You have probably memorised the rule: multiply an inequality by a negative number, and the inequality sign flips. Most students treat this as an arbitrary piece of bookkeeping — a magic "remember to flip" step you perform because the teacher said so. It is not arbitrary at all. It is a direct visual consequence of what multiplication by -1 does to the number line: it reflects every point across zero. If your solution set was the right ray x > 3 before, mirroring sends it to the left ray x < -3. The inequality sign flips because the geometry forces it to.
The widget below shows this in slow motion. A number line runs from -10 to 10. You set a threshold a with the slider. The original inequality x > a shades everything to the right of a. Hit "Multiply by −1" and watch the shading mirror across zero: every point x in the shaded set moves to -x. The new shaded set is what you get when you multiply both sides of the original inequality by -1, and the inequality that describes it — once you rewrite it in terms of the original variable — has the sign flipped.
The widget
Why the flip happens — the geometric argument
On the real line, multiplying by -1 is a reflection through the origin. The point 3 maps to -3. The point -4.2 maps to 4.2. Importantly, the map x \mapsto -x reverses order: if a < b before the reflection, then -a > -b after.
You can feel this with two pebbles. Put one pebble at 2 and another at 5. The pebble at 5 is to the right of the one at 2 — this is the statement 2 < 5. Now reflect both across zero. The pebble that was at 2 lands at -2; the pebble that was at 5 lands at -5. But -5 is now to the left of -2 — the reflection swapped which one is bigger. So -2 > -5. The order reversed.
A shaded region is just "all the pebbles satisfying some condition." When you reflect the entire line, the shaded region reflects too. If the original condition was "to the right of a" — i.e., x > a — the reflected condition is "to the left of -a" — i.e., x < -a. The inequality sign flipped because the geometric direction "to the right of" became "to the left of."
That is the whole story. The rule is not an algebraic accident. It is what reflection does.
The algebra, step by step
Start with x > 3. You want to multiply both sides by -1.
- Before: x > 3.
- Multiply both sides by -1: -x \text{ ? } -3 — what goes in the slot?
- If you leave the sign as >, you claim -x > -3. Test x = 4: the original says 4 > 3 (true), and your proposed version says -4 > -3 — but -4 is to the left of -3 on the number line, so this is false. The sign cannot stay.
- Flip the sign: -x < -3. Test x = 4: -4 < -3 — true. The flip fixes it.
So x > 3 \iff -x < -3. And if you prefer to state the solution in terms of x, multiply both sides of -x < -3 by -1 again (flipping once more): x > 3. You are back where you started, which is a good consistency check.
But here is the productive direction: the purpose of multiplying by -1 is usually to clean up a negative coefficient. Suppose you are solving -x < -3 and you want the answer in the standard form "x is greater/less than something." Multiply by -1 and flip: x > 3. Done.
Example 1: x > 3 mirrors to x < -3
Run the widget with the slider at a = 3. The top state shades everything to the right of 3 — the set (3, \infty). Press the button. The shading flows across zero and lands on everything to the left of -3 — the set (-\infty, -3). The inequality x > 3 became -x < -3, which rewrites to x < -3.
Check a specific value: x = 4 is in the original set (4 > 3, yes). After multiplying by -1: -x = -4, and the claim is -4 < -3. True. So the reflected set — described in the variable -x — correctly contains the reflected point.
Example 2: solving 2x - 5 > 7 two ways
Now the serious payoff. You are solving a real inequality, and you want to see why the "flip" rule makes the two solution paths agree.
Path A — isolate x with only positive multipliers.
Add 5 (no flip — addition never flips an inequality):
Divide by 2 (positive — no flip):
Solution: x \in (6, \infty).
Path B — introduce a negative on purpose, then flip.
Start again from 2x - 5 > 7. Subtract 2x from both sides:
Subtract 7:
Divide both sides by -2 — and flip:
Which is the same as x > 6. Solution: x \in (6, \infty).
Both paths agree. Path B deliberately introduces a negative coefficient on x and then cleans it up with the flip. If you had forgotten the flip, Path B would have given 6 > x, i.e., x < 6 — the interval (-\infty, 6), which is the exact opposite of the correct answer. The flip is the difference between "right answer" and "wrong half of the number line."
A common trap: flipping twice
If you multiply by -1 twice in the same manipulation, the inequality is unchanged — two flips cancel. This sounds obvious, but students sometimes flip the sign once when they multiply by -2 and then flip it again when they divide by -1 as a separate step. Don't double-count. One multiplication by a negative number is one flip, regardless of how many separate steps you think of it as.
Where to next
- Intervals and Inequalities Preview — the parent article, covering all four interval types, the unbounded cases, and full problem-solving technique for linear and absolute-value inequalities.
- Absolute Value Inequalities — where the "distance on the number line" interpretation replaces the mirror-and-flip rule, and the answers become bounded intervals instead of rays.
- Operations and Properties — the foundational rules of <, \le, >, \ge on \mathbb{R}, including the order-reversing property of multiplication by negatives, proved from the axioms.
The "mirror and flip" picture is not just a mnemonic. It is the fact that the real line has an orientation, and negating a number reverses it. Every inequality rule involving negative multipliers ultimately reduces to this one geometric move.