Inequalities and absolute value

Rules for operating with inequalities

Solving an inequality means transforming it, step by step, into a simpler one with the same solutions. Only operations that do not change the set of solutions are valid, and to see which ones are, you need to know what happens to the direction of the sign.

Rule 4 is easy to see on the number line: multiplying by $-1$ is a reflection about zero, which reverses the order. Since $2<5$, reflecting gives $-2>-5$. The same happens with $\ge$, $<$ and $\le$.

These rules let you solve a linear inequality like an equation, with one caution. For $7-2x<15$: subtract $7$ to get $-2x<8$ and divide by $-2$, reversing the sign, which gives $x>-4$. The solution is the interval $]-4,+\infty [$.

Interval notation

In this course, the brackets point outwards for endpoints that are not part of the interval: $]a,b[$ means $a<x<b$, and $[a,b]$ means $a\le x\le b$. Also $]a,b]$ means $a<x\le b$. The symbol $\infty$ always takes an open bracket, because it is not a number. The union $A\cup B$ gathers the solutions of both sets.

Proving an inequality with a square

A square is never negative: $t^2\ge 0$ for every real number $t$. Many inequalities are proved starting from this fact. For example, for any real numbers $a$ and $b$:

$$(a-b{)}^2\ge 0⟺a^2-2ab+b^2\ge 0⟺a^2+b^2\ge 2ab$$

Equality holds only if $a=b$. If in addition $a$ and $b$ are positive, dividing by $ab>0$ gives

$$\frac{a}{b}+\frac{b}{a}\ge 2$$

which says, for example, that a positive fraction plus its reciprocal never adds up to less than $2$. The technique is always the same: move everything to one side, recognise a square and conclude. Each step is an equivalence (a double arrow) or an implication that can be read backwards, and that is why the chain proves what was asked.

Non-linear inequalities: the sign table

To solve $x^2-x-6\ge 0$ you cannot just "isolate" $x$. Proceed as follows:

For $x^2-x-6\ge 0$: $x^2-x-6=(x-3)(x+2)$, with roots $-2$ and $3$. To the left of $-2$ both factors are negative and their product is positive; between $-2$ and $3$ the factor $x+2$ is positive and $x-3$ negative, so the product is negative; to the right of $3$ both are positive. As $\ge 0$ is required and the endpoints work, the solution is $]-\infty ,-2]\cup [3,+\infty [$.

A common trap with fractions: $\frac{x+1}{x-3}\ge 2$ is not solved by multiplying by $x-3$, because you do not know whether that number is positive or negative, and you would have to split into cases. Subtract $2$ and study the sign of a single fraction.

Absolute value

The result is never negative: $|-7|=7$ and $|7|=7$. Notice that $-x$ is positive when $x$ is negative, which is why the definition makes sense. An equivalent form is $|x|=\sqrt{x^2}$, because the square root is the positive one.

When there is an expression inside, you have to work out its sign before removing the bars. For example, in $|1-\sqrt{5}|$: since $\sqrt{5}>1$, the inside is negative and the value is $-(1-\sqrt{5})=\sqrt{5}-1$. With a variable, the definition is applied to the inside:

$$|2x-8|=\{\begin{array}{ll}2x-8& \text{if }x\ge 4\\ 8-2x& \text{if }x<4\end{array}$$

The first five follow from $|x|=\sqrt{x^2}$; for example, $|xy|=\sqrt{x^2{y}^2}=\sqrt{x^2}\sqrt{y^2}=|x||y|$.

The triangle inequality has a clear intuition: going directly from one point to another is never longer than taking a detour. If $x=5$ and $y=-3$, then $|x+y|=2$ but $|x|+|y|=8$: when adding numbers with opposite signs there is "cancellation", and the sum ends up closer to zero. Equality holds only when $x$ and $y$ have the same sign (or one of them is $0$). To prove it, it is enough to compare squares: ${\left(|x|+|y|\right)}^2=x^2+2|xy|+y^2\ge x^2+2xy+y^2=(x+y{)}^2$, because $|xy|\ge xy$, and then you take square roots, which preserve the order between non-negative numbers.

Absolute value equations

The first says that there are two numbers at distance $c$ from zero. The second, that two numbers have the same absolute value when they are equal or opposite. To solve $|2x-1|=|x+4|$ consider the two cases $2x-1=x+4$, which gives $x=5$, and $2x-1=-(x+4)$, which gives $x=-1$.

When the equation does not have such a simple form, for example $|x-3|=2x$, you have to split according to the sign of the inside and, at the end, check each result in the original equation: a solution that does not satisfy the condition of its case, or that makes the right-hand side negative, is rejected.

Absolute value inequalities

The important thing is to read it as a distance. $|x-4|<3$ means that $x$ is less than $3$ units away from $4$, that is, $1<x<7$. That is why "less than" gives one interval (an "and" condition) and "greater than" gives two half-lines (an "or" condition).

To solve an inequality with an expression inside, substitute $x$ by that expression: $|2x+1|\le 5$ is equivalent to $-5\le 2x+1\le 5$, that is, $-6\le 2x\le 4$ and $-3\le x\le 2$. Nothing is reversed here because we divide by $2>0$. If the absolute value is compared with an expression that depends on $x$, or there are two absolute values, the usual approach is the third property (squaring) or splitting into cases according to the sign of each inside.