \r\n[latex]5^{2} = 25[\/latex]
    \r\n[latex]5^{3} = 125[\/latex]
    \r\n[latex]5^{4} = 625[\/latex]
    \r\nEtc\u2026<\/p>\r\n

    Since [latex]348[\/latex] is smaller than [latex]625[\/latex], but bigger than [latex]125[\/latex], we see that [latex]5^{3} = 125[\/latex] is the highest power of five present in [latex]348[\/latex]. So we divide [latex]125[\/latex] into [latex]348[\/latex] to see how many of them there are:<\/p>\r\n

    [latex]348 \\div 125 = \\textcolor{red}{2}[\/latex] with remainder [latex]98[\/latex]<\/p>\r\n

    We write down the whole part, [latex]2[\/latex], and continue with the remainder. There are [latex]98[\/latex] left over, so we see how many [latex]25[\/latex]s (the next smallest power of five) there are in the remainder:<\/p>\r\n

    [latex]98 \u00f7 25 = \\textcolor{red}{3}[\/latex] with remainder [latex]23[\/latex]<\/p>\r\n

    We write down the whole part, [latex]2[\/latex], and continue with the remainder. There are [latex]23[\/latex] left over, so we look at the next place, the [latex]5[\/latex]s:<\/p>\r\n

    [latex]23 \u00f7 5 = \\textcolor{red}{4}[\/latex] with remainder [latex]\\textcolor{red}{3}[\/latex]<\/p>\r\n

    This leaves us with [latex]3[\/latex], which is less than our base, so this number will be in the \u201cones\u201d place. We are ready to assemble our base-five number:<\/p>\r\n

    [latex]348 = (\\textcolor{red}{2}\u00a0\u00d7 5^{3}) + (\\textcolor{red}{3} \u00d7 5^{2}) + (\\textcolor{red}{4} \u00d7 5^{1}) + (\\textcolor{red}{3}\u00a0\u00d7 1)[\/latex]<\/p>\r\n

    Hence, our base-five number is [latex]\\textcolor{red}{2343}[\/latex]. We\u2019ll say that [latex]348_{10}=2343_{5}[\/latex].<\/p>\r\n<\/section>\r\n

    [ohm2_question hide_question_numbers=1]3163[\/ohm2_question]<\/section>","rendered":"

    Converting from Base [latex]10[\/latex] to Other Bases<\/h2>\n

    Converting from an unfamiliar base to the familiar decimal system is not that difficult once you get the hang of it. It\u2019s only a matter of identifying each place and then multiplying each digit by the appropriate power. However, going the other direction can be a little trickier. Suppose you have a base-ten number and you want to convert to base-five.<\/p>\n

    \n

    How to: Convert from Base-[latex]10[\/latex] to Base-[latex]b[\/latex]<\/strong><\/p>\n

      \n
    1. Find the highest power of the base-[latex]b[\/latex] that will divide into the given number at least once and then divide.<\/li>\n
    2. Write down the whole number part, then use the remainder from the division in the next step.<\/li>\n
    3. Repeat step two, dividing by the next highest power of the base-[latex]b[\/latex], writing down the whole number part (including [latex]0[\/latex]), and using the remainder in the next step.<\/li>\n
    4. Continue until the remainder is smaller than the base. This last remainder will be in the \u201cones\u201d place.<\/li>\n
    5. Collect all your whole number parts to get your number in base-[latex]b[\/latex] notation.<\/li>\n<\/ol>\n<\/section>\n
      \n

      Remember, the remainder is not the decimal place when dividing a number. A remainder is the amount left over after dividing a number by another number. In division, the remainder is the part that’s left after dividing as evenly as possible. <\/p>\n

      For example, when you divide [latex]7[\/latex] by [latex]3[\/latex], you get the number [latex]2.33[\/latex]. The remainder is not [latex]33[\/latex]; the quotient is [latex]2[\/latex], and the remainder is [latex]1[\/latex], because [latex]
      7=3\u00d72+1[\/latex]. Think of the remainder as what’s “remaining” after you’ve done the division!<\/p>\n<\/section>\n

      Let\u2019s start with a worked example before you try it on your own.<\/p>\n

      Convert the base-ten number [latex]348[\/latex] to base-five.<\/p>\n

      The powers of five are:<\/p>\n

      [latex]5^{0} = 1[\/latex]
      \n[latex]5^{1} = 5[\/latex]
      \n[latex]5^{2} = 25[\/latex]
      \n[latex]5^{3} = 125[\/latex]
      \n[latex]5^{4} = 625[\/latex]
      \nEtc\u2026<\/p>\n

      Since [latex]348[\/latex] is smaller than [latex]625[\/latex], but bigger than [latex]125[\/latex], we see that [latex]5^{3} = 125[\/latex] is the highest power of five present in [latex]348[\/latex]. So we divide [latex]125[\/latex] into [latex]348[\/latex] to see how many of them there are:<\/p>\n

      [latex]348 \\div 125 = \\textcolor{red}{2}[\/latex] with remainder [latex]98[\/latex]<\/p>\n

      We write down the whole part, [latex]2[\/latex], and continue with the remainder. There are [latex]98[\/latex] left over, so we see how many [latex]25[\/latex]s (the next smallest power of five) there are in the remainder:<\/p>\n

      [latex]98 \u00f7 25 = \\textcolor{red}{3}[\/latex] with remainder [latex]23[\/latex]<\/p>\n

      We write down the whole part, [latex]2[\/latex], and continue with the remainder. There are [latex]23[\/latex] left over, so we look at the next place, the [latex]5[\/latex]s:<\/p>\n

      [latex]23 \u00f7 5 = \\textcolor{red}{4}[\/latex] with remainder [latex]\\textcolor{red}{3}[\/latex]<\/p>\n

      This leaves us with [latex]3[\/latex], which is less than our base, so this number will be in the \u201cones\u201d place. We are ready to assemble our base-five number:<\/p>\n

      [latex]348 = (\\textcolor{red}{2}\u00a0\u00d7 5^{3}) + (\\textcolor{red}{3} \u00d7 5^{2}) + (\\textcolor{red}{4} \u00d7 5^{1}) + (\\textcolor{red}{3}\u00a0\u00d7 1)[\/latex]<\/p>\n

      Hence, our base-five number is [latex]\\textcolor{red}{2343}[\/latex]. We\u2019ll say that [latex]348_{10}=2343_{5}[\/latex].<\/p>\n<\/section>\n