Algebra Know-It-ALL

592 Worked-Out Solutions to Exercises: Chapters 1 to 9

9. The set {1, {2, 3}} has only two elements: the number 1 and the set {2, 3}. You
   can’t break {2, 3} down and have it remain an element of the original set {1, {2, 3}}.
   Therefore, the set of all possible subsets of {1, {2, 3}} is

∅

{1}

{{2, 3}}

{1, {2, 3}}

10. The set {1, {2, {3}}} has two elements: the number 1 and the set {2, {3}}. You cannot
    break {2, {3}} down and have it remain an element of the original set {1, {2, {3}}}.
    Therefore, the set of all possible subsets of {1, {2, {3}}} is

∅

{1}

{{2, {3}}}

{1, {2, {3}}}

When writing down complicated sets like these, always be sure that the number of

opening braces is the same as the number of closing braces. If they’re different, you’ve

made a mistake somewhere.

Chapter 3

1. This is the ultimate trivial question. In the number-building systems described in this
   chapter, nothing doesn’t represent any number.


2. No. The number 6 is divisible by 3 without a remainder:

6/3= 2

Of course, the quotient here, 2, is even. There are plenty of other even numbers that can

be divided by 3 leaving no remainder.

3. When you multiply an odd number by 3, you always get an odd number as the product.
   The reason for this is similar to the reason why any even number times 7 gives you
   another even number. (Proving that was one of the “challenge” problems in this chapter.)
   For the first few odd numbers, multiplication by 3 always produces an odd number:

3 × 1 = 3

3 × 3 = 9

3 × 5 = 15

3 × 7 = 21

3 × 9 = 27