Inequality Basics: Additional Practice (Set 2)

For $a < x < b$ to describe any numbers, the bounds must satisfy $a < b$. Here $4 < 1$ is false.

$4 < x < 1 \;\Rightarrow\; \text{no such } x$

Nothing is both greater than $4$ and less than $1$. (Note $0 \le x \le 0$ does have the solution $x = 0$.)

The set runs left forever, so the left end is $-\infty$ with a parenthesis, and the excluded boundary $-3$ also takes a parenthesis.

$x < -3 \quad\Longleftrightarrow\quad (-\infty, -3)$

Both ends are open here.

Only a negative multiplier reflects the numbers across zero and reverses their order.

$\text{multiply by } -2 \;\Rightarrow\; \text{direction reverses}$

Multiplying by the positive numbers $5$ or $\tfrac{1}{3}$ keeps the direction, and multiplying by $0$ destroys the comparison entirely.

The symbol $>$ is strict, so $x$ must be larger than $-1$, not equal to it.

$-1 > -1 \;\text{ is false}$

The values $0$, $3$, and $5$ are all greater than $-1$, but $-1$ itself is not.

'Nonnegative' means not negative, which allows zero and every positive number.

$x \ge 0 \;\Rightarrow\; x \text{ is nonnegative}$

'Strictly positive' would exclude $0$, giving $x > 0$ instead.

The bracket includes $0$, and the set runs right to infinity.

$[0, \infty) \quad\Longleftrightarrow\quad x \ge 0$

That is zero together with all the positive numbers.

Divide both sides by $-1$. The divisor is negative, so reverse $<$ to $>$, and $-4 \div (-1) = 4$.

$\frac{-x}{-1} > \frac{-4}{-1} \;\Rightarrow\; x > 4$

A negative divided by a negative is positive.

Transitivity chains the two comparisons, and it carries the inclusive symbol through.

$p \le q \text{ and } q \le r \;\Rightarrow\; p \le r$

Equality can hold throughout, so the conclusion stays $\le$.

The circle on $-2$ is open, so $-2$ is excluded, which means a strict symbol. The shading runs left toward the smaller numbers, so the relation is 'less than'.

$\text{open at } -2, \text{ shade left} \;\Rightarrow\; x < -2$

A filled circle would give $x \le -2$, and shading right would give a 'greater' relation.

Divide both sides by the positive number $2$. Dividing by a positive keeps the direction, even with a negative on the right.

$\frac{2x}{2} > \frac{-10}{2} \;\Rightarrow\; x > -5$

No flip occurs, because $2$ is positive.

The symbol $\ge$ is inclusive, so $-5$ is included and gets a filled circle. 'Greater than or equal to' shades toward the larger numbers, to the right.

$x \ge -5 \;\Rightarrow\; \text{filled circle on } -5, \text{ shade right}$

An open circle would exclude $-5$.

'Strictly between' excludes both endpoints, so both fences are parentheses.

$-3 < x < 3 \quad\Longleftrightarrow\quad (-3, 3)$

Any bracket would wrongly include an endpoint.