# Inequalities

Here is a review of the ideas behind inequalities, especially compound inequalities.

## Compound statements

Notice that the equation t2 = 1 has two solutions: t = 1 and t = −1. This equation is not equivalent to the equation t = 1, nor it equivalent to t = −1, because each of these equations is missing one of the solutions. Instead, it's equivalent to this compound statement:
t = 1 or t = −1.
So that's how I've been writing the solutions of equations with multiple solutions.

In general, a compound statement consists of simpler statements (like equations or inequalities) joined by either the word ‘and’ or the word ‘or’. (If you study Logic, you'll see that there are other words that might be used here, like ‘if’ or ‘but not’. However, we won't need such connectives in this course.) For example:

• t = 1 or t = −1;
• x = 2 or x = 3;
• x = 2 or y = 3;
• x < 2 or x ≥ 3;
• x > 2 and x ≤ 3;
• p = 1 or p = 5 or p = 7;
• x > 2 and x ≤ 3, or x > 4.
The last couple of examples are extra-compound; for example, to form the last statement, you first join ‘x > 2’ to ‘x ≤ 3’ with ‘and’, then join this compound statement to ‘x > 4’ with ‘or’. But statements that complicated are pretty much never going to come up in this course.

Particularly common are compound inequalities, each of which consists of two inequalities joined with ‘and’ and sharing one expression. An example is

x > 2 and x ≤ 3
above. This example can be summarised simply as
2 < x ≤ 3;
that is its form as a compound inequality. (Notice that x > 2 is equivalent to 2 < x, and note that a bunch of statements run together always have implicit ‘and’s between them.)

Here are some more examples:

Compound inequality:Expanded meaning:
2 < x ≤ 3; x > 2 and x ≤ 3;
−1 < y < 6; y > −1 and y < 6;
5 > a ≥ 3; a < 5 and a ≥ 3;
2 < 5x − 9 ≤ 3; 5x − 9 > 2 and 5x − 9 ≤ 3.
In this last example, I used a complicated algebraic expression, just to show that you can do this; however, it's not considered to be solved. Even the example before it, 5 > a ≥ 3, is not really solved in the way that we usually like; it's preferable to write it as 3 ≤ a < 5.

The words ‘and’ and ‘or’ are pretty fundamental English words, but for purposes of Mathematics, I should tell you exactly what they mean. A compound statement with ‘and’ is true only if both statements are true, and false if either statement (or both!) is false. Conversely, a compound statement with ‘or’ is true if either statement (or both!) is true, and false only if both statements are false.

In particular, mathematicians always use ‘or’ in an inclusive sense, so that if both statements are true, then the compound statement is true. For example, consider the compound statement x ≤ 2 or x > 1; this statement is simply True. Sometimes this is because the first part (x ≤ 2) is true, sometimes it's because the last part (x > 1) is true, and sometimes it's because they are both true (which happens when 1 < x ≤ 2). But one way or another, no matter what value x takes, at least one of the two parts is true, so the compound statement is true.

## Interval notation

For equations, the solution sets can usually just be given as a list: {2}, {2, 3}, etc. (The main exceptions are when an equation is always False or always True; then the solution set is ∅ or ℝ, respectively.) For inequalities, you can instead use interval notation, which comes in various flavours:
• If there is a limited range of values, as we have with a compound inequality, give the first value and the last value, separated by a comma; put round parentheses or square brackets around this pair, depending on whether these values are or are not included. For example, the solution set for x of −3 < x ≤ 2 is (−3, 2]; the round parenthesis around −3 indicates that −3 is not included, while the square bracket around 2 indicates that 2 is included.
• If there is an unlimited range of values in the negative direction, then use the symbol ‘−∞’ (pronounced ‘minus infinity’) as the first value; always use a round parenthesis there (since −∞ itself is not a real number, so it can't possibly be a solution). For example, the solution set for y of y < 3 is (−∞, 3), while the solution set for y of y ≤ 3 is (−∞, 3].
• If there is an unlimited range of values in the positive direction, then use the symbol ‘∞’ (pronounced ‘infinity’) as the last value; again always use a round parenthesis there. For example, the solution set for y of y > 3 is (3, ∞), while the solution set for y of y ≥ 3 is [3, ∞).
• If there are two or more disjoint ranges of values, then list them all, separated by the symbol ‘∪’ (pronounced ‘union’); it's helpful to list these in increasing order. For example, the solution set for x of x ≤ −3 or x > 2 is (−∞, −3] ∪ (2, ∞).
These continuous ranges (as I've been calling them) are intervals in the real line; that's why writing solutions sets this way is called interval notation.

Compare interval notation to graphs; you'll see that (except for −∞ and ∞) the round parentheses and square brackets match up perfectly; they're used in the same places, facing in the same directions. This is no coincidence, of course; the notation is designed to work this way!

## Solved examples

Here's a linear inequality; the basic technique is the same as for a linear equation:
• 5y + 4 < 2y − 2 — original inequality;
• 3y + 4 < −2 — subtract 2y from both sides;
• 3y < −6 — subtract 4 from both sides;
• y < −2 — divide both sides by 3.
To check, 5(−2) + 4 = −6, and 2(−2) − 2 = −6, so the original inequality is just barely false when y = −2 (as it should be). To check the direction of the inequality, I'll try y = −3; then 5(−3) + 4 = −11, and 2(−3) − 2 = −8, and −11 < −8. So this checks. In other words, the solution set for y is (−∞, −2).

Here's an example where I have to change the direction of the inequality:

• 2c + 5 ≥ 3c + 7 — original inequality;
• c + 5 ≥ 7 — subtract 3c from both sides;
• c ≥ 2 — subtract 5 from both sides;
• c ≤ −2 — take the opposite of both sides.
To check, 2(−2) + 5 = 1, and 3(−2) + 7 = 1, so the original inequality is just barely true when c = −2. To check the direction, try c = −4 this time; then 2(−4) + 5 = −3, and 3(−4) + 7 = −5, and −3 ≥ −5. So this checks; the solution set for c is (−∞, −2].

To solve compound statements, you can just treat each statement separately. However, if you have a compound inequality, like

2 < x + 1 ≤ 5,
then instead of breaking this up into 2 < x + 1 and x + 1 ≤ 5, then solving each of these by doing the operations to both sides of both inequalities, you simply do those operations to all sides of the original compound inequality. So if I subtract 1 from all sides, then I get
1 < x ≤ 4;
that's the answer. To check when x = 1 (which should be barely false), 1 + 1 = 2, 2 = 2, and 2 < 5, so the original inequality is just barely false. To check when x = 4 (which should be barely true), 4 + 1 = 5, 2 < 5, and 5 = 5, so the original inequality is just barely true. Finally, I'll try x = 2 (which should be quite true); 2 + 1 = 3, 2 < 3, and 3 < 5, so the original inequality is quite true. So this checks; the solution set for x is (1, 4].
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This web page was written by Toby Bartels, last edited on 2020 November 24. Toby reserves no legal rights to it.

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