Inequalities: describe every admissible value

You have 560 currency units for materials. Delivery costs 80 and each kit costs 60. How many kits can you order? An equation finds exact spending; an inequality describes every order that fits the budget.

Start with real numbers and equations. All prices are hypothetical.

From a limit to a set

Let $n$ count the kits. The condition $80+60n\le560$ means spending at most 560. Subtract 80 and divide by positive 60: $n\le8$ . The model also requires an integer $n\ge0$ . The answer is $\{0,1,2,\ldots,8\}$ ; zero means no order and no actual delivery. If an order must contain at least one kit, only 1–8 remain. Nine kits would cost 620.

For a real variable, $x\le8$ instead describes $(-\infty,8]$ . A square bracket includes an endpoint; a parenthesis excludes it. Infinity is not a number on the line, so its end always takes a parenthesis. The strict inequality $x<8$ excludes 8.

Why a negative multiplier reverses the sign

Adding the same number shifts both points equally and preserves their order. Positive scaling also preserves it; multiplying by −1 reflects the line: $2<5$ , yet $-2>-5$ .

Thus $7-2x>1$ gives $-2x>-6$ , and division by −2 yields $x<3$ . Check: $x=2$ gives $3>1$ ; the endpoint $x=3$ gives an excluded equality. A variable multiplier with unknown sign requires separate sign cases. Multiplication by zero destroys information altogether. Inequality properties, OpenStax §2.7.

“And” and “or” produce different sets

The conditions $x>1$ and $x\le4$ hold together on $(1,4]$ , their intersection. With $x>1$ or $x\le4$ , at least one condition suffices, so the union is the entire real line. A system of inequalities means “and”; $1<x\le4$ writes the same system compactly.

Intersection: both conditions

(1, ∞) ∩ (−∞, 4] = (1, 4]

Union: at least one

(1, ∞) ∪ (−∞, 4] = ℝ

An open point excludes 1; a filled point includes 4. Thin grey lines show the real line; thick colored lines show selected values. Uniform scale.

Another example: $x<0$ or $x\ge3$ gives $(-\infty,0)\cup[3,\infty)$ . The gap from 0 to 3 is not filled. Requiring $x<0$ and $x\ge3$ together instead gives an empty intersection, $\varnothing$ .

Branch · Proving a boundA square is always nonnegative.

How to prove an estimate for every x

Every real $x$ satisfies $(x-3)^2\ge0$ . Expanding gives $x^2-6x+9\ge0$ , hence $x^2-6x+11\ge2$ . Equality occurs at $x=3$ , so 2 is the attained lower bound, or minimum. Checking ten individual values would not prove this for every $x$ .

Branch · MeansFor two nonnegative numbers a and b only.

The arithmetic mean is at least the geometric mean

From $(\sqrt a-\sqrt b)^2\ge0$ we get $a+b-2\sqrt{ab}\ge0$ . Therefore $\frac{a+b}{2}\ge\sqrt{ab}$ , with equality only when $a=b$ . For 4 and 9, the arithmetic mean is 6.5 and the geometric mean is 6. This is the two-number AM–GM inequality; $a,b\ge0$ is already required for the real square roots.

Read the answer against every original condition: is an endpoint included, must all conditions hold together, and does the problem permit arbitrary real values? These questions turn an inequality sign into a precise description of what is possible.