What is Fair? 1.) Martha and Ray want to divide the last piece of cake their mother baked last night. Propose a method for dividing the piece of cake that will seem fair to both of Martha and Ray. 2.) Juan and Mary are the only heirs to their mother’s estate. The only object of significant value is the house in which they were raised. Propose a method for fairly dividing the house. 3.) The sophomore, junior, and senior classes of Central High School have 333, 288, and 279 students, respectively. The school’s student council is composed of 30 members divided among the three classes. Determine a fair number of seats for each class.

Estate Division (Who’s getting what?) Upon her death, a mother leaves a farmhouse, a townhouse, and a 20 acres of land. The four heirs cannot decide on who gets what and a fair price to sell the items at, so they must be distributed in the fairest way possible. The following chart shows each heir and their bid on all items. State each person’s fair share and their final settlement. Ruth Steve Tina Uma Farmhouse $380,000 $350,000 $325,000 $400,000 Townhouse $250,000 $260,000 $300,000 $280,000 Land $100,000 $140,000 $125,000 $130,000

Total of Bids Fair Share Ruth Steve Tina Uma

Amount Fair Share - Item = owed/needed Ruth Steve Tina Uma

BANK amount of money in bank at 0 beginning of problem amount Tina owes + amount Uma owes amount Ruth needs amount Steve needs left over money (give this to each person)

Ruth Steve Tina Uma Final Settlement:

Estate Division (Who’s getting what?) The Jonas Brothers have decided to spilt and venture out on their solo careers. They have employed you to award their four most prized possessions and $5,000 from their Holiday savings account. The chart below shows the possessions and their bids for each item. Find each brothers fair share and their final settlement. Hair Care Camp Rock Grammy Teen Choice Products T-shirts Award Surf boards Nick $4,000 $2,000 $3,000 $12,000 Joe $3,500 $1,500 $2,750 $13,000 Kevin $5,500 $2,500 $2,000 $10,000

Apportionment – the distribution of an item fairly among a group. Ideal ratio - Total population Number of seats Quota - individual population ideal ratio Truncate – eliminate the decimal portion of the number (this is NOT rounding)

a.) Find the ideal ratio (round to the nearest hundredth) Apportionment A school has the following student populations: senior – 435, juniors 525, sophomores – 581, and freshmen – 653. There are 40 seats in the student council. a.) Find the ideal ratio (round to the nearest hundredth) b.) Find the quota for each class c.) Find the apportioned number of seats for each of the four methods (Hamilton, Jefferson, Webster, and Hill)

Methods of Apportionment Hamilton Method – award the remaining seat(s) to the choice with the highest decimal in the quota. If there are more than one seat to award, then highest decimal first and work your way in descending order until all seats are awarded.

Hamilton’s Method Class Students Quota Truncated Hamilton Sr 435 ( / ) Jr 525 ( / ) So 581 ( / ) Fr 653 ( / ) 2194 need Ideal Ratio = Hamilton uses the highest decimal from the quota to decide who receives the next seat.

Jefferson Method – Find the adjusted ratio for each choice and the adjusted ratio closest to the ideal ratio receives the seat. If more than one seat is being awarded, then start with choice whose adjusted ratio is closest to ideal ratio and work your way out until all available seats have been awarded. adjusted ratio = Individual population truncated quota + 1

Jefferson’s Method Adjusted Class Students Quota Truncated Ratio Jefferson Sr 435 ( / ) Jr 525 ( / ) So 581 ( / ) Fr 653 ( / ) 2194 Ideal Ratio = need Divide the class population by the truncated value + 1

Methods of Apportionment Webster Method - 1.) Find the quota for each state. 2.) Round each quota. 3.) Decide whether the house is okay, overfilled, or underfilled.

Find Adjusted Ratio: adjusted ratio = Individual population arithmetic mean (In this case, the arithmetic mean will be the same as truncated quota + .5) Overfilled – The state with the smallest adjusted ratio will lose a seat. Underfilled – The state with the largest adjusted ratio will gain a seat.

Webster’s Method Adjusted Class Students Quota Rounded Ratio Webster Sr 435 ( / ) Jr 525 ( / ) So 581 ( / ) Fr 653 ( / ) Ideal Ratio = Divide the class population by the truncated value + . 5

Hill Method – 1.) Find the quota for each state. 2.) Round each quota, using Geometric mean. 3.) Decide whether the house is okay, overfilled, or underfilled.

adjusted ratio = Individual population geometric mean ______________________ The geometric mean is: √ (truncate) X (truncate + 1) Overfilled – The state with the smallest adjusted ratio will lose a seat. Underfilled – The state with the largest adjusted ratio will gain a seat.

Hill’s Method Geo Adj Class Students Quota mean Rounded Ratio Seats Sr 435 ( / ) Jr 525 ( / ) So 581 ( / ) Fr 653 ( / ) Ideal Ratio = Divide the class population by the Square root of truncated value times the truncated value plus one

Class Students Quota Trunc Ham. Jeff Web Hill Sr 435 Jr 522 So 581 Fr 653 Ideal Ratio =

b.) Find the quota for each county Apportionment The new island state of Bennettland has 30 seats to award the 6 counties that make up the state’s House of Representatives. Here are the populations of the 6 counties: A – 1000, B – 255, C – 2339, D – 311, E – 634, F – 531. a.) Find the ideal ratio b.) Find the quota for each county c.) Find the number of apportioned seats in all four methods.

Induction Steps for solution: 1.) Plug in a value for the variable and check to see if the equation works for a value. 2.) Write the original equation 3.) Rewrite equation and change the variable to the variable k. 4.) Write the equation in terms of k+1 5.) Substitute the right side of the equation in terms of k into the beginning of the left side of the equation in terms of k+1. Solve (Make both sides look the same, show your work)

Solve by Mathematical Induction 1.) 1 + 2 + 3 + … + n = n (n+1) 2

Step 1 1 + 2 + 3 + … + n = n ( n+1) 2 Plug in any value for n to see if the equation works for a value. Set n = 4 So, 1 + 2 + 3 + 4 = 4 (4 + 1) 2 10 = 20 10 = 10 √ works On the left side the value for n represent the number of terms you are going to use to find the value, I chose n=4, so I used the first four terms. The last term on the left side tells you how the pattern is formed.

Steps 2 and 3 1 + 2 + 3 + … + n = n ( n+1) 2 1 + 2 + 3 + … + k = k ( k+1) 2

Steps 4 1 + 2 + 3 + … + n = n ( n+1) 2 1 + 2 + 3 + … + k = k ( k+1) 2 1 + 2 + 3 + … + k + (k+1) = (k+1) ( (k+1) +1) 2 Notice the left side the k+1 is placed at the end and on the right side the variables k are changed to k+1

Step 5 1 + 2 + 3 + … + n = n ( n+1) 2 1 + 2 + 3 + … + k = k ( k+1) 2 1 + 2 + 3 + … + k + (k+1) = (k+1) ( (k+1) +1) 2 k ( k+1) + (k+1) = (k+1) ( (k+2) 2 2 Solve…

6 Solve by using Mathematical Induction. 1 + 4 + 9 + … + n2 = n(n + 1)(2n + 1) 6