Chapter 1.1, Problem 85E

(a)

To determine

Whether: subtraction is commutative.

Answer to Problem 85E

Subtraction is not commutative.

Explanation of Solution

Calculation:

Try to find a counterexample to the statement “Subtraction is commutative.” If we can find even example in which subtraction is not commutative, we can say that the statement is false. Let’s assume that subtraction is commutative. This means that for any numbers $a$ and $b$ , the order of the subtraction can be switched and the result will be the same. That is,

  $a-b=b-a$

Let’s take two numbers. Let

  $a=2$ and $b=1$ ,

Let’s perform the subtractions.

  $\begin{array}{l}a-b=2-1\\ a-b=1\end{array}$

Now, reverse the numbers,

  $\begin{array}{l}b-a=1-2\\ b-a=-1\end{array}$

Since 1 and -1 are not equal, the subtraction order cannot be switched to produce the same results. This provides a counterexample for our argument. Therefore, subtraction is not commutative.

(b)

To determine

Whether the division of nonzero real numbers commutative.

Answer to Problem 85E

For nonzero real numbers division is not commutative.

Explanation of Solution

Calculation:

Try to find a counterexample to the statement “divison is commutative.” If we can find even example in which subtraction is not commutative, we can say that the statement is false.

Let’s assume that division is commutative. This means that for any nonzero numbers $a$ and $b$ , the order of the division can be switched and the result will be the same. That is,

  $\frac{a}{b}=\frac{b}{a}$ .

Let’s take two nonzero real numbers. Let

  $a=2$ and $b=1$ ,

Let’s perform the divisions.

  $\begin{array}{l}\frac{a}{b}=\frac{2}{1}\\ \frac{a}{b}=2\end{array}$

Now, reverse the numbers,

  $\frac{b}{a}=\frac{1}{2}$

Since 2 and $\frac{1}{2}$ are not equal, the subtraction order cannot be switched to obtain the same results. This provides a counterexample for our argument. Therefore, for nonzero real numbers division is not commutative.