Refute a claim—or justify every case
What can a finite search establish about a universal statement?
Entering midway? A universal statement asserts a property for every element of a domain. One failing case disproves it.
Start with a concrete question
Inspect integer claims, then write an argument that distinguishes evidence from proof.
Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.
Go directly to the experiment · Review prerequisite mathematics and return here
What do you expect?
Predict before calculating: What can a finite search establish about a universal statement?
Change one thing. Inspect what happens.
- Predict whether a counterexample exists in the displayed range.
- Inspect the prime expression and then its supplied n=40 case.
- Repair the false square expansion.
- Write a proof for n²≥n with the domain and both cases stated.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Finite tests can find failure; passing dots do not prove a universal claim
Horizontal: nonnegative integer n. Vertical: failure indicator (0 passes this test; 1 is a counterexample).
- 1. Inspected cases 0 through 4 (points)
Axes use the displayed ranges.
Read plotted values as a table
| Series | nonnegative integer n | failure indicator (0 passes this test; 1 is a counterexample) |
|---|---|---|
| Inspected cases 0 through 4 | 0 | 0 |
| Inspected cases 0 through 4 | 1 | 0 |
| Inspected cases 0 through 4 | 2 | 0 |
| Inspected cases 0 through 4 | 3 | 0 |
| Inspected cases 0 through 4 | 4 | 0 |
Follow the calculation
- Claim over nonnegative integers: n²≥n.
- Inspected n=0 through 4: no counterexample in this finite range.
- General argument: n=0 or n≥1; then n(n−1)≥0, so n²≥n.
- A single valid counterexample refutes a universal claim. Passing finitely many cases does not prove a statement about all integers.
- The plot shows the bounded search only, plus the separately supplied n=40 witness for the prime claim. A row of passing dots cannot establish every untested case.
Use examples to seek a counterexample or suggest an argument. Then write a proof whose logic covers the entire stated domain.
Write a proof and a counterexample
Replace the flawed argument. Prove n²≥n for every nonnegative integer by covering n=0 and n≥1. Then give a valid counterexample to the prime claim. State why finite testing alone is insufficient.
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Compare with my earlier draft
I tested n=0,1,2,3,4. Every result worked, so both claims are proved for all integers.
Explain what you changed and why in your observation notes. Text differences do not establish quality.
Inspect an example solution after your attempt
For n=0, n²−n=0. For an integer n≥1, n and n−1 are nonnegative, so n²−n=n(n−1)≥0. Thus n²≥n for every nonnegative integer. The prime claim fails at n=40: 1681=41×41. A finite list does not cover all nonnegative integers.
This is one worked example. Test another input and explain its limitations.
Captures keep your inputs, prediction and observation. Results are recalculated from those inputs. Resetting inputs keeps your captures and challenge history.
Explain the mechanism
Refute a claim—or justify every case
Inspect integer claims, then write an argument that distinguishes evidence from proof.
Concepts to recall
A universal statement asserts a property for every element of a domain. One failing case disproves it.
Understand the mechanism
n²≥n holds for every nonnegative integer because n(n−1)≥0: n=0 gives zero; n≥1 gives two nonnegative factors.
The prime expression survives the small search but fails at n=40, where it equals 41².
A program can exhaust a finite domain, but an unbounded integer claim requires an argument covering all remaining cases.
Symbols and calculation
n²−n=n(n−1); (n+1)²=n²+2n+1. A counterexample must belong to the stated domain.
Worked example
Claim over nonnegative integers: n²≥n.
Inspected n=0 through 4: no counterexample in this finite range.
General argument: n=0 or n≥1; then n(n−1)≥0, so n²≥n.
A single valid counterexample refutes a universal claim. Passing finitely many cases does not prove a statement about all integers.
The plot shows the bounded search only, plus the separately supplied n=40 witness for the prime claim. A row of passing dots cannot establish every untested case.
Your investigation
Predict whether a counterexample exists in the displayed range.
Inspect the prime expression and then its supplied n=40 case.
Repair the false square expansion.
Write a proof for n²≥n with the domain and both cases stated.
Limits and interpretation
- These are bounded teaching models with explicit assumptions, not evidence of a real-world intervention.
Try a fresh case
At n=40, n²+n+41 equals 41 squared. What is its value?
This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.
Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Explain a fresh small numerical case with a diagram or table and an argument.
College investigation
Implement the method, test a boundary or counterexample, and justify its assumptions.
Workplace investigation
Use a small decision scenario, state units and constraints, and test how the recommendation changes.
Keep the evidence, choose your next step
Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.