Salvatore Mosaico · 2026
Contents
- 1. AI answers. Is the problem already over?
- 2. The formula contains another route
- 3. An example that suggests a new question
- 4. Halve first, double afterwards
- 5. Change the problem, keep the approach
- 6. “Percent of what?” Not always the right question
- 7. The initial value is unnecessary for finding the relative change
- 8. What if we reverse the order?
- 9. Two opposite changes: the answer needs completing
- 10. From example to formula
- 11. A teaching insight dating back to 2009
- 12. The same structure, two different problems
- Conclusion: who asks the next question?
A simulated lesson on the value of the next question.
One morning I see a video on Facebook: two parallel solutions to the same problem, one attributed to a mathematician and the other to an engineer.
The problem is simple: a right triangle has hypotenuse 25 and one leg 15. How long is the other leg?
I am not interested in deciding who wins this imaginary contest between professions. The alternative method intrigues me: I want to understand it, explore it and see where it leads. So I start a conversation with artificial intelligence.
The following lesson reworks that journey as a dialogue between teacher, students and AI. It is a teaching simulation, not a literal transcript, but it preserves the sequence of the decisive steps.
1. AI answers. Is the problem already over?
Teacher: We have a right triangle with hypotenuse 25 and one leg 15. Let us ask AI to calculate the other leg.
AI: Apply the Pythagorean theorem:
b = √(25² − 15²)
= √(625 − 225) = √400 = 20
Student: Correct answer. We are done.
Teacher: We have found the result. That does not necessarily mean we have finished thinking. Let us ask: “Without Pythagoras?”
AI: We can recognise the triple 3, 4, 5. Multiplying every side by 5 gives 15, 20, 25. The other leg is therefore 20.
Student: This time we did not calculate any squares.
Teacher: True. But this is still not the method I wanted to explore. Let us suggest: “Square root of sum times difference.”
2. The formula contains another route
AI: A difference of two squares can be factored:
c² − a² = (c + a)(c − a)
Therefore:
b = √[(c + a)(c − a)]
In our case:
b = √[(25 + 15)(25 − 15)]
= √(40 · 10) = 20
Student: Have we found a formula different from Pythagoras?
Teacher: A different form, not a different principle. The geometric relationship is still Pythagoras. We have transformed the calculation using an algebraic identity.
We can state an operational rule:
A leg is the square root of the product of the sum and the difference of the hypotenuse and the other leg.
Calling it the sum-and-difference method for teaching purposes helps us recognise it. It does not mean we have discovered a new theorem.
Student: So the name helps us remember a procedure.
Teacher: Exactly. But we still need to ask when that procedure is actually useful.
3. An example that suggests a new question
The conversation also recalls the formulas for generating Pythagorean triples:
a = m² − n²
b = 2mn
c = m² + n²
where m > n > 0 are integers. Choosing m = 8 and n = 7:
a = 64 − 49 = 15
b = 2 · 8 · 7 = 112
c = 64 + 49 = 113
This gives us a new case to explore: hypotenuse 113 and leg 15.
Teacher: Show us an easier calculation without a calculator.
AI:
b = √[(113 + 15)(113 − 15)]
= √(128 · 98)
Observe that:
128 = 64 · 2 = 8² · 2
98 = 49 · 2 = 7² · 2
Therefore:
√(128 · 98)
= √[(8² · 2)(7² · 2)]
= √(8² · 7² · 4)
= 8 · 7 · 2 = 112
Student: AI has found a convenient way to calculate that square root.
Teacher: Yes. Now examine its answer: the same 2 appears in both factors. Can we turn this observation into a rule?
An important step happens here: a calculation supplied by AI suggests a new question to the person.
There is more than a machine answering and a human reading. An intermediate result is examined and becomes an opportunity to move forward.
4. Halve first, double afterwards
Teacher: If the sum and difference are even, could we halve both and then take the factor 2 outside the square root?
AI: Yes. Let S = c + a and D = c − a. Then:
b = √(S · D)
= √[(2 · S/2)(2 · D/2)]
= √[4 · (S/2) · (D/2)]
Therefore:
b = 2√[((c + a)/2) · ((c − a)/2)]
Student: In the previous example this becomes:
b = 2√(64 · 49) = 2 · 8 · 7 = 112
Teacher: Exactly. We can call it the half-sum and half-difference method:
Calculate the half-sum and half-difference, multiply them, take the square root and double the result.
Try hypotenuse 65 and leg 33:
(65 + 33)/2 = 49
(65 − 33)/2 = 16
b = 2√(49 · 16) = 2 · 7 · 4 = 56
Student: Does it work only when the hypotenuse and leg have the same parity?
Teacher: Be careful: the formula is valid for all lengths c > a > 0. When the two numbers are integers with the same parity, the half-sum and half-difference are integers, which may make the calculation easier.
We must not confuse a transformation’s validity with its convenience. For example, with hypotenuse 17 and leg 8, this is enough:
b = √[(17 + 8)(17 − 8)]
= √(25 · 9) = 15
Halving offers no obvious advantage. First observe the numbers, then choose the procedure.
5. Change the problem, keep the approach
Teacher: Let us try another example. An item increases in price by 10%, then decreases by 15%. What is the overall percentage change?
AI: Suppose the initial value is 100 euros. After the increase it is 110 euros; after the decrease:
110 · 0.85 = 93.50
The overall change is therefore −6.5%.
Student: This answer is correct too.
Teacher: Certainly. Choosing 100 helps illustrate the problem. But do we need to choose an initial price to solve it?
Before continuing, we need to clarify what the percentage symbol means.
6. “Percent of what?” Not always the right question
Teacher: What is 5% equal to?
Student: Five percent of what?
Teacher: I did not ask for 5% of a quantity. I asked what the number 5% is equal to.
The percentage symbol means division by one hundred:
5% = 5/100 = 0.05
We do not need a price, a population or a length to understand this equality.
Student: So when do we need to know “of what”?
Teacher: When we need to apply that number to a quantity. For example, 5% of 200 is calculated as follows:
(5/100) · 200 = 10
These are two different questions:
- What is 5% equal to? It is 0.05.
- What is 5% of a quantity? We need to know the quantity.
The misunderstanding arises when a base is demanded even to give meaning to the percentage number.
It is not true that a percentage means nothing on its own. It represents a number. The base comes into play when that number is used to calculate a part or a specific change.
Consequently:
100% = 1
100% + 10% = 110% = 1.10
100% − 15% = 85% = 0.85
This is where the factors we use in calculations come from.
7. The initial value is unnecessary for finding the relative change
Teacher: A 10% increase multiplies the value by 110%. A 15% decrease multiplies the new value by 85%. The combined factor is:
110% · 85% = 1.10 · 0.85
= 0.935 = 93.5%
Student: So the change is 93.5%?
Teacher: No. That is the ratio of the final value to the initial value. To find the change, we must subtract the relative initial value, namely 1 = 100%:
93.5% − 100% = −6.5%
In a single expression:
(100% + 10%)(100% − 15%) − 100%
= −6.5%
We did not need to assign a price to the item. If the initial value is V > 0, then:
V_f = V · 1.10 · 0.85
The relative change is:
(V_f − V)/V
= (V · 1.10 · 0.85 − V)/V
= 1.10 · 0.85 − 1
The initial value cancels out. This is why choosing 100 is a presentational convenience, not a mathematical necessity.
8. What if we reverse the order?
Teacher: Look at the formula: what happens if we first decrease by 15% and then increase by 10%?
Student: We swap the factors:
0.85 · 1.10 = 1.10 · 0.85
The result does not change.
Teacher: We have derived another property without repeating the problem. Intermediate values change, but the final result is identical because multiplication is commutative.
This holds for purely multiplicative changes, each applied to the current value, without intermediate rounding or additional operations.
A formula is not only for calculating: it makes properties visible that a single example may leave in the background.
9. Two opposite changes: the answer needs completing
Teacher: Now the item increases by 30% and subsequently decreases by 30%.
AI: Starting from 100, we first reach 130, then 91. The change is −9%. We can also write:
(1 + 0.30)(1 − 0.30)
= 1 − 0.30² = 0.91
Teacher: Stop there. The numerical result −9% is correct. But the formula has taken us to 0.91, the final factor. To express the change directly, one step is still missing.
Student: Subtract 1!
(1 + 0.30)(1 − 0.30) − 1
= 1 − 0.09 − 1
= −0.09 = −9%
Teacher: Exactly. We are not correcting the example’s result: we are asking the formalisation to distinguish clearly between final factor and change.
Now we can ask the next question: what happens with an arbitrary percentage?
10. From example to formula
Let a and b be the numbers expressing two signed percentage changes: positive for increases, negative for decreases.
The combined factor is:
(1 + a/100)(1 + b/100)
Subtracting 1:
(1 + a/100)(1 + b/100) − 1
= a/100 + b/100 + ab/10000
Rewriting the result in percentage notation gives the total change:
(a + b + ab/100)%
Student: In the first example:
(10 − 15 + 10 · (−15)/100)%
= (10 − 15 − 1.5)% = −6.5%
Teacher: Now take two opposite changes: a = x and b = −x.
(x − x + x · (−x)/100)%
= (−x²/100)%
Here is the rule we were looking for:
Two successive changes +x% and −x%, in either order, produce an overall change of (−x²/100)%.
For a positive price we normally consider 0 ≤ x ≤ 100. If x > 0, the change is negative; if x = 0, it is zero.
For example:
| Successive changes | Overall change |
|---|---|
| +10%, −10% | −1% |
| +20%, −20% | −4% |
| +30%, −30% | −9% |
| +50%, −50% | −25% |
We no longer need to invent an initial price each time and repeat every calculation.
11. A teaching insight dating back to 2009
I had already approached this problem with the same intention in 2009.
I encountered it in various contexts, often presented as a difficult or tricky question. The solutions I saw repeatedly used the procedure: “Suppose the initial value is 100.”
That procedure was correct, but it seemed incomplete as a teaching destination. Why stop at an example when a rule could be formalised for an entire family of problems?
The aim was not to claim a new theorem. It was to explore a solution: generalise it, identify a special case and make a useful consequence explicit.
Today, dialogue with AI makes it possible to revisit that way of working, compare it with other examples and turn it into a readable journey.
12. The same structure, two different problems
Teacher: What connects the triangle to percentage changes?
Student: Both contain a sum times a difference:
(A + B)(A − B) = A² − B²
In the triangle:
(c + a)(c − a) = c² − a²
In opposite percentage changes:
(1 + x/100)(1 − x/100)
= 1 − x²/10000
Teacher: Exactly. The contexts differ, but the algebraic structure is the same.
Exploration lets us recognise it, give it an operational formulation and reuse it. Naming the rule may help us remember it; proving it allows us to understand and apply it correctly.
Conclusion: who asks the next question?
AI’s first answer was correct:
√(25² − 15²) = 20
Had the dialogue stopped there, we would have solved an exercise.
By continuing, however, we encountered a factorisation, explored favourable examples, formulated a half-sum and half-difference rule and recognised the same identity in percentage changes.
The journey did not come from only one side. A human question guided AI; a factorisation proposed by AI suggested a further refinement; teaching experience introduced another problem and called for clearer formalisation.
Competence is not only about obtaining an answer, but knowing what to do with it.
Check it. Transform it. Generalise it. Ask about its conditions of validity. Notice when an intermediate result does not yet answer the precise question asked.
AI can help calculate, explain and organise. But in this journey, experience, curiosity and teaching direction are not incidental details: they turn a sequence of answers into a lesson.
An answer can close an exercise. A new question can open a mathematical journey.