Chapter 27
Solutions for ninth week’s assignments

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
• distributing multiplication over addition
• adding fractions with a common denominator
• multiplying fractions

Problem 12

• First subtraction of 32, then multiplication by the inverse of {9\over 5}.

• Celsius	-40	-20	0	10	20	45	100
  Fahrenheit	-40	-13	32	50	68	104	212

• As seen in the table above, both agree at - 40. If - F = C, then - F = {5\over 9}(F - 32), {-9\over 5} F = F - 32, 32 = {14\over 5} F, F = {80\over 7} ≈ 11.