Chapter 17
Solutions for fourth week’s assignments

Also available as PDF.

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

17.1 Problem set 2.2

17.1.1 Problem 1

17.1.2 Problem 2

17.1.3 Problem 6

17.1.4 Problem 13

The drawing will have

17.1.5 Problem 21

17.1.6 Problem 23

For (a) and (b), the table is Pascal’s triangle, which we already have in the notes. For (c), we want the entry {P}_{6,3} in our earlier notation. There are 20 such ways.

17.1.7 Why answering problem 32 would be a bad idea.

Wow. Never, ever give someone a list of all your identifying numbers. Identity theft is a serious problem. While quite often all these numbers are available for a little work, at least make a criminal work for them.

17.2 Problem set 2.3

17.2.1 Problem 2

Substituting into |A ∪ B| = |A| + |B|-|A ∩ B|, we see that 10 = 5 + 8 -|A ∩ B|, or that |A ∩ B| = 3.

17.2.2 Problem 5

I’m just going to show these as lines. Illustrating them on a “number line” is equivalent. A better diagram for the last would wrap portions of the result in parentheses to show how the line extended.

17.2.3 Problem 11

Again, I’m just going to show these as lines. Illustrating them on a “number line” is equivalent.

17.2.4 Problem 24

Draw with two shadings and show that the intersection is shaded twice.

17.3 Write 2 + 3 using disjoint sets.

If we let 2 ≡\{a,b\} and 3 ≡\{c,d,e\}, then 2 + 3 ≡\{a,b\} ∪\{c,d,e\} = \{a,b,c,d,e\}.

17.4 Illustrate 2 + 3 using Peano arithmetic.

We defined addition with

\eqalignno{ a + 0 & = a,\text{ and} & & \cr a + S(b) & = S(a + b). & & }

Here, 3 = S(2), so

\eqalignno{ 2 + 3 & = 2 + S(2) & & \cr & = S(2 + 2) & & \cr & = S(2 + S(1)) & & \cr & = S(S(2 + 1)) & & \cr & = S(S(2 + S(0))) & & \cr & = S(S(S(2 + 0))) & & \cr & = S(S(S(2))) & & \cr & = 5. & & }

17.5 Problem set 2.4

17.5.1 Problem 5

17.5.2 Problem 10

Each product is equal to the appropriate shaded area, and the sum is equal to the entire rectangle. The letters correspond directly to the positions.

I had completely forgotten about the FOIL mnemonic. After long enough, it’s just a part of what you do.

17.5.3 Problem 26

The operation is closed because no new shapes are introduced in the operator’s table of results.

The operation is commutative because the operation table is symmetric across the diagonal axis. Thus a ⋆ b = b ⋆ a.

Because ∘ ⋆x = x, the symbol is ’s identity.

After the identity, there are only two symbols remaining. Thus the commutative property combined with the identity means that this operator must be associative.

17.6 Illustrate 2 ⋅ 3 using Peano arithmetic. You do not need to expand addition.

We defined multiplication with

\eqalignno{ a ⋅ 0 = 0,\text{ and} & & \cr a ⋅ S(b) = a + (a ⋅ b). & & }

So

\eqalignno{ 2 ⋅ 3 & = 2 ⋅ S(2) & & \cr & = 2 + (2 ⋅ 2) & & \cr & = 2 + (2 ⋅ S(1)) & & \cr & = 2 + (2 + (2 ⋅ 1)) & & \cr & = 2 + (2 + (2 ⋅ S(0))) & & \cr & = 2 + (2 + (2 + 2 ⋅ 0)) & & \cr & = 2 + 2 + 2 = 6. & & }

17.7 Illustrate (1 ⋅ 2) ⋅ 3 = 1 ⋅ (2 ⋅ 3) using a volume of size six.

Draw two 1 × 2 × 3 boxes made of 1 × 1 × 1 boxes. To illustrate (1 ⋅ 2) ⋅ 3, show that the 1 × 2 portion is stacked 3 times. To illustrate 1 ⋅ (2 ⋅ 3), show that the 2 × 3 portion is stacked across 1 time.