LESSON 1 · 17 MIN
Quantitative comparison
Determine whether one quantity is greater, the quantities are equal, or the relationship varies by using exact differences, safe bounds, or admissible counterexamples.
ESSENTIAL QUESTIONCan structure prove one relationship for every allowed value—or can two valid cases force different conclusions?
Prove one relationship without calculating more than the claim needs.
A negative exact difference proves that Quantity B is greater. No decimal approximation is needed.
The ranges never touch, so every allowed value makes Quantity A greater.
| Quantity A | Quantity B | A − B | Certified relation |
|---|---|---|---|
| 34 | 51 | −17 | Quantity B is greater |
| Quantity | Minimum | Maximum | What the interval proves |
|---|---|---|---|
| Quantity A = x + 5 | 11 | 14 | Every value lies above 10 |
| Quantity B = 10 | 10 | 10 | The fixed comparison ceiling is 10 |
Overlapping ranges alone prove neither equality nor variation. Switch to structure or admissible cases before drawing a conclusion.
Break a universal claim with two cases the prompt actually allows.
Text equivalent: the reciprocal comparison permits the nonzero integers negative four, negative three, negative two, negative one, one, two, three, and four. Zero is excluded.
Complete reciprocal domain: {−4, −3, −2, −1, 1, 2, 3, 4}. Zero is excluded before any value is tested.
| Allowed input | Quantity A | Quantity B | Outcome | Evidence role |
|---|---|---|---|---|
| x=−3 | 9 | −3 | Quantity A is greater | Selected witness |
| x=−2 | 4 | −1 | Quantity A is greater | Supporting case |
| x=−1 | 1 | 1 | The quantities are equal | Supporting case |
| x=1 | 1 | 5 | Quantity B is greater | Selected witness |
| x=2 | 4 | 7 | Quantity B is greater | Supporting case |
| x=3 | 9 | 9 | The quantities are equal | Supporting case |
| Allowed input | Quantity A | Quantity B | Outcome | Evidence role |
|---|---|---|---|---|
| x=−4 | −1/4 | −4 | Quantity A is greater | Selected witness |
| x=−3 | −1/3 | −3 | Quantity A is greater | Supporting case |
| x=−2 | −1/2 | −2 | Quantity A is greater | Supporting case |
| x=−1 | −1 | −1 | The quantities are equal | Supporting case |
| x=1 | 1 | 1 | The quantities are equal | Supporting case |
| x=2 | 1/2 | 2 | Quantity B is greater | Selected witness |
| x=3 | 1/3 | 3 | Quantity B is greater | Supporting case |
| x=4 | 1/4 | 4 | Quantity B is greater | Supporting case |
Witness rows say “Selected witness” in text as well as using a visual highlight. Each selected pair contains one case where A is greater and one where B is greater.
Subtract before solving
The sign of Quantity A minus Quantity B identifies the relationship when the difference can be simplified safely.
- A positive difference means Quantity A is greater.
- Do not divide by an expression whose sign or nonzero status is unknown.
Use the tightest bound
If the smallest possible value of one quantity exceeds the largest possible value of the other, the comparison is settled without exact evaluation.
- Carry every domain restriction into the bound.
- Overlapping bounds do not prove equality or indeterminacy by themselves.
Disprove with admissible cases
When a fixed relationship is not guaranteed, two allowed inputs that produce different outcomes prove that no single relationship follows.
- Every witness must satisfy the complete stated domain.
- Two agreeing examples are not a universal proof.
Worked example
For x in {−3, −2, −1, 1, 2, 3}, compare Quantity A = x² with Quantity B = 2x + 3.
- 1
The domain is finite and explicitly excludes 0, so only the six listed values are admissible.
- 2
At x = −3, Quantity A is 9 and Quantity B is −3, so Quantity A is greater.
- 3
At x = 1, Quantity A is 1 and Quantity B is 5, so Quantity B is greater. The two valid outcomes disagree.
ConclusionThe relationship varies across the allowed domain; no single greater-than or equality statement is determined.
Calculator decision ledger
- Model mentally
- Simplify the difference, identify usable bounds, and write every allowed-value restriction before entering numbers.
- Estimate
- Predict the sign or locate a likely sign-changing boundary before exact evaluation.
- Calculator
- Use arithmetic only when exact simplification, bounds, or small admissible cases do not settle the relation.
- Audit
- Check denominator signs and confirm that every counterexample satisfies the complete stated domain.
Close the notes first
Retrieve the proof.
01What does the sign of A − B prove?
Subtracting places the comparison on one exact sign test.
02When do bounds prove Quantity A is greater?
Then every admissible A lies above every admissible B.
03What proves that a relationship varies?
A single fixed conclusion cannot survive both valid cases.