Exercise 4.1

Main Problem.

(a)

If $x+y=x$, then $y=0$.

Proof. Clearly by (3) we have $x+0=x=x+y$ so that it has to be that $y=0$ by Lemma 1 . □

(b)

$0\cdot x=0$. [Hint: Compute $(x+0)\cdot x$.]

Proof. We have

$$\begin{array}{lllllll}\hfill x\cdot x+0\cdot x& =x\cdot x+x\cdot 0& \hfill \text{(since }0\cdot x=x\cdot 0\text{ by (2))}& & \hfill & & \hfill \\ \hfill & =x\cdot (x+0)& \hfill \text{(by (5))}& & \hfill & & \hfill \\ \hfill & =x\cdot x.& \hfill \text{(since }x+0=x\text{ by (3))}& & \hfill & & \hfill \end{array}$$

Thus it must be that $0\cdot x=0$ by part (a). □

(c)

$-0=0$.

Proof. By (4) we have $0+(-0)=0$ so that it has to be that $-0=0$ by part (a). □

(d)

$-(-x)=x$.

Proof. We have

$$\begin{array}{lllllll}\hfill -(-x)& =-(-x)+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =-(-x)+(x+(-x))& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =-(-x)+((-x)+x)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(-(-x)+(-x))+x& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =((-x)+(-(-x)))+x& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =0+x& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =x+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =x& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

as desired. □

(e)

$x(-y)=-(\mathit{xy})=(-x)y$.

Proof. First we have

$$\begin{array}{lllllll}\hfill x(-y)& =x(-y)+0& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =x(-y)+(\mathit{xy}+(-(\mathit{xy}))& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =(x(-y)+\mathit{xy})+(-(\mathit{xy}))& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =x(-y+y)+(-(\mathit{xy}))& \hfill \text{(by (5))}& & \hfill & & \hfill \\ \hfill & =x(y+(-y))+(-(\mathit{xy}))& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =x\cdot 0+(-(\mathit{xy}))& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =0\cdot x+(-(\mathit{xy}))& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =0+(-(\mathit{xy}))& \hfill \text{(by part(b))}& & \hfill & & \hfill \\ \hfill & =-(\mathit{xy})+0& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =-(\mathit{xy}).& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & & \hfill & & \hfill \end{array}$$

We also have

$$\begin{array}{lllllll}\hfill (-x)y& =y(-x)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =-(\mathit{yx})& \hfill \text{(by what was just shown)}& & \hfill & & \hfill \\ \hfill & =-(\mathit{xy})& \hfill \text{(by (2))}& & \hfill & & \hfill \end{array}$$

so that the result follows since equality is transitive. □

(f)

$(-1)x=-x$.

Proof. We have

$$\begin{array}{lllllll}\hfill (-1)x& =-(1\cdot x)& \hfill \text{(by part(e))}& & \hfill & & \hfill \\ \hfill & =-(x\cdot 1)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =-x& \hfill \text{(since }x\cdot 1=x\text{ by (3))}& & \hfill & & \hfill \end{array}$$

as desired. □

(g)

$x(y-z)=\mathit{xy}-\mathit{xz}$.

Proof. We have

$$\begin{array}{lllllll}\hfill x(y-z)& =x(y+(-z))& \hfill \text{(by the definition of subtraction)}& & \hfill & & \hfill \\ \hfill & =\mathit{xy}+x(-z)& \hfill \text{(by (5))}& & \hfill & & \hfill \\ \hfill & =\mathit{xy}+(-(\mathit{xz}))& \hfill \text{(by part(e))}& & \hfill & & \hfill \\ \hfill & =\mathit{xy}-\mathit{xz}& \hfill \text{(by the definition of subtraction)}& & \hfill & & \hfill \end{array}$$

as desired. □

(h)

$-(x+y)=-x-y$; $-(x-y)=-x+y$.

Proof. We have

$$\begin{array}{lllllll}\hfill -(x+y)& =(-1)(x+y)& \hfill \text{(by part (f))}& & \hfill & & \hfill \\ \hfill & =(-1)x+(-1)y& \hfill \text{(by (5))}& & \hfill & & \hfill \\ \hfill & =-x+(-y)& \hfill \text{(by part (f) twice)}& & \hfill & & \hfill \\ \hfill & =-x-y& \hfill \text{(by the definition of subtraction)}& & \hfill & & \hfill \end{array}$$

and

$$\begin{array}{lllllll}\hfill -(x-y)& =-(x+(-y))& \hfill \text{(by the definition of subtraction)}& & \hfill & & \hfill \\ \hfill & =-x-(-y))& \hfill \text{(by what was just shown)}& & \hfill & & \hfill \\ \hfill & =-x+(-(-y))& \hfill \text{(by the definition of subtraction)}& & \hfill & & \hfill \\ \hfill & =-x+y& \hfill \text{(by part (d))}& & \hfill & & \hfill \end{array}$$

as desired. □

(i)

If $x\ne 0$ and $x\cdot y=x$, then $y=1$.

Proof. By (3) we have $x\cdot 1=x=x\cdot y$ so that it has to be that $y=1$ by Lemma 2 , noting that this applies since $x\ne 0$. □

(j)

$x∕x=1$ if $x\ne 0$.

Proof. By the definition of division we have $x∕x=x\cdot (1∕x)=1$ by (4) since $x\ne 0$ and $1∕x$ is defined as the reciprocal (i.e. the multiplicative inverse) of $x$. □

(k)

$x∕1=x$.

Proof. First, we have by (4) that $1\cdot (1∕1)=1$, where $1∕1$ is the reciprocal of $1$. We also have that $1\cdot (1∕1)=(1∕1)\cdot 1=1∕1$ by (2) and (3). Therefore $1∕1=1\cdot (1∕1)=1$ so that $1$ is its own reciprocal. Then, by the definition of division, we have $x∕1=x\cdot (1∕1)=x\cdot 1=x$ by (3). □

(l)

$x\ne 0$ and $y\ne 0\Rightarrow \mathit{xy}\ne 0$.

Proof. Suppose that $x\ne 0$ and $y\ne 0$. Also suppose to the contrary that $\mathit{xy}=0$. Since $y\ne 0$ it follows from (4) that $1∕y$ exists. So, we have $(\mathit{xy})\cdot (1∕y)=0\cdot (1∕y)=0$ by part (b). We also have

$$\begin{array}{lllllll}\hfill (\mathit{xy})\cdot \frac{1}{y}& =x\left(y\cdot \frac{1}{y}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =x\cdot 1& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =x& \hfill \text{(by (3))}& & \hfill & & \hfill \end{array}$$

so that $x=(\mathit{xy})\cdot (1∕y)=0$, which is a contradiction since we supposed that $x\ne 0$. Hence it must be that $\mathit{xy}\ne 0$ as desired. □

(m)

$(1∕y)(1∕z)=1∕(\mathit{yz})$ if $y,z\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill (\mathit{yz})\left(\frac{1}{y}\cdot \frac{1}{z}\right)& =(\mathit{yz})\left(\frac{1}{z}\cdot \frac{1}{y}\right)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\left((\mathit{yz})\cdot \frac{1}{z}\right)\frac{1}{y}& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =\left(y\left(z\cdot \frac{1}{z}\right)\right)\frac{1}{y}& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =\left(y\cdot 1\right)\frac{1}{y}& \hfill \text{(by (4))}& & \hfill & & \hfill \\ \hfill & =y\cdot \frac{1}{y}& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =1& \hfill \text{(by (4))}& & \hfill & & \hfill \end{array}$$

so that $(1∕y)(1∕z)$ is a multiplicative inverse of $\mathit{yz}$. Since this inverse is unique by (4), however, it has to be that $(1∕y)(1∕z)=1∕(\mathit{yz})$ as desired. □

$$

(n)

$(x∕y)(w∕z)=(\mathit{xw})∕(\mathit{yz})$ if $y,z\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{x}{y}\cdot \frac{w}{z}& =\left(x\cdot \frac{1}{y}\right)\left(w\cdot \frac{1}{z}\right)& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =\left(x\cdot \frac{1}{y}\right)\left(\frac{1}{z}\cdot w\right)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\left(\left(x\cdot \frac{1}{y}\right)\frac{1}{z}\right)w& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =\left(x\left(\frac{1}{y}\cdot \frac{1}{z}\right)\right)w& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =\left(x\cdot \right)w& \hfill \text{(by part (m) since }y,z\ne 0\text{)}& & \hfill & & \hfill \\ \hfill & =\left(\cdot x\right)w& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(\mathit{xw})& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =(\mathit{xw})& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\frac{\mathit{xw}}{\mathit{yz}}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \end{array}$$

as desired. □

(o)

$(x∕y)+(w∕z)=(\mathit{xz}+\mathit{wy})∕(\mathit{yz})$ if $y,z\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{x}{y}+\frac{w}{z}& =\frac{x}{y}\cdot 1+\frac{w}{z}\cdot 1& \hfill \text{(by (3))}& & \hfill & & \hfill \\ \hfill & =\frac{x}{y}\cdot \frac{z}{z}+\frac{w}{z}\cdot \frac{y}{y}& \hfill \text{(by part (j))}& & \hfill & & \hfill \\ \hfill & =\frac{\mathit{xz}}{\mathit{yz}}+\frac{\mathit{wy}}{\mathit{zy}}& \hfill \text{(by part(n))}& & \hfill & & \hfill \\ \hfill & =(\mathit{xz})\frac{1}{\mathit{yz}}+(\mathit{wy})\frac{1}{\mathit{zy}}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =(\mathit{xz})\frac{1}{\mathit{yz}}+(\mathit{wy})\frac{1}{\mathit{yz}}& \hfill \text{(by Lemma }\text{3<!-- tex4ht:ref: lem:intreal:recom -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =\frac{1}{\mathit{yz}}(\mathit{xz})+\frac{1}{\mathit{yz}}(\mathit{wy})& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\frac{1}{\mathit{yz}}\left(\mathit{xz}+\mathit{wy}\right)& \hfill \text{(by (5))}& & \hfill & & \hfill \\ \hfill & =\left(\mathit{xz}+\mathit{wy}\right)\frac{1}{\mathit{yz}}& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =\frac{\mathit{xz}+\mathit{wy}}{\mathit{yz}}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \end{array}$$

as desired. □

(p)

$x\ne 0\Rightarrow 1∕x\ne 0$.

Proof. Suppose that $x\ne 0$ but $1∕x=0$. Then we first have that $x\cdot (1∕x)=x\cdot 0=0\cdot x=0$ by (2) and part (b). However, we also have $x\cdot (1∕x)=1$ by (4). Hence we have $0=x\cdot (1∕x)=1$, which is a contradiction since we know that $0$ and $1$ are distinct by (3). So, if we accept that $x\ne 0$, then it must be that $1∕x\ne 0$ also. □

(q)

$1∕(w∕z)=z∕w$ if $w,z\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{w}{z}\cdot \frac{z}{w}& =\frac{\mathit{wz}}{\mathit{zw}}& \hfill \text{(by part (n) since }w,z\ne 0\text{)}& & \hfill & & \hfill \\ \hfill & =(\mathit{wz})\frac{1}{\mathit{zw}}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =(\mathit{wz})\frac{1}{\mathit{wz}}& \hfill \text{(by Lemma }\text{3<!-- tex4ht:ref: lem:intreal:recom -->}\text{ since }w,z\ne 0\text{)}& & \hfill & & \hfill \\ \hfill & =1& \hfill \text{(by (4))}& & \hfill & & \hfill \end{array}$$

so that by definition $z∕w$ is the reciprocal of $w∕z$. Since this is unique by (4) we then have $z∕w=1∕(w∕z)$ as desired. □

(r)

$(x∕y)∕(w∕z)=(\mathit{xz})∕(\mathit{yw})$ if $y,w,z\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{x∕y}{w∕z}& =\frac{x}{y}\cdot \frac{1}{w∕z}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =\frac{x}{y}\cdot \frac{z}{w}& \hfill \text{(by part (q) since }w,z\ne 0\text{)}& & \hfill & & \hfill \\ \hfill & =\frac{\mathit{xz}}{\mathit{yw}}& \hfill \text{(by part (n) since }y,w\ne 0\text{)}& & \hfill & & \hfill \end{array}$$

as desired. □

(s)

$(\mathit{ax})∕y=a(x∕y)$ if $y\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{\mathit{ax}}{y}& =(\mathit{ax})\cdot \frac{1}{y}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =a\left(x\cdot \frac{1}{y}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =a\cdot \frac{x}{y}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \end{array}$$

as desired. □

(t)

$(-x)∕y=x∕(-y)=-(x∕y)$ if $y\ne 0$.

Proof. We have

$$\begin{array}{lllllll}\hfill \frac{-x}{y}& =(-x)\cdot \frac{1}{y}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =((-1)x)\cdot \frac{1}{y}& \hfill \text{(by part (f))}& & \hfill & & \hfill \\ \hfill & =(-1)\left(x\cdot \frac{1}{y}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =(-1)\frac{x}{y}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =-\left(\frac{x}{y}\right).& \hfill \text{(by part (f))}& & \hfill & & \hfill \end{array}$$

Now, we have $(-1)(-1)=-(-1)=1$ by parts (f) and (d) so that $-1$ is its own reciprocal, since the reciprocal is unique, i.e. $1∕(-1)=-1$. We also have

$$\begin{array}{lllllll}\hfill \frac{-x}{y}& =(-x)\cdot \frac{1}{y}& \hfill \text{(by the definition of division)}& & \hfill & & \hfill \\ \hfill & =((-1)x)\cdot \frac{1}{y}& \hfill \text{(by part (f))}& & \hfill & & \hfill \\ \hfill & =(x(-1))\cdot \frac{1}{y}& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =x\left((-1)\frac{1}{y}\right)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =x\left(\frac{1}{-1}\cdot \frac{1}{y}\right)& \hfill \text{(by what was just shown above)}& & \hfill & & \hfill \\ \hfill & =x\frac{1}{(-1)y}& \hfill \text{(part (m) since }y\ne 0\text{)}& & \hfill & & \hfill \\ \hfill & =x\frac{1}{-y}& \hfill \text{(by part (f))}& & \hfill & & \hfill \end{array}$$

so that $-(x∕y)=(-x)∕y=x∕(-y)$ as desired. □