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Quantitative Wisdom Numerical literacy from the perspective of Biblical wisdom

Dr. Ryan Johnson

Section 6.1 Apportionment

Subsection 6.1.1 Historical Context for This Chapter

The U.S. government was officially formed in 1787. During the Constitutional Convention, the framers were tasked with making decisions on the three branches of government and how people (or those considered to be people) would be represented. At that time only white men with property were allowed to vote.
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In creating the Legislative Branch of government, it was decided that Congress would be made up of the Senate and the House of Representatives. Representation for the Senate would be equal with two senators per state. To account for the different sizes of the states, the representation in the House of Representatives would be proportional to the state’s population, but there was a controversy over how to count the population.
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Representatives and direct Taxes shall be apportioned among the several States which may be included within this Union, according to their respective Numbers, which shall be determined by adding to the whole Number of free Persons, including those bound to Service for a Term of Years, and excluding Indians not taxed, three fifths of all other Persons. The actual Enumeration shall be made within three Years after the first Meeting of the Congress of the United States, and within every subsequent Term of ten Years, in such Manner as they shall by Law direct. The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at Least one Representative; and until such enumeration shall be made, the State of New Hampshire shall be entitled to chuse three, Massachusetts eight, Rhode Island and Providence Plantations one, Connecticut five, New-York six, New Jersey four, Pennsylvania eight, Delaware one, Maryland six, Virginia ten, North Carolina five, South Carolina five, and Georgia three.
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―United States Constitution, Article I, Section 2, Clause 3
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Subsection 6.1.2 What is Apportionment?

The number of representatives each state gets is based on its population, so once the framers decided how to count the population, they had to figure out how to divide up the representatives. This math problem is called apportionment.
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Apportionment is the problem of dividing up a fixed number of things among groups of different sizes. In the United States, there is a certain number of representatives as stated in the constitution, currently 435, and they need to be divided fairly among the 50 states. Since the states are different sizes, and we cannot use fractions of people, this is not an easy task. In addition, the population in each state may change over time. Every 10 years after the census is taken, the representatives are reapportioned.
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The apportionment problem comes up in a variety of non-political areas too. Here are the rules for apportionment in general.
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Apportionment rules.

  1. The things being divided up can exist only in whole numbers.
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  2. We must use all of the things being divided up, and we cannot use any more.
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  3. Each group must get at least one of the things being divided up.
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  4. The number of things assigned to each group should be at least approximately proportional to the population of the group. (Exact proportionality isn’t possible because of the whole number requirement, but we should try to be close.)
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In terms of the apportionment of the United States House of Representatives, these rules imply:
  1. We can only have whole representatives (a state can’t have 3.4 representatives).
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  2. We can only use the 435 representatives available.
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  3. Every state gets at least one representative.
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  4. The number of representatives each state gets should be approximately proportional to the state population. This way, the number of constituents each representative has should be approximately equal.
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We will look at four ways of solving the apportionment problem, developed by different people: Hamilton, Jefferson, Webster, and the Huntington-Hill method that is used today. We will continue to look at U.S. history throughout this section including Hamilton, Jefferson and Webster’s relationship to slavery.
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Subsection 6.1.3 Hamilton’s Method

Alexander Hamilton (1755 - 1804) was raised in St. Croix in the U.S. Virgin Islands by a poor family. He was motivated by his low socioeconomic status to work himself into a higher social standing. He eventually came to the colonies and worked himself into circles of wealth and influence. While Hamilton wasn’t pro-slavery and considered himself an abolitionist, when choosing between his societal status and moral obligation, he chose the former.
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He wed Elizabeth Schuyler, who was from a prominent family who owned slaves. He was involved with the transactions of slaves for his in-laws which further muddled his anti-slavery stance. Furthermore, Hamilton also traded and sold slaves as part of his duties for the Continental Army.
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Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791 but was vetoed by President Washington. It was later adopted in 1852 and used through 1910. He begins by determining, to several decimal places, how many people each representative should represent (the divisor).
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Hamilton’s Method.

  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 divisor.
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  2. Divide each state’s 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.
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    Since we can only allocate whole representatives, Hamilton resolved the whole number problem as follows:
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  3. Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). This is the initial apportionment and will always be less than or equal to the total number of representatives. Add up the whole numbers.
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  4. If the total from Step 3 is less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.
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    Make sure that each state ends up with at least one representative.
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Let’s see how this works in an example.
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Example 6.1.1.

A new state called Graceland has three counties: A, B and C. The State House of Representatives has 41 members. If the legislature wants to divide this representation along county lines (which is not required, but let’s pretend they do), let’s use Hamilton’s method to apportion them. The populations of the counties are as follows:
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County Population
A 162,310
B 538,479
C 197,145
Total 897,934
Step 1: First, we divide the total population by the number of representatives to find the divisor: \(897934/41 = 21900.82927\text{.}\)
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Step 2: Now we determine each county’s quota by dividing the county’s population by the divisor: For example, for A, we take \(162,310/21,900.82927\) which equals 7.4111.
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County Population Quota
A 162,310 7.4111
B 538,479 24.5872
C 197,945 9.0017
Total 897,934
Step 3: Removing the decimal parts of the quotas gives our initial apportionment and we add those numbers up.
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County Population Quota Initial
A 162,310 7.4111 7
B 538,479 24.5872 24
C 197,945 9.0017 9
Total 897,934 40
Step 4: We need 41 representatives, and right now we only have 40. The remaining one goes to the county with the largest decimal part, which is B:
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County Population Quota Initial Final
A 162,310 7.4111 7 7
B 538,479 24.5872 24+1 25
C 197,945 9.0017 9 9
Total 897,934 40 41
Our final apportionment is A: 7, B: 25, C: 9, for a total of 41 representatives.
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Example 6.1.2.

We will use Hamilton’s method again, to apportion 75 seats for a new state of Lewis, which has five counties. Lewis and its counties are named for civil rights leaders John Lewis, Rosa Parks, Dr. Martin Luther King Jr., Ella Baker, Daisy Bates and Roy Wilkins.
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Solution.
Step 1: The divisor is \(1,052,567/75 = 14,034.22667\text{.}\)
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Step 2: Determine each county’s quota by dividing its population by the divisor:
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County Population Quota
Parks 49,875 3.5538
King 166,158 11.8395
Baker 82,888 5.9061
Bates 626,667 44.6528
Wilkins 126,979 9.0478
Total 1,052,567
Step 3: Remove the decimal part of each quota and add up the initial apportionment:
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County Population Quota Initial
Parks 49,875 3.5538 3
King 166,158 11.8395 11
Baker 82,888 5.9061 5
Bates 626,667 44.6528 44
Wilkins 126,979 9.0478 9
Total 1,052,567 72
Step 4: We need 75 representatives and we only have 72, so we assign the remaining three, one each, to the three counties with the largest decimal parts, which are Baker, King, and Bates in that order:
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County Population Quota Initial Final
Parks 49,875 3.5538 3 3
King 166,158 11.8395 11+1 12
Baker 82,888 5.9061 5+1 6
Bates 626,667 44.6528 44+1 45
Wilkins 126,979 9.0478 9 9
Total 1,052,567 72 75
Our final apportionment is Parks: 3, King: 12, Baker: 6, Bates: 45, and Wilkins: 9 for a total of 75 representatives.
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Note that even though Parks County’s decimal part is greater than .5, it isn’t big enough to get an additional representative, because three other counties have greater decimal parts.
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Hamilton’s method obeys something called the Quota Rule. The Quota Rule isn’t a law, but an idea that some people think is a good one.
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Quota Rule.

The Quota Rule says that the final number of representatives a state gets should be within one of that state’s quota. Since we’re dealing with whole numbers for our final answers, that means that each state should either go up to the next whole number above its quota, or down to the next whole number below its quota.
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Subsection 6.1.4 Problems with Hamilton’s Method

After using Hamilton’s method for many years, three paradoxes happened, on separate occasions, where unfair things happened in new apportionments. This led to other methods being needed.
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The Alabama Paradox is named for an incident that happened during the apportionment that took place after the 1880 census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-1880 apportionment had been completed, using Hamilton’s method and the new population numbers from the census. Then it was decided that because of the country’s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton’s method and the same 1880 census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren’t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process, even though no populations were changed, and the total number of representatives increased.
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The New States Paradox happened when Oklahoma became a state in 1907. Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another thing also happened: Maine gained a representative (from New York).
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The Population Paradox happened between the apportionments after the census of 1900 and of 1910. In those ten years, Virginia’s population grew at an average annual rate of 1.07%, while Maine’s grew at an average annual rate of 0.67%. Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn’t mean that Virginia should gain representatives or Maine shouldn’t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine.
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Subsection 6.1.5 Jefferson’s Method

Thomas Jefferson proposed a different method for apportionment. After Washington vetoed Hamilton’s method, Jefferson’s 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’s. Primarily, his method favors larger states, and his own home state of Virginia was the largest at the time. He would also argue that it’s the ratio of people to representatives that is the critical thing, and apportionment methods should be based on that.
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The first three steps of Jefferson’s method are the same as Hamilton’s. He found the same divisor and the same quota and cut off the decimal parts in the same way, giving the same initial apportionment that is less than the required total.
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What changes is how Jefferson assigned the remaining representatives. He said that since we ended up with an answer that is too small, our divisor must have been too big. He changed the divisor by making it smaller and looked at the new total. He would raise or lower the divisor until he found one that produced the required total. This is a trial and error process that takes some patience.
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Jefferson’s Method.

The first three steps are the same as Hamilton’s
  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 divisor.
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  2. Divide each state’s 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.
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  3. Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). This is the initial apportionment and will always be less than or equal to the total number of representatives. Add up the whole numbers.
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  4. If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If you lower it too far, increase it. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor you end up using is called the modified divisor. This is a trial-and-error process.
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Example 6.1.3.

We’ll return to the state of Graceland and apply Jefferson’s method. We begin, as we did with Hamilton’s method, by finding the quotas with the original divisor, 21,900.82927.
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Steps 1-3 are the same as Hamilton’s Method. Since we will be using different divisors, we will write the divisor at the bottom of the population column to be clear what we are dividing each number in that column by. You can also write it above that column if you prefer.
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County Population Quota Initial
A 162,310 7.4111 7
B 538,479 24.5872 24
C 197,145 9.0017 9
Total 897,934 40
Divisor ÷21,900.82927
We need 41 representatives, and this divisor gives only 40. We must reduce the divisor until we get 41 representatives. Let’s try 21,500 as the divisor
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Step 4:
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County Population Quota Initial
A 162,310 7.5493 7
B 538479 25.0455 25
C 197,145 9.1695 9
Total 897,934 41
Divisor ÷21,500
That worked and our final apportionment is A: 7, B: 25, and C: 9.
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Notice that with the new, lower divisor, the quota for B County (the largest county in the state) increased by much more than those of A County or C County.
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In this example, we got lucky and found the modified divisor on the first try. If we still had 40, we would reduce the divisor more. If we had more than 41, we would need to raise it. We will show how to do the trial-and-error part in the next example.
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Example 6.1.4.

We’ll apply Jefferson’s method for Lewis. The original divisor of 14,034.22667 gave these results:
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County Population Quota Initial
Parks 49,875 3.5538 3
King 166,158 11.8395 11
Baker 82,888 5.9061 5
Bates 626,667 44.6528 44
Wilkins 126,979 9.0478 9
Total 1,052,567 72
Divisor ÷14,034.22667
We need 75 representatives and we only have 72, so we need to use a smaller divisor. Let’s try lowering it to 13,500:
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County Population Quota Initial
Parks 49,875 3.6944 3
King 166,158 12.3080 12
Baker 82,888 6.1399 6
Bates 626,667 46.4198 46
Wilkins 126,979 9.4059 9
Total 1,052,567 76
Divisor ÷13,500
We got a total of 76 representatives which is too many, so we lowered it too far. We need a divisor that’s greater than 13,500 but less than 14,034.22667. Let’s try 13,700:
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County Population Quota Initial
Parks 49,875 3.6405 3
King 166,158 12.1283 12
Baker 82,888 6.0502 6
Bates 626,667 45.7421 45
Wilkins 126,979 9.2685 9
Total 1,052,567 75
Divisor ÷13,700
Using a modified divisor of 13,700 gives us exactly 75 representatives. Note there is usually more than one modified divisor that will work.
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This can take a lot of writing, so in practice, we can write this all out in one table. With each try we are dividing the population by the new divisor to get the new quotas.
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County Population Quota Initial 2nd Quota 2nd Try 3rd Quota Final
Parks 49,875 3.5538 3 3.6944 3 3.6405 3
King 166,158 11.8395 11 12.3080 12 12.1283 12
Baker 82,888 5.9061 5 6.1399 6 6.0502 6
Bates 626,667 44.6528 44 46.4198 46 45.7421 45
Wilkins 126,979 9.0478 9 9.4059 9 9.2685 9
Total 1,052,567 72 76 75
Divisor ÷14,034.22667 ÷13,500 ÷13,700
Notice, in comparison to Hamilton’s method, that although the results were the same, they came about in a different way, and the outcome was almost different. Bates County (the largest) almost went up to 46 representatives before King (which is much smaller) got to 12. Although that didn’t happen here, it can. Divisor-adjusting methods like Jefferson’s are not guaranteed to follow the quota rule.
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Subsection 6.1.6 Webster’s Method

Daniel Webster (1782-1852) was a lawyer, congressman, Senator of Massachusetts and also served as the U.S. Secretary of State. He was from the North and did not own slaves. He was an opponent of slavery extension and he spoke against annexing Texas and against going to war with Mexico. He argued, however, that no law was needed to prevent the further extension of slavery in new states and he supported the Compromise of 1850, which disappointed his abolitionist supporters (History.com Editors, 2018b).
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Webster proposed a method similar to Jefferson’s in 1832. It was adopted by Congress in 1842 but replaced by Hamilton’s method in 1852. It was then adopted again in 1911. The difference is that Webster rounded the quotas to the nearest whole number rather than dropping the decimal parts. If that didn’t produce the exact number of representatives, he adjusted the divisor like in Jefferson’s method. (In Jefferson’s case, the first adjustment will always be to make the divisor smaller. That is not always the case with Webster’s method because some numbers may be rounded up.)
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Webster’s Method.

Steps 1-2 are the same as Hamilton and Jefferson
  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 divisor.
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  2. Divide each state’s 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.
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  3. Round all the quotas to the nearest whole number (but don’t forget what the decimals were). This is the initial apportionment. Add up the whole numbers.
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  4. If the total from Step 3 is less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 3 is larger than the total number of representatives, increase 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. This is a trial-and-error process.
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Let’s see how Webster’s method works in this example:
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Example 6.1.5.

We will look at Graceland again, with an initial divisor of 21,900.82927 and 41 representatives:
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County Population Quota Initial
A 162,310 7.4111 7
B 538,479 24.5872 25
C 197,145 9.0017 9
Total 897,934 41
Divisor ÷21,900.82927
This time B is the only county with a decimal of 0.5 or higher, so it gets rounded up. This gives the required total, so we’re done.
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Example 6.1.6.

Let’s look at Lewis again, with an initial divisor of 14,034.22667 and 75 representatives:
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County Population Quota Initial
Parks 49,875 3.5538 4
King 166,158 11.8395 12
Baker 82,888 5.9061 6
Bates 626,667 44.6528 45
Wilkins 126,979 9.0478 9
Total 1,052,567 76
Divisor ÷14,034.22667
This is one too many, so we need to increase the divisor. Let’s try 14,100:
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County Population Quota Initial
Parks 49,875 3.5372 4
King 166,158 11.7843 12
Baker 82,888 5.8786 6
Bates 626,667 44.4445 44
Wilkins 126,979 9.0056 9
Total 1,052,567 75
Divisor ÷14,100
This gives us exactly 75, so we’re done. This is how it would look all in one table:
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County Population Quota Initial 2nd Quota 2nd Try
Parks 49,875 3.5538 4 3.5371 4
King 166,158 11.8395 12 11.7843 12
Baker 82,888 5.9061 6 5.8786 6
Bates 626,667 44.6528 45 44.4445 44
Wilkins 126,979 9.0478 9 9.0056 9
Total 1,052,567 76 75
Divisor ÷14,034.22667 ÷14,100
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Like Jefferson’s method, Webster’s method carries a bias in favor of larger states but rounding the quotas to the nearest whole number greatly reduces this bias. Notice that Bates County, the largest, is the one that gets a representative trimmed because of the increased quota.
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Also, like Jefferson’s method, Webster’s method does not always follow the quota rule, but it follows the quota rule much more often than Jefferson’s method does. In fact, if Webster’s method had been applied to every apportionment of Congress in all of U.S. history, it would have followed the quota rule every single time.
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Subsection 6.1.7 Land Rights, Citizenship and Voting Rights

In 1830 the Indian Removal Act was signed by President Jackson. The Act relocated Native Americans to land west of the Mississippi and the forced journey is known as the Trail of Tears. To learn more, here is a link to an online museum exhibit with a timeline of Native American history and artifacts
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http://recordsofrights.org/themes/4/rights-of-native-americans
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Toward the end of the Civil War, slaves were freed by Abraham Lincoln’s Emancipation Proclamation on January 1st, 1863, but it was not until June 19th, 1865 that the last group of slaves in Texas learned they had been freed. This is now celebrated as Juneteenth
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https://www.juneteenth.com/history.htm
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https://www.ferris.edu/htmls/news/jimcrow/timeline/slavery.htm
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Although they were technically free, Black people still faced unequal and unjust treatment. After Black men were granted the right to vote in 1869, voter suppression laws like poll taxes and literacy tests were implemented barriers to voting (American Civil Liberties Union, 2020).
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In 1887, the Dawes Act was passed, which allotted Native American land to individuals and 60 million acres were taken by the government or non-Indian homesteaders (Indian Land Trust Tenure Foundation, 2020). This act also provided citizenship under certain restrictions which theoretically allowed native men to vote. All women were granted the right to vote in 1920, but the barriers enacted for men of color applied to women of color as well. You can read more about voting rights and suppression in this online timeline
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https://www.carnegie.org/topics/topic-articles/voting-rights/voting-rights-timeline/
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Subsection 6.1.8 Huntington-Hill Method

In 1920, no new apportionment was done, because Congress couldn’t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster’s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the method of apportionment still used today.
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The Huntington-Hill Method is similar to Webster’s method, but attempts to minimize the difference in the percentage of how many people each representative will represent.
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Huntington-Hill Method.

The first two steps are the same as the previous methods
  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 divisor.
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  2. Divide each state’s 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.
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  3. Cut off the decimal part of the quota to obtain the lower quota, which we’ll call \(n\text{.}\) Compute \(\sqrt{n(n+1)}\text{,}\) which is the geometric mean of the lower quota and one value higher.
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  4. Instead of using 0.5 to round like we are used to, we use the geometric mean. If the quota is as large or larger than the geometric mean, round up; if the quota is smaller than the geometric mean, round down. This is the initial allocation. Add up the resulting whole numbers.
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  5. If the total from Step 4 is less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 is larger than the total number of representatives, increase the divisor and recalculate. Continue doing this until the total is equal to the exact number of representatives. The divisor we end up using is called the modified divisor. This is a trial-and-error process.
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Subsection 6.1.9 What is the Geometric Mean?

Let’s look at why Hill and Huntington decided to use the geometric mean for rounding instead of the arithmetic mean of 0.5, which is halfway between two numbers. By calculating some geometric means and looking at them in a table we can see a pattern:
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\(n\) \(\sqrt{n(n+1)}\)
1 \(\sqrt{(1(1+1)}=\sqrt{1(2)}\approx 1.414\)
2 \(\sqrt{(2(2+1)}=\sqrt{2(3)}\approx 2.449\)
3 \(\sqrt{(3(3+1)}=\sqrt{3(4)}\approx 3.464\)
4 \(\sqrt{(4(4+1)}=\sqrt{4(5)}\approx 4.472\)
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10 \(\sqrt{(10(10+1)}=\sqrt{10(11)}\approx 10.488\)
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100 \(\sqrt{(100(100+1)}=\sqrt{100(101)}\approx 100.499\)
Table 6.1.7. Geometric Mean Pattern
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Notice that each geometric mean is between \(n\) and \(n+1\text{.}\) For smaller numbers, the decimal part is less than 0.5 and as n gets larger, the decimal part gets closer and closer to 0.5. This gives smaller states a chance to round up before larger states. That’s how the Hill-Huntington method reduces the bias in favor of larger states.
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Now let’s see how to use the geometric mean in an example.
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Example 6.1.8.

Again, we will practice with Graceland, with an initial divisor of 21,900.82927 and 41 representatives:
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County Population Quota Lower Quota Geometric Mean Initial
A 162,310 7.4111 7 \(\sqrt{7(8)} \approx 7.48\) 7
B 538,479 24.5872 24 \(\sqrt{24(25)} \approx 24.49\) 25
C 197,945 9.0017 9 \(\sqrt{9(10)} \approx 9.49\) 9
Total 897,934 41
Divisor ÷21,900.82927
Notice that for B, the quota of 24.5872 is above the geometric mean of 24.49, so we round up to 25. This gives the required total, so we’re done.
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Example 6.1.9.

Again, here is Lewis, with an initial divisor of 14,034.22667:
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County Population Quota Lower Quota Geometric Mean Initial
Parks 49,875 3.5538 3 \(\sqrt{3(4)} \approx 3.46\) 4
King 166,158 11.8395 11 \(\sqrt{11(12)} \approx 11.49\) 12
Baker 82,888 5.9061 5 \(\sqrt{5(6)} \approx 5.48\) 6
Bates 626,667 44.6528 44 \(\sqrt{44(45)} \approx 44.50\) 45
Wilkins 126,979 9.0478 9 \(\sqrt{9(10)} \approx 9.49\) 9
Total 1,052,567 76
Divisor ÷14,034.22667
We end up with 76 which is too many, so we need to increase the divisor. Let’s try 14,100:
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County Population Quota Lower Quota Geometric Mean 2nd Try
Parks 49,875 3.5372 3 \(\approx 3.46\) 4
King 166,158 11.7843 11 \(\approx 11.49\) 12
Baker 82,888 5.8786 5 \(\approx 5.48\) 6
Bates 626,667 44.4445 44 \(\approx 44.50\) 44
Wilkins 126,979 9.0056 9 \(\approx 9.49\) 9
Total 1,052,567 75
Divisor ÷14,100
This time Bates had a quota less than its geometric mean, so it did not get rounded up and we have exactly 75 representatives.
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In both these cases, the apportionment produced by the Huntington-Hill method was the same as those from Webster’s method, but that will not always be the case.
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In 1980, two mathematicians, Peyton Young and Mike Balinski, proved what we now call the Balinski-Young Impossibility Theorem.
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Balinski-Young Impossibility Theorem.

The Balinski-Young Impossibility Theorem shows that any apportionment method which always follows the quota rule will be subject to the possibility of paradoxes like the Alabama, New States, or Population paradoxes. In other words, we can choose a method that avoids those paradoxes, but only if we are willing to give up the guarantee of following the quota rule.
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Example 6.1.10.

Consider a small country with 5 states, two of which are much larger than the others. We need to apportion 70 representatives. Apportion the representatives using both Webster’s method and the Huntington-Hill method.
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State Population
A 300,500
B 200,000
C 50,000
D 38,000
E 21,500
Solution.
Step 1: The total population is 610,000. Dividing this by the 70 representatives gives the divisor: 8714.286.
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Step 2: Dividing each state’s population by the divisor gives the quotas.
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State Population Quota
A 300,500 34.48361
B 200,000 22.95082
C 50,000 5.737705
D 38,000 4.360656
E 21,500 2.467213
Total 610,000
Divisor ÷8714.286
Webster’s Method
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Step 3: Using Webster’s method, we round each quota to the nearest whole number using the rounding rule of 0.5 or higher to round up.
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State Population Quota Initial
A 300,500 34.48361 34
B 200,000 22.95082 23
C 50,000 5.737705 6
D 38,000 4.360656 4
E 21,500 2.467213 2
Total 610,000 69
Divisor ÷8714.286
Step 4: Adding these up only gives us 69 representatives, so we adjust the divisor down. We try 8,700, which gives us 70 representatives. Notice that State A, the largest state, is the one that got rounded up the second time.
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State Population Quota Initial
A 300,500 34.54023 35
B 200,000 22.98851 23
C 50,000 5.747126 6
D 38,000 4.367816 4
E 21,500 2.471264 2
Total 610,000 70
Divisor ÷8,700
Huntington-Hill Method
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Step 3: Using the Huntington-Hill method, we cut off the decimal to find the lower quota, then calculate the geometric mean based on each lower quota. If the quota is less than the geometric mean, we round down; if the quota is more than the geometric mean, we round up.
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State Population Quota Lower
Quota		 Geometric
Mean		 Initial
A 300,500 34.48361 34 34.49638 34
B 200,000 22.95082 23 22.49444 23
C 50,000 5.737705 6 5.477226 6
D 38,000 4.360656 4 4.472136 4
E 21,500 2.467213 2 2.44949 3
Total 610,000 70
Divisor ÷8714.286
These allocations add up to 70. Notice that this allocation is different than that produced by Webster’s method. In this case, state E, which is smaller, got one more seat and state A got one less.
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In this section we have learned four different methods of apportionment used in U.S. history, intertwined with the U.S. treatment of women, Native Americans and Black people. In the next section we will continue with our brief history of voting rights and look at different methods used for voting.
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Exercises 6.1.10 Exercises

Exercise Group.

In exercises 1-14, determine the apportionment using
  1. Hamilton’s Method
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  2. Jefferson’s Method
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  3. Webster’s Method
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  4. Huntington-Hill Method
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1.
A college offers tutoring in Math, English, Chemistry, and Biology. The number of students enrolled in each subject is listed below. If the college can only afford to hire 15 tutors, determine how many tutors should be assigned to each subject.
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Math English Chemistry Biology
330 265 130 70
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2.
Reapportion the previous problem if the college can hire 20 tutors.
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3.
The number of salespeople assigned to work during a shift is apportioned based on the average number of customers during that shift. Apportion 20 salespeople given the information below.
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Shift Morning Midday Afternoon Evening
Avereage number
of customers		 95 305 435 515
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4.
Reapportion the previous problem if the store has 25 salespeople.
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5.
Three people invest in a treasure dive, each investing the amount listed below. The dive results in 36 gold coins. Apportion those coins to the investor.
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Aisha Basir Carlos
$7,600 $5,900 $1,400
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6.
Reapportion the previous problem if 37 gold coins are recovered.
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7.
A small country consists of four states, whose populations are listed below. If the legislature has 116 seats, apportion the seats.
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A B C D
33,700 559,500 141,300 89,100
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8.
Reapportion the previous problem with 124 seats.
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9.
A small country consists of five states, whose populations are listed below. If the legislature has 119 seats, apportion the seats.
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A B C D E
810,000 473,000 292,000 594,000 211,000
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10.
Reapportion the previous problem with 126 seats.
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11.
A small country consists of six states, whose populations are listed below. If the legislature has 200 seats, apportion the seats.
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A B C D E F
3,411 2,421 11,586 4,494 3,126 4,962
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12.
Reapportion the previous problem with 180 seats
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13.
A small country consists of six states, whose populations are listed below. If the legislature has 250 seats, apportion the seats.
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A B C D E F
82,500 84,600 96,000 98,000 356,500 382,500
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14.
Reapportion the previous problem with 240 seats.
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15.

A small country consists of three states, whose populations are listed below.
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A B C
6,000 6,000 2,000
  1. If the legislature has 10 seats, use Hamilton’s method to apportion the seats.
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  2. If the legislature grows to 11 seats, use Hamilton’s method to apportion the seats.
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  3. Explain what happened in part b. What do you think would be a fair solution?
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  4. Try Jefferson’s method for 11 seats. Does that solve the problem?
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16.

A small country consists of three states, whose populations are listed below.
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A B C
10,000 10,000 1,000
  1. If the legislature has 10 seats, use Hamilton’s method to apportion the seats. Which rule is not met in this case?
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  2. If the legislature grows to 11 seats, use Hamilton’s method to apportion the seats.
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  3. If there could only be 10 seats, what do you think would be a fair solution?
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17.

A state with five counties has 50 seats in their legislature. Using Hamilton’s method, apportion the seats based on the 2010 census, then again using the 2020 census. Explain what happened in 2020 apportionment. Do you think it is fair?
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County 2000 Population 2010 Population
Douglass 60,000 60,000
Parks 31,200 31,200
King 69,200 72,400
Du Bois 81,600 81,600
Lewis 118,000 118,400
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18.

A state with five counties has 62 seats in their legislature. Using Hamilton’s method, apportion the seats based on the 2010 census, then again using the 2010 census. Explain what happened in 2020 apportionment. Do you think it is fair?
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County 2000 Population 2010 Population
Candycane Forest 75,000 83,200
Licorice Castle 89,000 89,000
Gumdrop Mountain 32,500 32,500
Molasses Swamp 153,000 153,000
Frosted Plains 109,000 112,000
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19.

A school district has two high schools: Clatsop, serving 1715 students, and Siletz, serving 7364. The district could only afford to hire 13 guidance counselors.
  1. Determine how many counselors should be assigned to each school using Hamilton’s method.
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  2. The following year, the district expands to include a third school, Cayuse, serving an additional 2989 students. Based on the divisor from above, how many additional counselors should be hired for Cayuse?
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  3. After hiring that many new counselors, the district recalculates the reapportion using Hamilton’s method. Determine the outcome.
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  4. Explain what happened in the new apportionment. Do you think the outcome from part c is fair? Why or why not.
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20.

A school district has two middle schools: Tubman, serving 451 students, and Blackshear, serving 176. The district could afford to hire 8 art teachers.
  1. Determine how many art teachers should be assigned to each school using Hamilton’s method.
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  2. The following year, the district expands to include a third school, Banneker, serving 231 additional students. Based on the divisor from above, how many additional art teachers should be hired for the new school?
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  3. After hiring that many new art teachers, the district recalculates the reapportion using Hamilton’s method. Determine the outcome.
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  4. Explain what happened in the new apportionment. Do you think the outcome from part c is fair? Why or why not.
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