## Chapter 27Solutions for ninth week’s assignments

Also available as PDF.

Note: These are my approaches to these problems. There are many ways to tackle each.

### 27.1 Diophantine equations

See the previous week’s solutions.

### 27.2 Problem set 6.1

Problem 2
I’m not going to draw this, but it should be fairly straight-forward. This was more an exercise in something handy for elementary classes.
Problem 6
• {20\over 60} = {1\over 3}
• {30\over 60} = {1\over 2}
• {5\over 7}
• {25\over 100} = {1\over 4}
• {25\over 100} = {1\over 4}
• {3\over 12} = {1\over 4}
• {2\over 3}
• {3\over 4} (A quart is made of quarters... There are four cups in a quart. Cooking is a source of bizarre but traditional units.)
Problem 11
• {4\over 5} = {6⋅4\over 6⋅5} = {24\over 30}
• {6\over 9} = {3⋅2\over 3⋅3} = {2\over 3}
• {-7\over 25} = {20⋅-7\over 20⋅25} = {-140\over 500}
• {18\over 3} = {3⋅6\over 3⋅1} = {6\over 1} = {-1⋅6\over -1⋅1} = {-6\over -1}
Problem 12
• {18\over 42} = {6⋅3\over 6⋅7}\mathbf{=}{3\over 7}.
• Here (18,49) = 1 and (5,14) = 1, so both are in lowest common terms. That form is unique, and these fractions differ, so these fractions cannot be equal. Following the text’s method, you want to compare {2⋅18\over 2⋅49} = {36\over 98} and {7⋅5\over 7⋅14} = {35\over 98}, because \mathop{lcm}\nolimits (49,14) = 98.
• {9\over 25} = {20⋅9\over 20⋅25} = {180\over 500}\mathbf{\mathrel{≠}}{140\over 500}.
• {24\over 144} = {2\over 12} = {1\over 6}, {32\over 96} = {1\over 3} = {2\over 6}. These are not equal.
Problem 32
Yes, this is a general property. But you need to provide some examples.

### 27.3 Problem set 6.2

Problem 6
• {5\over 7}
• {10\over 5} = 2
• {20\over 24} = {5\over 6}
• {56\over 65}
• {76\over 60} = {19\over 15}
• {100\over 200} = {1\over 2}
• {-31\over 100}
• {321\over 450} = {107\over 150}
Problem 18
• 1
• {1\over 4}
• 1
Problem 13
• 3 ⋅{5\over 2} = {15\over 2}
• {2\over 3} ⋅{3\over 2} = 1
• {3\over 4} ⋅ 2 = {3\over 2}
Problem 25
For each, you proceed by solving the row, column, or diagonal that has only one open spot. Repeating suffices to fill the squares.
•  {1\over 2} {1\over 12} \mathbf{ {5\over 12}} {1\over 4} \mathbf{{1\over 3}} \mathbf{ {5\over 12}} \mathbf{{1\over 4}} \mathbf{ {7\over 12}} \mathbf{{1\over 6}}
•  \mathbf{ {8\over 15}} \mathbf{{1\over 5}} \mathbf{ {4\over 15}} \mathbf{ {1\over 15}} {1\over 3} {3\over 5} \mathbf{{2\over 5}} \mathbf{ {7\over 15}} {2\over 15}

### 27.4 Problem set 6.3

Problem 2
• Reassociate to add the first two terms with the common denominator of 5 first.
• Commute the terms in the parenthesis and reassociate to add terms with common denominator of 4 first.
• Use the distributive property to pull out the {2\over 3}, then add the eighths.
• Commute and reassociate to multiply {3\over 4} ⋅{4\over 3} = 1 first.
Problem 8
• {2\over 3} ⋅{4\over 7} + {2\over 3} ⋅{3\over 7} = {2\over 3} ⋅ ({4\over 7} + {3\over 7}) = {2\over 3} ⋅ 1 = {2\over 3}
• {4\over 5} ⋅{2\over 3} - {3\over 10} ⋅{2\over 3} = ({4\over 5} - {3\over 10}) ⋅{2\over 3} = {1\over 2} ⋅{2\over 3} = {1\over 3}
• {4\over 7} ⋅{3\over 2} -{4\over 7} ⋅{6\over 4} = {4\over 7} ⋅ ({3\over 2} -{6\over 4}) = {4\over 7} ⋅ 0 = 0
• ({4\over 7} ⋅{2\over 5})∕{2\over 7} = {4⋅2\over 7⋅5} ⋅{7\over 2} = {4\over 5}
Problem 9
• adding fractions with a common denominator
• multiplying fractions