Study least elements, contradictions, and induction alternatives clearly. Check sets, bounds, and counterexamples with confidence. Turn abstract number arguments into concrete, teachable proof workflows.
The page stays single-column overall. The input controls use three columns on large screens, two on medium screens, and one on mobile screens.
| Mode | Example input | Least element or failure | Interpretation |
|---|---|---|---|
| Finite set | {12, 5, 19, 7, 4} | Least element = 4 | A nonempty finite set always has a smallest value. |
| Progression | a = 5, d = 3, lower bound = 20 | Least qualifying term = 20 | The first term meeting the bound anchors the proof. |
| Theorem explorer | f(n) = n² - 3n + 2, test f(n) ≥ 10 | Least counterexample found by scan | The first failure becomes the candidate minimal counterexample. |
| Negative progression | a = 20, d = -2, lower bound = 0 | Infinite set not well-ordered here | Descending terms warn that the proof setup needs revision. |
Well-ordering principle: Every nonempty subset of the natural numbers has a least element.
Counterexample set: \( C = \{ n \in \mathbb{N} : P(n)\text{ is false} \} \). If \( C \neq \varnothing \), then \( C \) has a least element \( m \).
Least element in a finite set: \( m = \min(S) \).
Arithmetic progression subset: \( S = \{ a + kd : k \in \mathbb{N}_0,\ a + kd \ge L \} \).
First qualifying progression index for positive d: \( k_0 = \max(0,\lceil(L-a)/d\rceil) \).
Least qualifying progression term: \( m = a + k_0d \).
Polynomial test expression: \( f(n) = An^2 + Bn + C \).
Proof strategy: Assume a counterexample exists, pick the least one by well-ordering, then contradict its minimality.
It does not replace a formal proof. It organizes sets, identifies least elements, finds minimal counterexamples, and drafts reasoning steps you can turn into a rigorous argument.
A least counterexample is the core contradiction tool. If a smaller case would force the statement at that least failure, the assumed counterexample cannot exist.
Yes. The finite set mode accepts negative integers. For infinite arguments, you usually need a subset translated or constrained so the proof fits a bounded-below framework.
A negative difference creates a descending infinite progression. That setup may fail the bounded-below requirement needed for a clean well-ordering argument.
No. Range checking gives evidence, not a universal proof. It helps you locate a likely least counterexample or build intuition before writing the symbolic argument.
Use it when you want to verify ordering quickly, confirm a least element, or illustrate how a nonempty finite integer set satisfies the least-element idea.
The graph displays sorted set values, sampled progression terms, or polynomial outputs across n. The horizontal reference line marks the chosen lower bound or threshold.
Yes. The tool is useful for lecture demos, worksheet preparation, student practice, and checking whether a proposed minimal-counterexample setup is sensible.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.