What is mathematics: from a question to a justified answer

First decide which problem we are solving

“How should we arrange the plot?” has many possible answers. Access, ease of work or proximity to a path might matter. Let’s begin with a narrower question: which of the proposed rectangles gives more area for the same boundary length?

Imagine a flat surface and a complete rectangle. Its angles are right angles, like the corner of a sheet of paper, and opposite sides are equal. All edging goes around the four sides, with no gate, obstacles or gaps. We ignore material thickness and joint allowances: the boundary on our plan is a thin line. We take lengths as exactly specified.

These are our assumptions, the conditions supporting the answer. We chose them to isolate the relationship between shape and area. Every dimension here is imaginary, not measured on a real plot.

Moving to quantities, geometric shapes and relationships between them creates a mathematical model of the situation. We must also check whether the conclusion makes sense for the original question. These elements of modeling are described in the SIAM and COMAP GAIMME guide, chapter 1.

Boundary length and the space inside

Walk around the 10-by-2-metre rectangle. The four sides add up to 10 + 2 + 10 + 2 = 24 metres. The total boundary length of a shape is its perimeter. This is how much edging our model needs.

To count the space inside, imagine squares with sides of 1 metre. Each has an area of 1 square metre, abbreviated 1 m². The long rectangle holds two rows of ten squares: 10 × 2 = 20 m². Multiplication shortens the counting of equal rows.

Now take sides of 7 and 5 metres. The boundary measures 7 + 5 + 7 + 5 = 24 metres, and inside there are five rows of seven squares: 7 × 5 = 35 m². This distinction between linear and square measure is explained in OpenStax, Prealgebra 2e, §9.4.

Both drawings use the same scale. Compare the outline and the number of cells: they answer different questions.

The second option gives 15 m² more, because 35 − 20 = 15. We now have a conclusion we can check in two ways: count cells, or multiply the number of rows by the number of squares per row.

What is mathematical here besides multiplication?

Multiplication gave us a number. Before that, however, we had to notice two different quantities, choose conditions and decide how to compare options. Afterwards, we explained why the count answered the question about area.

Mathematics investigates quantities, shapes, structures and patterns, and seeks reasons for claims about them. This is a useful orientation for getting to know the field. It also includes questions about change, data and uncertainty. For example, the educational OECD PISA mathematics framework distinguishes quantity, space and shape, change and relationships, and uncertainty and data.

Questions can also extend beyond immediate practical needs: what properties does a structure have, and which statements follow from given rules? Investigating these is mathematics too. The edging problem simply gives us a concrete starting point.

We can already see a relationship: the boundary length can stay constant while the area changes. To examine other dimensions, it helps to write this more compactly.

Letters appear when they save work

Call the lengths of two adjacent sides a and b, the perimeter P, and the area S. We measure sides and perimeter in metres, and area in square metres. Here a and b are variables: we can choose different values when comparing rectangles.

For example, the wider plot has a = 7 m and b = 5 m. A letter does not conceal a new operation; it lets us discuss a side without choosing a particular length yet. Variables are explained in OpenStax, §2.1.

The four sides form two equal pairs. This gives the compact expression:

First add two adjacent sides inside the parentheses. Then multiply by 2, because that pair occurs twice. For sides of 7 and 5 metres: 2 × (7 m + 5 m) = 2 × 12 m = 24 m.

The area expression is even shorter:

Here ab means a multiplied by b. Substitute our lengths: 7 m × 5 m = 35 m². This is the same counting of rows of squares we already performed. Rectangle formulas are given in OpenStax, §9.4.

The = sign says that both sides represent the same value. So “7 + 5 = 12” and “12 = 7 + 5” are equally correct. In our problem, a formula is a compact description of a relationship between quantities.

Drawing, table and formula describe the same relationship

With a perimeter of 24 metres, two adjacent sides total 12 metres: the same pair occurs on the opposite side. We can therefore choose one side and use what remains of 12 for the other.

Compare a few more options:

The 7-and-5 row matches the second drawing. For 8 and 4, imagine four rows of eight squares, or substitute into the formula: 8 × 4 = 32 m². The table lets us compare results quickly; the drawing shows where they come from.

In these options, side lengths get closer together while area grows. The largest area among the listed options is 36 m². A 6-by-6-metre square is also a rectangle: all its angles are right angles and opposite sides are equal; its adjacent sides are equal too.

But we have checked only a few dimensions. Others could involve half-metres or smaller fractions. “What is largest in this table?” and “What is largest among all possible shapes?” require different justifications.

When is an example enough, and when do we need proof?

The pattern in the table can suggest a conjecture, a statement still requiring justification. Proving it for all permitted dimensions requires reasoning that covers all those cases. Simply adding more rows does not provide that.

Compare this with the perimeter formula. Opposite sides of any rectangle are equal. Following the boundary gives a + b + a + b. Group the equal pairs to obtain 2(a + b). We did not substitute special numbers, so the reasoning works for arbitrary positive side lengths. This is a short general justification based on the shape’s properties.

An example has another useful role. The claim “equal perimeter always means equal area” has already been disproved by our two plots: the condition holds, but their areas differ. Such an example is a counterexample. One contradicting case is enough to disprove “always.”

The distinction between general proof and disproof by counterexample is explained in Richard Hammack’s Book of Proof, §4.3 and §9.1. Our original problem already has a sufficient verified conclusion: 7 by 5 metres encloses more space than 10 by 2. We have not proved which shape is best among all possibilities.

What changes on a real plot?

An ideal rectangle with sides of exactly 7 and 5 metres has exactly 35 m² of area. But measuring “about 7 metres” no longer specifies a length exactly. Angles other than right angles require a different geometric model. If the edging has noticeable thickness, we must clarify where our 7 and 5 metres run: along the inside or outside edge.

The calculation remains correct under the stated conditions. Now we must check how well those conditions fit the real object. This is why GAIMME, pp. 12–13 and 18 includes analyzing results and revisiting assumptions as parts of modeling.

The question itself can change too. If we need a path, the greatest total area may not give the most usable space. We can add it to the plan and calculate the available area separately. That extends the model to match what we actually want to know.

An exact answer is useful alongside its conditions: “35 m², assuming a complete rectangle measuring 7 by 5 metres.” This makes both the strength of the conclusion and its limits understandable.

Choose tools to suit the question

Within one problem, we have used several mathematical tools. They connect with one another and can help in other situations.

This is a short map, not a complete division of the field. Some programming and scientific problems need several fields at once. A concrete first problem can be understood through accessible operations and a drawing; new tools are introduced when we need them.

How to approach the next problem

Mathematics helped us move from a vague “what is better?” to a conclusion with conditions and an explanation. We chose an ideal rectangle, distinguished boundary length from area, checked two options and found that, with the same 24 metres of edging, 7 by 5 metres encloses 15 m² more than 10 by 2.

For the next question, start here:

1. What is known, and what do I want to find? Name quantities and units.
2. What will I treat as essential? State assumptions and what simplifies the model.
3. Which representation will help? A drawing, table or formula can reveal different aspects of one relationship.
4. Why does the conclusion follow? Check operations, units and reasoning, then return to the original situation.

Try drawing an 8-by-4-metre rectangle on paper, treating each cell as a 1-by-1-metre square. Before counting, predict whether the area will exceed 20 m². Then check: four rows of eight give 32 m², and following 8 + 4 + 8 + 4 around the boundary gives the same 24 m. The drawing, count and formula should tell the same story.