Chapter 1.1, Problem 37E

(a)

To determine

To write: the statement “ $x$ is positive” in terms of inequality.

Answer to Problem 37E

  $x>0$

Explanation of Solution

Given information:

  $x$ is positive

Calculation:

Consider the given statement.

  $x$ is positive

Now, positive number is always greater than 0.

So, $x>0$ .

(b)

To determine

To write: the statement “ $t$ is less than 4” in terms of inequality.

Answer to Problem 37E

  $t<4$

Explanation of Solution

Given information:

  $t$ is less than 4

Calculation:

Consider the given statement.

  $t$ is less than 4

Now, less than sign is $\text{'}<\text{'}$ .

So, $t$ is less than 4, in terms of an inequality can be written as

  $t<4$ .

(c)

To determine

To write: the statement “ $a$ is greater than or equal to $\pi$ ” in terms of inequality.

Answer to Problem 37E

  $a\ge \pi$

Explanation of Solution

Given information:

  $a$ is greater than or equal to $\pi$ .

Calculation:

Consider the given statement.

  $a$ is greater than or equal to $\pi$

Now, greater than equal sign is $\text{'}\ge \text{'}$ .

So, $a$ is greater than or equal to $\pi$ , in terms of an inequality can be written as

  $a\ge \pi$ .

(d)

To determine

To write: the statement “ $x$ is greater than $-5$ and less than $\frac{1}{3}$ ” in terms of inequality.

Answer to Problem 37E

  $-5<x<\frac{1}{3}$

Explanation of Solution

Given information:

  $x$ is greater than $-5$ and less than $\frac{1}{3}$ .

Calculation:

Consider the given statement.

  $x$ is greater than $-5$ and less than $\frac{1}{3}$

Now, greater than sign is $\text{'}>\text{'}$ and less than symbol is $\text{'}<\text{'}$ .

So, $x$ is greater than $-5$ and less than $\frac{1}{3}$ , in terms of an inequality can be written as

  $-5<x<\frac{1}{3}$ .

(e)

To determine

To write: the statement “the distance between $p$ and 3 is at most $5$ ” in terms of inequality.

Answer to Problem 37E

  $\left|p-3\right|\le 5$

Explanation of Solution

Given information:

The distance between $p$ and 3 is at most $5$ .

Calculation:

Consider the given statement.

The distance between $p$ and 3 is at most $5$ .

Distance between two points a and b can be written as

  $d\left(a,b\right)=\left|a-b\right|$

Now, for the word “at most” the required symbol is $\text{'}\le \text{'}$ .

So, the distance between $p$ and 3 is at most $5$ , in terms of an inequality can be written as

  $\begin{array}{l}d\left(p,3\right)\le 5\\ \left|p-3\right|\le 5\end{array}$ .