{"id":8900,"date":"2023-10-11T20:10:23","date_gmt":"2023-10-11T20:10:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=8900"},"modified":"2024-10-18T20:58:37","modified_gmt":"2024-10-18T20:58:37","slug":"apportionment-learn-it-3","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/apportionment-learn-it-3\/","title":{"raw":"Apportionment: Learn It 3","rendered":"Apportionment: Learn It 3"},"content":{"raw":"

Jefferson\u2019s Method<\/h2>\r\n

Thomas Jefferson proposed a different method for apportionment. After Washington vetoed Hamilton\u2019s method, Jefferson\u2019s method was adopted, and used in Congress from 1791 through 1842. Jefferson, of course, had political reasons for wanting his method to be used rather than Hamilton\u2019s. Primarily, his method favors larger states, and his own home state of Virginia was the largest in the country at the time. He would also argue that it\u2019s the ratio of people to representatives that is the critical thing, and apportionment methods should be based on that. But the paradoxes we saw also provide mathematical reasons for concluding that Hamilton\u2019s method isn\u2019t so good, and while Jefferson\u2019s method might or might not be the best one to replace it, at least we should look for other possibilities.<\/p>\r\n

\r\n
\r\n

Jefferson's method<\/h3>\r\n
    \r\n\t
  1. Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor<\/strong>.<\/li>\r\n\t
  2. Divide each state\u2019s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota<\/strong>.<\/li>\r\n\t
  3. Cut off all the decimal parts of all the quotas (but don\u2019t forget what the decimals were). These are the lower quotas<\/strong>. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives.<\/li>\r\n\t
  4. If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor we end up using is called the modified divisor<\/strong> or adjusted divisor<\/strong>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section>\r\n
    \r\n

    The first steps of Jefferson\u2019s method are the same as Hamilton\u2019s method. He finds the same divisor and the same quota, and cuts off the decimal parts in the same way, giving a total number of representatives that is less than the required total. The difference is in how Jefferson resolves that difference. He says that since we ended up with an answer that is too small, our divisor must have been too big. He changes the divisor by making it smaller, finding new quotas with the new divisor, cutting off the decimal parts, and looking at the new total, until we find a divisor that produces the required total.<\/p>\r\n<\/section>\r\n

    \r\n

    We\u2019ll return to Delaware and apply Jefferson\u2019s method. As a reminder, the state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members.<\/p>\r\n

    The populations of the counties are as follows (from the 2010 Census):<\/p>\r\n

    [latex]\\begin{array}{lr} \\text { County } & \\text { Population } \\\\ \\hline \\text { Kent } & 162,310 \\\\ \\text { New Castle } & 538,479 \\\\ \\text { Sussex } & 197,145 \\\\ \\textbf{ Total } & \\bf{ 897,934 }\\end{array}[\/latex]<\/p>\r\n [reveal-answer q=\"4331\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"4331\"]\r\n\r\n\r\n

    We begin, as we did with Hamilton\u2019s method, by finding the quotas with the original divisor, [latex]21,900.82927[\/latex].<\/p>\r\n

    [latex]\\begin{array}{lrrc} \\text { County } & \\text { Population } & \\text{ Quota } & \\text{ Initial } \\\\ \\hline \\text { Kent } & 162,310 & 7.4111 & 7 \\\\ \\text { Newport } & 538,479 & 24.5872 & 24 \\\\ \\text { Providence } & 197,145 & 9.0017 & 9 \\\\ \\textbf{ Total } & \\bf{ 897,934 } & & \\bf{ 40 }\\end{array} [\/latex]<\/p>\r\n

    We need [latex]41[\/latex] representatives, and this divisor gives only [latex]40[\/latex]. We must reduce the divisor until we get [latex]41[\/latex] representatives. Let\u2019s try [latex]21,500[\/latex] as the divisor:<\/p>\r\n

    [latex]\\begin{array}{lrrc} \\text { County } & \\text { Population } & \\text{ Quota } & \\text{ Initial } \\\\ \\hline \\text { Kent } & 162,310 & 7.5493 & 7 \\\\ \\text { Newport } & 538,479 & 25.0455 & 25 \\\\ \\text { Providence } & 197,145 & 9.1695 & 9 \\\\ \\textbf{ Total } & \\bf{ 897,934 } & & \\bf{ 41 }\\end{array} [\/latex]<\/p>\r\n

    This gives us the required [latex]41[\/latex] representatives, so we\u2019re done. If we had fewer than [latex]41[\/latex], we\u2019d need to reduce the divisor more. If we had more than [latex]41[\/latex], we\u2019d need to choose a divisor less than the original but greater than the second choice.<\/p>\r\n

    Notice that with the new, lower divisor, the quota for New Castle County (the largest county in the state) increased by much more than those of Kent County or Sussex County.<\/p>\r\n\r\n\r\n[\/hidden-answer]<\/section>\r\n

    \r\n

    [ohm2_question hide_question_numbers=1]13231[\/ohm2_question]<\/p>\r\n<\/section>\r\n

    Notice, in comparison to Hamilton\u2019s method, that although the results were the same, they came about in a different way, and the outcome was almost different. Although that didn\u2019t happen here, it can. Divisor-adjusting methods like Jefferson\u2019s are not guaranteed to follow the quota rule!<\/p>","rendered":"

    Jefferson\u2019s Method<\/h2>\n

    Thomas Jefferson proposed a different method for apportionment. After Washington vetoed Hamilton\u2019s method, Jefferson\u2019s method was adopted, and used in Congress from 1791 through 1842. Jefferson, of course, had political reasons for wanting his method to be used rather than Hamilton\u2019s. Primarily, his method favors larger states, and his own home state of Virginia was the largest in the country at the time. He would also argue that it\u2019s the ratio of people to representatives that is the critical thing, and apportionment methods should be based on that. But the paradoxes we saw also provide mathematical reasons for concluding that Hamilton\u2019s method isn\u2019t so good, and while Jefferson\u2019s method might or might not be the best one to replace it, at least we should look for other possibilities.<\/p>\n

    \n
    \n

    Jefferson’s method<\/h3>\n
      \n
    1. Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor<\/strong>.<\/li>\n
    2. Divide each state\u2019s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota<\/strong>.<\/li>\n
    3. Cut off all the decimal parts of all the quotas (but don\u2019t forget what the decimals were). These are the lower quotas<\/strong>. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives.<\/li>\n
    4. If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor we end up using is called the modified divisor<\/strong> or adjusted divisor<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<\/section>\n
      \n

      The first steps of Jefferson\u2019s method are the same as Hamilton\u2019s method. He finds the same divisor and the same quota, and cuts off the decimal parts in the same way, giving a total number of representatives that is less than the required total. The difference is in how Jefferson resolves that difference. He says that since we ended up with an answer that is too small, our divisor must have been too big. He changes the divisor by making it smaller, finding new quotas with the new divisor, cutting off the decimal parts, and looking at the new total, until we find a divisor that produces the required total.<\/p>\n<\/section>\n

      \n

      We\u2019ll return to Delaware and apply Jefferson\u2019s method. As a reminder, the state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has [latex]41[\/latex] members.<\/p>\n

      The populations of the counties are as follows (from the 2010 Census):<\/p>\n

      [latex]\\begin{array}{lr} \\text { County } & \\text { Population } \\\\ \\hline \\text { Kent } & 162,310 \\\\ \\text { New Castle } & 538,479 \\\\ \\text { Sussex } & 197,145 \\\\ \\textbf{ Total } & \\bf{ 897,934 }\\end{array}[\/latex]<\/p>\n