Examples, explanations and counterexamples

An example describes one case

Define the model: a whole number of separate full groups, including zero, four stones per group, and no loose stones. Let g count groups and T count stones, so T = 4 × g. Our earlier arrangement with two loose stones instead used T = 4 × g + 2.

At g = 2, eight stones is a particular result. “Every allowed arrangement contains at least four stones” covers every group count the model allows. Words such as “every” and “always” require checking the stated scope.

Explain why, rather than only showing a result

An explanation connects what you know to what follows. Three full groups contain twelve stones because each contributes four, none overlap, and all stones are counted once. The multiplication records this structure. Saying “I calculated twelve” gives a result; explaining the groups shows why the calculation fits.

A general explanation must use features shared by all cases in its scope. For “Adding one full group adds four stones,” the reason is that the new group contains exactly four additional stones. It does not depend on whether you started with two or twenty groups. Keep the original groups unchanged.

A counterexample shows an allowed failure

Test the claim that every allowed arrangement has at least four stones. Group counts 1, 2 and 3 give totals 4, 8 and 12, which agree. Now try g = 0. Zero groups are allowed, and no loose stones remain, so T = 0. Zero is less than four.

This counterexample meets the conditions but violates the conclusion. One valid counterexample refutes a claim about every allowed case. The cases that agreed still agree. What fails is “every,” not the arithmetic for one, two or three groups.

Check both parts of a counterexample

First ask whether the proposed case satisfies the assumptions. Then ask whether its result fails the conclusion. A group containing only three stones is not a counterexample here: it is not a full group of four. A negative group count also falls outside this counting model.

If the model instead requires at least one full group, zero groups are outside its conditions too. Do not quietly change the model while testing it. In the earlier model with two fixed loose stones, zero groups give two stones, not zero; use that model’s conditions and formula together.

Repair a claim by stating a useful condition

A useful repair is: “An arrangement with at least one full group of four contains at least four stones.” We keep no loose stones and whole group counts, but require g to be at least one. Zero no longer contradicts this revised claim because it is outside the revised scope.

One full group already supplies four stones. Other full groups add stones without removing those four. This reason covers every allowed positive group count. In another problem, merely excluding the one failing case you found would not automatically justify a revised claim.

Say exactly what your evidence establishes

Several successful checks help you explore a claim, but they are not by themselves a proof for every possible group count. A list can settle a claim if it covers every allowed case and each check is sound. Checking three counts does not cover our unlimited whole-number counts.

A proof gives a complete mathematical reason for the stated scope. Here an explanation using groups can suffice; formal methods come later. Distinguish a supporting example, a counterexample, and a reason covering all allowed cases. Finding no counterexample is not itself a proof.

Worked examples

Try it yourself

1. For the original no-loose-stone claim, which refutes it: zero full groups, or one group containing three stones?

   Show the answer and reasoning

   Zero full groups is a valid counterexample. The three-stone group is outside the conditions.

   Zero is allowed and gives fewer than four stones. A group of three does not meet the required group size.

2. In T = 4 × g + 2, g can be zero or a positive whole number. Explain why every allowed total is at least two stones.

   Show the answer and reasoning

   The two loose stones are always present. Any full groups add stones without removing them.

   This reason covers every allowed group count. The zero case gives exactly two; positive group counts give more.

3. Repair the multiplication claim by requiring a positive whole starting number and a whole multiplier of at least two. Explain why the result is larger.

   Show the answer and reasoning

   There are at least two copies of a positive count, so the result is larger than one copy.

   One copy supplies the original count; at least one extra positive copy increases it. This explanation uses the revised conditions.

What to carry forward

• State a claim’s scope before testing or explaining it.
• A valid counterexample meets the assumptions and fails the conclusion.
• Supporting cases help exploration; a general conclusion needs a reason covering its whole scope.