How to Solve Mathematics Problems With Super Intelligence

1. The goal is an answer you can defend

To solve mathematics problems with Super Intelligence, first make your own representation of the question, then use AI for a bounded hint or proposed method, and independently verify the result. Check the original conditions, each important transformation, the units and the completeness of the answer. Finish by solving a related problem without help. A convincing explanation on screen becomes learning only when you can explain and reproduce the mathematical decisions yourself.

This is Article 057 in the eduKateSG Super Intelligence Learning series. Here, “Super Intelligence” or “SI” is the series label for working with contemporary AI tools. It does not mean that current systems have been proved to possess artificial superintelligence, that their mathematics is infallible, or that a particular prompt guarantees a correct answer. An AI response is a proposed piece of mathematical work. Its validity comes from the definitions, assumptions and reasoning that support it.

The course is for students, parents supporting study, and adult learners who want useful assistance without surrendering mathematical judgment. The examples move from percentages and ratios to algebra, geometry, probability, proof and elementary optimisation. You do not need to know every topic before starting. Choose a problem at a level you can partly understand, identify a specific difficulty, and study the prerequisite rather than rushing through increasingly advanced solutions.

All problems, practice questions and deliberately incorrect responses in this guide are original teaching examples. They are not official examination questions, actual transcripts from a named AI product, or evidence about a model’s benchmark performance. The fictional settings keep the mathematics inspectable. For assessed work, follow the teacher’s or institution’s rules about calculators, AI assistance, disclosure and independent work. Use practice problems when outside help is prohibited on the actual assignment.

A useful working cycle has six actions: formulate, represent, attempt, inspect, verify and transfer. Formulate means deciding precisely what must be found. Represent means turning the situation into quantities, diagrams or relationships. Attempt means doing something yourself before seeing a complete solution. Inspect means understanding the proposed method. Verify means testing the answer against independent evidence. Transfer means using the idea when the numbers, wording or conditions change.

Keep two small records. On the mathematics page, write the solution that another learner could follow. On the learning page, write what you initially misunderstood, what assistance you used and what you can now do unaided. A copied solution can look excellent while leaving the second page empty. That is a useful warning: the task may be finished, but the learning is not yet demonstrated.

The IES practice guide on organising instruction and study recommends spacing learning, alternating worked examples with independent problem solving, and using retrieval and explanatory questions. This article turns those broad recommendations into a practical study routine. The particular examples and suggested timings are editorial choices, not a validated intervention or a promise of a particular examination result.

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2. Write a mathematical problem contract

Before asking an AI system to solve anything, write a short statement of the problem in your own words. Identify what is given, what is unknown, the allowed values of the unknowns, the relationships between quantities, and the required form of the answer. This is a mathematical problem contract: a compact description of what would count as a valid solution.

Consider the request, “Find the best ticket price.” It cannot be solved as written. Best could mean highest revenue, largest attendance, maximum profit or widest access. Demand may depend on price. Capacity may be limited. A venue fee may matter for profit but not revenue. An assistant that immediately produces a number has silently invented parts of the problem. The right first step is to identify the missing objective and information.

Now consider a complete version: “A club sells adult tickets for $12 and student tickets for $7. It sells 40 tickets and collects $365. How many of each type were sold? Assume every ticket is one of these two types, every ticket is paid for, and there are no other fees or discounts.” Let a be the number of adult tickets and s the number of student tickets. Both are non-negative integers. The total-count relationship is a + s = 40. The money relationship is 12a + 7s = 365.

The contract immediately catches several possible mistakes. The unknowns count tickets, so an answer of 17.5 adult tickets would be inadmissible. The coefficients 12 and 7 are dollars per ticket. The right side of the money equation is dollars, while 40 belongs to the ticket-count equation. A solution that swaps those totals is solving a different model, even if the subsequent algebra is flawless.

Substituting s = 40 − a gives 12a + 7(40 − a) = 365. Expanding yields 5a + 280 = 365, so a = 17 and s = 23. Verify both relationships independently: 17 + 23 = 40, and 17 × 12 + 23 × 7 = 204 + 161 = 365. A second method starts with 40 student tickets costing $280. The extra $85 must come from adult tickets, each adding $5, so there are 17 adult tickets.

Change the collection to $367 while keeping every other condition. The equations give a = 17.4. Do not round that to 17 or 18. Neither rounded value satisfies the stated receipts. Instead, conclude that the information is inconsistent under the model. Perhaps a number was copied incorrectly or an unmentioned discount exists, but those are possibilities to investigate, not facts to add. Inconsistency can be the correct mathematical finding.

A well-scoped request to AI is therefore: “Here is my definition of the variables and my two equations. Check whether they represent the question. Do not solve them yet. If a condition is missing, name it.” This separates understanding the problem from carrying out a procedure. It also gives you a chance to correct the model before a long chain of calculations makes the wrong starting point harder to notice.

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3. Choose a representation that exposes the relationship

The same problem can be expressed in words, a diagram, a table, an equation or a graph. Different representations make different errors visible. The purpose of changing representation is to reveal a relationship, not to decorate the page. If a picture introduces assumptions that were absent from the wording, it can make the solution worse rather than better.

For the ticket problem, a two-row table separates count, unit price and total money. An equation captures each total compactly. A graph of the two linear equations would show their intersection, but the graph must still be interpreted under the integer-count restriction. If a plotted intersection lies between whole numbers, the situation cannot be repaired by choosing the nearest point. Representation does not remove the original conditions.

For ratios, equal-sized bars often expose the unit being compared. Suppose red and blue counters are initially in the ratio 3:5. Drawing three equal red units and five equal blue units is meaningful only if each unit represents the same number of counters. If a learner draws three red circles and five blue circles while the actual total is 64, the picture must be understood as ratio units, not individual counters. Write what one mark represents.

For geometry, label what is given and what is inferred. A sloping side that looks longer than another side is not a measured inequality unless the problem says the diagram is to scale or gives enough information. A corner that looks square is not a right angle without a mark or a mathematical reason. Ask AI to list the facts actually supplied by the diagram before suggesting a theorem. If an image is blurred, transcribe the relevant labels yourself or seek a clearer copy.

For changing quantities, draw a timeline. A tank may fill at one rate before a tap closes and another rate afterwards. A journey may contain equal distances but unequal travel times. A timeline prevents a single average from being applied across stages that have different conditions. Label the start, each change and the event being asked about. Decide whether an endpoint belongs to one stage or the next when that affects the calculation.

An effective self-check is to translate your representation back into ordinary language. Read 2w + l = 60 as “two widths and one length use sixty metres of fencing.” If the original question requires fencing on all four sides, your equation is wrong. Read 0.8P as “eighty per cent of the original price.” If the discount is applied to a different subtotal, the expression is wrong. This reverse translation is often more revealing than asking a second system to solve the same equation.

When requesting help, ask for a comparison: “Show how a bar model and an equation describe the same quantity. Explain which feature makes the equation valid.” Then cover the response and redraw the representation. If you cannot reconstruct the labels, reduce the difficulty. A simpler problem with a fully understood diagram is a better foundation than an elaborate solution whose symbols remain mysterious.

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4. Keep domains, assumptions and units visible

A domain is the set of values under consideration. It may be specified directly, such as real numbers, or arise from the situation, such as whole numbers of buses. An algebraic expression can have additional restrictions. A denominator cannot be zero. A real square root requires a non-negative radicand. A length in a non-degenerate rectangle must be positive. Write these restrictions before transformations that might conceal them.

For example, (x² − 9)/(x − 3) simplifies to x + 3 only for x ≠ 3. Factoring the numerator gives (x − 3)(x + 3). Cancelling the common factor is valid wherever that factor is non-zero. The simplified expression has a value at x = 3, but the original fraction does not. A computer algebra system may display the shorter expression while the learner remains responsible for the original restriction.

Units are another form of meaning. Distance divided by time has units of speed. Litres per minute multiplied by minutes gives litres. Adding 5 metres to 3 square metres is not meaningful as an ordinary measurement sum. A calculation may have numerically reasonable digits and still be dimensionally wrong. Keep the units attached through at least the setup and the final interpretation, rather than adding a unit as an afterthought.

Suppose a rectangular container has an internal base 40 cm by 25 cm. Water enters at 2 litres per minute for 3 minutes, with no leak and enough capacity. The added volume is 6 litres, or 6,000 cm³. The base area is 1,000 cm². The water rises by 6,000 cm³ ÷ 1,000 cm² = 6 cm. Dividing 6 litres by 1,000 cm² without converting produces a mixed-unit expression, not a height of 0.006 cm.

Now inspect the assumptions. The base dimensions are internal. The container has vertical sides over the relevant height. There is no overflow. The inflow is constant. The question concerns the increase in height, so the starting height is unnecessary unless capacity might be reached. If a sloping-sided vessel replaces the rectangular container, volume divided by the original base area need not give the height increase. A familiar formula has a domain of application too.

Precision needs the same care. A question may specify exact dimensions for a mathematical exercise or rounded measurements for a real object. If a length is reported to the nearest centimetre, its true value is not known to unlimited decimal places. A solver that prints twelve digits has not created twelve digits of measurement accuracy. State whether you are using exact givens, rounded inputs or an approximation, and round only as required by the task.

A useful AI request is: “List the domain restrictions, unit conversions and modelling assumptions before any calculation. Mark each assumption as stated or introduced.” Do not accept a long list mechanically. Decide which assumptions affect the answer. The skill is to connect a restriction to a specific step: x ≠ 3 permits cancellation; constant cross-sectional area permits division by area; whole-number counts prevent arbitrary rounding.

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5. Estimate before calculating precisely

An estimate gives you an independent expectation. It may be an order of magnitude, a range, a sign or a comparison. Establishing it before reading the proposed solution reduces the risk of adjusting your intuition to fit whatever answer appears. Estimation is especially useful when an arithmetic slip changes a result by a factor of ten or when an average must lie between known extremes.

If 48 items each cost $19.80, the total should be close to 50 × $20 = $1,000 and slightly lower. The exact total is $950.40: 48 × (20 − 0.20) = 960 − 9.60. A proposed $95.04 fails the scale check immediately. A proposed $960 is plausible in scale but misses the discount from $20 per item, so an estimate alone cannot certify it. Estimation screens; exact calculation resolves.

Bounds can be stronger than one rounded guess. Since 19 < 19.80 < 20, multiplying by 48 gives 912 < total < 960. The exact total fits this interval. If the result concerns a probability, it must lie from zero to one inclusive. If a rectangle has positive side lengths, its area must be positive. If the requested value is the shorter side, it cannot exceed the longer side. These checks are inexpensive and often decisive.

Before calculating a percentage reduction, ask whether the answer must be above or below the starting value. Before solving a rate problem, ask whether combining two positive filling rates should make the filling time shorter than either tap alone. Before solving a mixture problem, ask whether the final concentration lies between the component concentrations, assuming a simple additive-volume model and no reaction. These expectations identify errors in the model as well as in the arithmetic.

Use an estimate honestly. Write “about $1,000” when it is a rough expectation. Do not present it as an exact upper bound unless your rounding choices justify that claim. Rounding both numerator and denominator in a fraction can move the result in competing directions. For a guaranteed lower or upper bound, reason about which change increases or decreases the expression.

The most useful question after an unexpected result is, “Which assumption or step explains why this differs from my expectation?” Sometimes the exact result is correct and the intuition was weak. Investigate the difference rather than forcing agreement. An estimate should make you curious enough to check, not so attached to your first impression that you reject valid mathematics.

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6. Worked problem: percentages with a changing base

Problem: A shop reduces a bag’s original price by 20%. It then applies a further 10% discount to the reduced price. The final price is $108. Find the original price and the overall percentage reduction. Assume no tax, delivery charge or other adjustment is included in the $108.

Start with the direction of the calculation. The original price must exceed $108. The second discount acts on the already reduced amount, so adding 20% and 10% will not describe the total reduction. Let P be the original price in dollars. After the first discount, the price is 0.8P. After the second, it is 0.9 × 0.8P = 0.72P. Therefore 0.72P = 108, giving P = 150.

Now find the overall reduction. The final price is 72% of the original, so the total reduction is 28%. Alternatively, the dollar saving is 150 − 108 = 42, and 42/150 × 100% = 28%. Both methods use the original price as the denominator. Dividing by the final price would answer a different question: the saving as a percentage of what was paid.

Verify forwards using dollars. Twenty per cent of $150 is $30, leaving $120. Ten per cent of $120 is $12, leaving $108. The two savings total $42. The second saving is smaller than ten per cent of the original price because its base is smaller. This forward check validates the interpretation and catches an inverse calculation performed on the wrong amount.

Deliberately wrong illustrative AI solution: “The discounts total 30%. Therefore the final price is 70% of the original, so the original price is 108/0.70, about $154.29.” The division is consistent with that model; the first sentence is the error. Repair the changing percentage base before repairing the final number. If you simply tell the assistant “the answer is wrong,” it may change the arithmetic while preserving the misconception.

A second deliberately wrong solution says, “Add 20% and then 10% back to $108: 108 × 1.2 × 1.1 = 142.56.” A percentage reduction is reversed by dividing by the retained fraction, not by adding the same percentage to a smaller base. For example, reducing $100 by 20% gives $80; increasing $80 by 20% gives $96. The operations do not undo each other.

For a changed-condition test, suppose the shop offers 20% off the original price and then a fixed $10 reduction, still ending at $108. The model becomes 0.8P − 10 = 108. Hence 0.8P = 118 and P = $147.50. The total saving is $39.50, approximately 26.78% of the original price. A learner who automatically multiplies by 0.9 has recognised the surface wording but missed the changed operation.

The useful AI assistance here is a targeted question: “Which amount is the second percentage calculated from?” The independent achievement is being able to identify the base, write the multiplier and reverse the process without seeing the model first.

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7. Worked problem: ratios after a fixed change

Problem: A box contains red and blue counters in the ratio 3:5. Eight red counters are added and no blue counters are added. The new ratio of red to blue is 1:1. How many counters were in the box originally?

Let each original ratio unit contain k counters. Then red = 3k and blue = 5k. Since adding eight red counters makes the counts equal, 3k + 8 = 5k. Subtracting 3k gives 8 = 2k, so k = 4. The original counts are 12 red and 20 blue, making 32 counters in total. After the addition there are 20 of each colour, which confirms the new ratio.

A bar representation gives the same argument without formal algebra. The blue collection initially exceeds the red collection by two equal ratio units. Eight added counters fill that gap. Each unit therefore represents four counters. There are eight original units altogether, giving 8 × 4 = 32. Notice that the number eight appears twice with different meanings: eight added counters and eight total ratio units. Label these meanings so a numerical coincidence does not hide the structure.

Deliberately wrong illustrative AI solution: “Add eight to the red part of the ratio, giving 11:5.” This treats a ratio number as an actual count. The ratio 3:5 could represent 3 and 5 counters, 12 and 20 counters, or many other pairs. A fixed addition cannot be applied directly to an unspecified ratio unit. The equation must connect the units to actual counters.

Change the final ratio to 4:5 while retaining the original 3:5 and adding eight red counters. The blue count is still 5k, while the new red count must be 4k. Therefore 3k + 8 = 4k, so k = 8. The original total is 8k = 64. This different answer is expected: the same addition closes one ratio unit rather than two.

If the final ratio were 2:5, the setup would require 3k + 8 = 2k, giving k = −8. That is not a valid count. Adding red counters while blue remains unchanged cannot lower the red-to-blue ratio. You can detect the impossibility before algebra. Retain that qualitative check even when a solver provides a formally derived negative value.

A good follow-up prompt asks the assistant to change one condition and hide the answer. You then identify which quantity remains fixed, define a common ratio unit and verify the final counts. Being able to explain why the impossible version is impossible is as important as solving the valid versions.

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8. Worked problem: solve an equation without losing its domain

Problem: Solve √(x + 6) = x over the real numbers. The square-root symbol means the principal, non-negative square root. First, x + 6 must be non-negative. More strongly, the right side x must be non-negative because it equals a non-negative square root. Thus any solution must satisfy x ≥ 0.

Squaring both sides gives x + 6 = x², or x² − x − 6 = 0. Factoring yields (x − 3)(x + 2) = 0, so the candidate values are 3 and −2. The word candidate matters: squaring can produce values that satisfy the squared equation without satisfying the original equation.

Check the original equation. At x = 3, √9 = 3, so the equation holds. At x = −2, the left side is √4 = 2 while the right side is −2, so it fails. The real solution set is therefore {3}. The initial restriction x ≥ 0 already excludes −2, and direct substitution provides a transparent second check.

Deliberately wrong illustrative AI solution: “After squaring and factorising, x = 3 or −2. Both roots are valid.” Its factorisation is correct. The failure is treating a necessary consequence as an equivalent statement without checking the lost sign information. If a = b, then a² = b². The converse does not always hold: 2² = (−2)², but 2 ≠ −2.

This distinction appears in many operations. Multiplying an equation by an expression that can be zero may introduce candidates. Dividing by an expression that might be zero can lose solutions. Taking a square root requires attention to signs. Applying a logarithm requires a suitable positive argument in real arithmetic. Ask which values or conditions make each transformation reversible.

Consider x(x − 4) = 0. Dividing both sides by x gives x − 4 = 0 and finds x = 4, but loses x = 0 because the division was invalid there. The zero-product property preserves both cases: x = 0 or x − 4 = 0. Substitution confirms both. The habit is not “never divide”; it is “record what the division assumes, and examine excluded cases separately.”

Now solve (x² − 9)/(x − 3) = 6. The original domain requires x ≠ 3. Cancelling gives x + 3 = 6 within that domain, whose only candidate is x = 3. Because the candidate is excluded, the original equation has no solution. An explicit statement of no solution, or an empty solution set, is correct. Replacing the original fraction by a continuous extension at the missing point would change the problem.

The OpenStax section on radical equations discusses checking for extraneous solutions after raising both sides to an even power. Use that reference to strengthen the general rule. The numerical examples here are original and should be worked through rather than treated as quotations from a textbook.

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9. Worked problem: geometry needs a reason for every length

Problem: A rectangle has perimeter 34 cm and diagonal 13 cm. Find its side lengths and area. Assume an ordinary Euclidean rectangle with positive side lengths. Let its sides be a cm and b cm. The perimeter gives 2a + 2b = 34, hence a + b = 17. The diagonal forms a right triangle, so a² + b² = 13² = 169.

We can find the product before finding either side. Expanding (a + b)² gives a² + 2ab + b². Therefore 17² = 169 + 2ab. Since 289 − 169 = 120, we obtain ab = 60. The area is 60 cm². The side lengths have sum 17 and product 60, so they are roots of t² − 17t + 60 = 0. Factoring gives (t − 5)(t − 12) = 0. The rectangle is 5 cm by 12 cm.

Check both original measurements. The perimeter is 2(5 + 12) = 34 cm. The diagonal length is √(5² + 12²) = √169 = 13 cm. The area is 5 × 12 = 60 cm². These are separate checks of different constraints. Verifying only the perimeter would not distinguish 5 by 12 from 6 by 11, whose perimeter is also 34 cm but whose diagonal is different.

An alternative method substitutes b = 17 − a into a² + b² = 169. Expanding gives 2a² − 34a + 120 = 0, or a² − 17a + 60 = 0 after division by two. The same pair results. Agreement is useful because the routes organise the information differently. Both still depend on the rectangle’s right angle and the correctly transcribed measurements.

Deliberately wrong illustrative AI solution: “The diagonal divides the perimeter in half, so each side is 8.5 cm.” A diagonal divides the rectangle into congruent triangles, but it does not imply that the adjacent sides are equal. The proposed square would have diagonal 8.5√2 cm, approximately 12.02 cm, not 13 cm. The false geometric inference appears before the arithmetic.

What if the perimeter were still 34 cm but the diagonal were 10 cm? From the same identity, ab = (289 − 100)/2 = 94.5. Yet sides with sum 17 have product at most 8.5² = 72.25, because (a − b)² ≥ 0. The measurements are inconsistent. Equivalently, a square gives the smallest diagonal among rectangles with that perimeter, and that diagonal is already about 12.02 cm.

Do not ask the assistant to invent a diagram that makes impossible measurements look plausible. Ask it to identify the inconsistent constraints. If a scanned question supplies a blurry 13 that could have been read as 10, return to the source image. A mathematical contradiction is sometimes a signal of transcription failure, not a difficult theorem waiting to be discovered.

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10. Worked problem: average speed and stage boundaries

Problem: A cyclist travels 12 km outwards at a constant 18 km/h and returns along the same 12 km route at a constant 12 km/h. There is no stop between the two stages. Find the average speed over the whole journey.

Average speed is total distance divided by total elapsed time. The outward time is 12/18 = 2/3 hour. The return time is 12/12 = 1 hour. The total distance is 24 km and the total time is 5/3 hours. The average speed is 24 ÷ (5/3) = 72/5 = 14.4 km/h.

The result lies between 12 and 18 km/h, as expected. It is lower than the arithmetic mean of 15 km/h because the cyclist spends longer at the slower speed. A second calculation in minutes confirms it: 40 minutes outwards plus 60 minutes back equals 100 minutes. Travelling 24 km in 100 minutes corresponds to 0.24 km per minute, or 14.4 km/h.

Deliberately wrong illustrative AI solution: “The average is (18 + 12)/2 = 15 km/h.” This would be correct for equal times at the two speeds, but the question specifies equal distances. The error concerns weighting. Before averaging rates, identify whether the pieces have equal times, equal distances or neither. An average is a relationship to a total, not an instruction to add every displayed number and divide by the count.

If the cyclist pauses for 20 minutes before returning, the average over the entire trip including the pause changes. The elapsed time becomes 120 minutes, or 2 hours, so the average is 12 km/h. The moving average remains 14.4 km/h. Both numbers can be meaningful, but the question must specify which time is included. Do not let an assistant silently choose the definition that produces a familiar answer.

A filling problem illustrates a different rate structure. Tap A fills an empty tank in 6 minutes and tap B fills it in 9 minutes. Assume constant rates and no leak. Together they fill 1/6 + 1/9 = 5/18 of the tank per minute, so they take 18/5 = 3.6 minutes. That is 3 minutes 36 seconds, not 3 minutes 6 seconds. The answer is shorter than either individual filling time, which passes the qualitative check.

If a leak empties a full tank in 18 minutes at a constant rate, the combined net rate while all three flows operate is 1/6 + 1/9 − 1/18 = 2/9 tank per minute. Starting empty, the tank fills in 9/2 = 4.5 minutes under this idealised model. A real leak may depend on water height; this example explicitly assumes a constant rate. Stating that assumption keeps the mathematical exercise separate from a claim about every physical tank.

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11. Worked problem: probability without replacement

Problem: A bag contains three red counters and two blue counters. Two counters are selected uniformly at random, one after the other, without replacement. What is the probability that the two counters have different colours?

Define the event before calculating. Different colours can occur as red then blue, or blue then red. These two ordered cases cannot happen simultaneously, so their probabilities can be added. The probability of red then blue is (3/5) × (2/4) = 3/10. The probability of blue then red is (2/5) × (3/4) = 3/10. The required probability is 3/10 + 3/10 = 3/5.

The second denominator is four because the first selected counter has been removed. The second numerator depends on the first colour. If the first counter was red, two blue counters remain. If it was blue, three red counters remain. A probability tree makes those conditions visible. Label each second branch with the state of the bag at that point rather than copying the original probabilities automatically.

Check by counting unordered pairs. Label the individual counters R1, R2, R3, B1 and B2. There are 5 × 4/2 = 10 equally likely unordered pairs. A mixed-colour pair can be chosen in 3 × 2 = 6 ways, giving 6/10 = 3/5. This second method relies on the sampling rule making all individual pairs equally likely. If the physical selection mechanism favours some counters, the model must change.

A third check uses the complement. Two reds have probability (3/5)(2/4) = 3/10. Two blues have probability (2/5)(1/4) = 1/10. The same-colour probability is 4/10, leaving 6/10 for different colours. The mutually exclusive colour outcomes account for the full probability of one. This is a useful completeness check, not just a third way to repeat the same arithmetic.

Deliberately wrong illustrative AI solution: “The probability is (3/5)(2/5) = 6/25.” It makes two errors: it treats the second draw as though the counter were replaced, and it counts only red then blue. Correcting only one error leaves the answer wrong. Name every dependency affected by the sampling process, then rebuild the event.

If the counter is replaced and the bag is mixed before the second draw, the answer becomes (3/5)(2/5) + (2/5)(3/5) = 12/25. This is 0.48, compared with 0.6 without replacement. Removing a red makes blue relatively more likely on the next draw, and removing a blue makes red relatively more likely. That explanation accounts for the direction of the difference.

A simulation could approximate either probability by repeating a correctly specified experiment many times. It would provide empirical numerical evidence about that simulation, subject to random variation and implementation mistakes. It would not replace the exact counting argument or prove that the real selection mechanism is uniform. Always distinguish a mathematical probability under stated assumptions from an observed relative frequency in a finite sample.

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12. Proof, examples and counterexamples do different jobs

A calculation answers a particular numerical question. A proof establishes a claim for every case covered by its assumptions. An example can illustrate a general statement, and a counterexample can refute a universal statement. Confusing these jobs is a common way to accept an attractive AI explanation that has not established what was asked.

Claim: the square of every odd integer leaves remainder one when divided by eight. Checking 1², 3², 5² and 7² supports the pattern, but it does not cover all odd integers. Write an arbitrary odd integer as 2k + 1, where k is an integer. Its square is 4k² + 4k + 1 = 4k(k + 1) + 1. Consecutive integers k and k + 1 include an even integer, so k(k + 1) is divisible by two. Therefore 4k(k + 1) is divisible by eight, proving the claim.

Each sentence has a role. The representation 2k + 1 covers every odd integer, including negative odd integers. The expansion is an algebraic identity. The consecutive-integer observation supplies the divisibility step. The final addition of one establishes the remainder. If an explanation merely says “it is obvious that the middle part is divisible by eight,” ask for the missing reason rather than accepting the tone of certainty.

Now consider the claim “n² + n + 41 is prime for every non-negative integer n.” Trying n = 0, 1 and 2 gives 41, 43 and 47, all prime. Nevertheless, n = 40 gives 1,600 + 40 + 41 = 1,681 = 41², which is composite. One admissible counterexample disproves the universal claim. Finding many supporting cases could never have ruled out that failure.

A counterexample must satisfy the assumptions. Using a negative input does not refute a claim restricted to positive inputs. Using a non-integer does not refute an integer claim. When asking AI to challenge a statement, include the exact quantifiers and domain: “Find a non-negative integer counterexample, or explain what remains unproved.” Do not accept an example from outside the agreed scope.

Another useful distinction is between a statement and its converse. “If an integer is divisible by four, then it is even” is true because 4k = 2(2k). “If an integer is even, then it is divisible by four” is false; six is a counterexample. An assistant may inadvertently reverse an implication when summarising a theorem. Write the direction explicitly and test the converse separately if the solution relies on it.

For a proof, inspect the first unsupported step, not just the conclusion. A correct theorem can be accompanied by an invalid proof. Conversely, a failed proof does not establish that the theorem is false. The appropriate response may be to repair the argument, find a valid counterexample, or say the question remains unresolved. This restraint is especially important when a generated proof uses a theorem you have not studied and cannot verify.

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13. Symbolic results and numerical evidence

An exact symbolic answer preserves a mathematical relationship, such as √2 or 3/5. A numerical approximation expresses a value to a stated precision, such as 1.414 or 0.600. Neither form is automatically superior for every task. The question determines whether an exact expression, a rounded measurement, a justified bound or a complete solution set is required.

Consider x² = 2 over the real numbers. The exact solutions are x = √2 and x = −√2. Reporting only 1.414 misses the negative solution and changes exact equality into an approximation. Substituting 1.414 gives 1.999396, not exactly two. That discrepancy is consistent with rounding and does not invalidate the exact solution. A good answer distinguishes the exact set from its decimal approximations.

A graph can help locate roots, but the visible window is limited. A coarse plot may miss a narrow feature or a repeated root that touches an axis without crossing it. For f(x) = (x − 1)², the values on either side of one are positive. A search looking only for sign changes can miss the root at x = 1. A graph is evidence to inspect alongside algebra and the stated domain.

A small residual means that substituting a candidate makes the equation’s two sides nearly equal. It does not always mean the candidate lies near an actual solution. For the equation 0.000001x = 0, the candidate x = 1 gives a residual of only 0.000001, yet its distance from the exact solution zero is one. The residual must be interpreted relative to the scale and behaviour of the function. Do not turn “the displayed error is small” into an unexplained accuracy guarantee.

For an original numerical-root example, let f(x) = x³ − x − 2. We have f(1) = −2 and f(2) = 4. Since a polynomial is continuous, there is at least one root between one and two. On that interval, f′(x) = 3x² − 1 is positive, so the function is strictly increasing and the root there is unique. These statements justify a bracketing approach rather than merely showing that a numerical solver printed a decimal.

Further calculation gives f(1.521) = −0.002256239 and f(1.522) = 0.003688648. The unique root in the interval therefore lies between 1.521 and 1.522. That bracket has width one thousandth, but it alone does not determine which side of 1.5215 the root occupies. Avoid claiming a three-decimal rounded answer until the bracket or additional calculation establishes the rounding decision.

The SymPy solving guidance explains how exact and inexact inputs affect symbolic solving. Its solveset documentation also makes the solution domain explicit. If you use a symbolic tool, preserve exact fractions when appropriate, specify the intended domain, and inspect the original equation yourself. A tool evaluates the mathematical input it receives; it cannot establish that you transcribed the intended problem correctly.

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14. Worked problem: optimise under the actual constraint

Problem: A rectangular garden is built against a straight wall. Sixty metres of fencing are available for the other three sides. Find the dimensions that maximise the area, assuming the wall is long enough, the ground is suitable and the fencing can be divided continuously into the required lengths.

Let w metres be each perpendicular side and l metres the side parallel to the wall. The fencing constraint is 2w + l = 60. Since the garden has positive dimensions, 0 < w < 30 and l = 60 − 2w. Its area is A = wl = w(60 − 2w) = 60w − 2w² square metres.

Complete the square: A = 450 − 2(w − 15)². A square is non-negative, so A ≤ 450 for every allowed w. Equality holds when w = 15, giving l = 30. The maximum area is therefore 450 m², achieved by a 15 m by 30 m rectangle. The equality case is inside the allowed interval, so it is a feasible garden.

This proof establishes a global maximum under the model. It does more than test a few dimensions. A numerical table might show areas for w = 10, 15 and 20, but it would leave infinitely many other widths unchecked. Completing the square directly explains why no allowed width can produce an area above 450 m².

A calculus check differentiates A to obtain A′ = 60 − 4w. Setting this to zero gives w = 15. The second derivative is −4, so the quadratic is concave down. Together with the domain, this confirms the same maximum. A stationary point alone would not be sufficient for every function or constrained problem; endpoints and other possible cases must also be considered where relevant.

Deliberately wrong illustrative AI solution: “A square maximises area for a fixed perimeter, so all sides should be 15 m.” The cited idea concerns a different constraint. Only three sides need fencing here. A 15 m square uses 45 m of fencing under this setup and has area 225 m², leaving available fencing unused. A true theorem can lead to a wrong answer when its assumptions do not match the problem.

Change the problem so that the wall segment available is at most 20 m long. Then l ≤ 20. To use all the fencing, w = (60 − l)/2, and A = 30l − l²/2 for 0 < l ≤ 20. This expression is increasing over that interval because its derivative is 30 − l, which is positive there. The largest area is achieved at l = 20 and w = 20, giving 400 m². The earlier 15 by 30 rectangle is no longer feasible.

The changed-condition test is the heart of the lesson. Before reusing a solution, inspect whether its optimum still belongs to the allowed set. If fencing comes only in indivisible panels, if a gate changes the available length, or if the wall cannot support the intended layout, those are new modelling conditions. Recompute within the changed model rather than repeating the old maximum as a universal recommendation.

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15. Inspect an AI proposal line by line

When an assistant supplies a solution, first check that it answered the same question. Match the unknown, domain, units and requested answer form. Then inspect the first line that transforms the problem. A correctly calculated answer to a substituted question can be harder to notice than a simple arithmetic error because the written solution may be internally consistent.

Mark each significant line as a definition, a given fact, an assumption, a transformation or a conclusion. Definitions introduce meaning. Given facts come from the question. Assumptions require disclosure. Transformations need mathematical justification. Conclusions must follow from the previous work. This small classification makes a hidden assumption stand out: “the events are independent” cannot be treated as a given if the experiment samples without replacement.

For algebra, name the operation and its restriction. Subtracting the same quantity from both sides preserves equality. Dividing by a non-zero constant preserves an equation’s solutions. Dividing an inequality by a negative constant reverses its direction. For example, −2x > 6 implies x < −3. Checking x = −4 confirms the inequality, while x = 0 refutes the incorrect proposal x > −3.

For geometry, connect the theorem to a marked or proved property. For probability, identify the sample space and whether cases overlap. For a word problem, explain what each term measures. For a proof, inspect quantifiers and any missing cases. A general request to “check carefully” is less useful than a specific challenge tied to a mathematical dependency.

Do not assess a solution by length, polished formatting or the assistant’s confidence. A short argument may be complete; a long argument may repeat an unsupported claim. Likewise, a changed answer after you object is not evidence that the new answer is correct. Require the correction to identify the faulty line and explain why the replacement is valid.

If two proposed solutions disagree, preserve both until you can locate the earliest difference in meaning or reasoning. Check the original data, then the representation, then the first non-equivalent transformation. A calculator can settle arithmetic; a definition or theorem may settle an inference. Asking the same model to vote between its answers does not provide the independent evidence needed to resolve the disagreement.

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16. Ask for help at the smallest useful level

Begin with an independent attempt, even if it is only a diagram, an equation or a statement of what confuses you. Then request the smallest help that would let you continue. This preserves a role for your own thinking and gives the assistant evidence about the actual difficulty. “I cannot do maths” is too broad to guide a repair; “I do not know why the second denominator changes” identifies a teachable step.

A useful hint ladder starts with orientation, then representation, then one next step, then a worked example. Orientation might identify that the problem compares equal distances rather than equal times. Representation might suggest a distance-time table. One next step might ask you to calculate the duration of the outward trip. A full solution is appropriate when the earlier hints still leave a prerequisite gap, but it should be followed by a fresh attempt with the explanation covered.

Try this request: “I have written the givens and this equation. Tell me whether the equation matches the wording. If it does, ask one question that helps me choose the next step. Do not reveal the final answer.” Another useful request is: “Explain why this transformation is valid, using the domain restrictions. Then give me one similar expression to simplify independently.” These prompts specify an instructional job rather than demanding a performance of confidence.

For error diagnosis, provide your actual working. Ask the assistant to locate the first incorrect or unsupported line and give a reason. Do not replace the whole solution immediately. If the first mistake is converting 3.6 minutes to 3 minutes 6 seconds, preserve the valid rate calculation and repair decimal-time conversion. Local repair helps you see what already works and prevents a small error from being mislabelled as failure of the entire topic.

Keep personal details out of uploaded worksheets or screenshots. Crop out names, student identifiers and unrelated information when they are unnecessary, and follow your school’s rules for approved tools. If you use an image, compare the assistant’s transcription against the source before trusting the answer. A missing minus sign, superscript or bracket can produce a fully correct solution to the wrong question.

Finally, lower the assistance level on the next attempt. If the first problem needed a full example, try the next with only a representation hint. If the next needed one hint, attempt another unaided. Progress is a change in what you can do, explain and check with less support. The number of pages generated by the assistant is not a reliable measure of that progress.

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17. Independent practice: attempt before opening the answers

Use a separate sheet for these eight exercises. For each one, write the domain or practical restrictions, your representation, the solution and at least one independent check. Where a question asks for an explanation, a final number alone is incomplete. The questions are original practice material and are not matched to any official examination’s mark allocation.

Exercise A: A jacket is reduced by 15%, then by a further 20% of the reduced price. It finally costs $136. Find the original price and the overall percentage reduction. Explain why adding the two discount percentages would be wrong.

Exercise B: A class box has green and yellow counters in the ratio 2:7. After ten green counters are added, the ratio becomes 4:7. How many counters were in the box at the start? Explain which quantity stayed fixed.

Exercise C: Solve √(2x + 3) = x over the real numbers. State the domain restriction that matters before squaring, find all candidates, and check them in the original equation.

Exercise D: A rectangle has perimeter 46 cm and diagonal 17 cm. Find both side lengths and its area. Verify the perimeter and diagonal separately.

Exercise E: A person travels 10 km at 5 km/h and then 10 km at 10 km/h. There are no stops. Find the average speed for the full journey. State one change to the wording that would make the arithmetic mean of the two speeds appropriate.

Exercise F: A bag contains four black counters and two white counters. Two counters are selected uniformly at random without replacement. Find the probability of one of each colour. Check using either a complement or a count of equally likely pairs.

Exercise G: An AI proposal says, “For every real x, √(x²) = x.” Decide whether the claim is true. If it is false, supply a counterexample and a corrected identity valid for every real x. Explain why restricting x to non-negative values changes the conclusion.

Exercise H: Eighty metres of fencing enclose three sides of a rectangle against a sufficiently long wall. Find the maximum possible area and the dimensions that achieve it. Then solve the changed problem when the side along the wall cannot exceed 30 m. State why a new feasibility check is necessary.

If you are stuck, record the exact blockage before using help: interpreting the wording, selecting a representation, performing an operation, checking the answer or explaining completeness. That record gives you a more useful next lesson than a simple total of right and wrong answers. Keep the answer section covered until you have made a genuine attempt at the chosen question.

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18. Full answers and verification notes

Answer A: Let P dollars be the original price. The retained fractions are 0.85 and 0.80, so 0.68P = 136 and P = $200. The final price is 68% of the original, giving an overall reduction of 32%. Check forwards: 15% of $200 is $30, leaving $170; 20% of $170 is $34, leaving $136. The total saving is $64. Adding 15% and 20% would incorrectly treat the second discount as a percentage of the original $200.

Answer B: Let each original ratio unit represent k counters. Initially there are 2k green and 7k yellow. The yellow count is unchanged, so the new ratio can be expressed as 4k green and 7k yellow. Thus 2k + 10 = 4k, giving k = 5. The original total is 9k = 45 counters: 10 green and 35 yellow. Afterwards there are 20 green and 35 yellow, and 20:35 simplifies to 4:7. The invariant is the actual yellow count, not the overall total.

Answer C: Since the principal square root is non-negative, x must be non-negative. Squaring gives 2x + 3 = x², hence x² − 2x − 3 = 0. Factorisation gives (x − 3)(x + 1) = 0, so the candidates are 3 and −1. Substitution at 3 gives √9 = 3, which works. Substitution at −1 gives √1 = 1, which does not equal −1. The complete real solution set is {3}. Merely checking that both candidates satisfy the squared equation would not be sufficient.

Answer D: Let the positive side lengths be a and b centimetres. Their sum is 23 and the sum of their squares is 289. Thus 23² = 289 + 2ab, so ab = (529 − 289)/2 = 120. The side lengths are roots of t² − 23t + 120 = 0, which factors as (t − 8)(t − 15) = 0. The rectangle is 8 cm by 15 cm and its area is 120 cm². Check: 2(8 + 15) = 46 cm and √(64 + 225) = 17 cm.

Answer E: The first 10 km takes 2 hours and the second takes 1 hour. Total distance is 20 km and total time is 3 hours, so the average speed is 20/3 km/h, approximately 6.67 km/h. It lies between the two speeds and is nearer the slower speed because more time is spent there. If the person instead travelled for the same duration at each speed, the arithmetic mean 7.5 km/h would apply. Equal distances do not supply that equal-time condition.

Answer F: Black then white has probability (4/6)(2/5) = 4/15. White then black has probability (2/6)(4/5) = 4/15. Their sum is 8/15. Counting gives the same answer: there are 6 × 5/2 = 15 unordered pairs and 4 × 2 = 8 mixed-colour pairs. Alternatively, the same-colour probabilities sum to (4/6)(3/5) + (2/6)(1/5) = 2/5 + 1/15 = 7/15, leaving 8/15. The check explicitly accounts for all colour outcomes.

Answer G: The claim is false. At x = −3, √(x²) = √9 = 3, while x = −3. The identity valid for every real x is √(x²) = |x|. The square root returns the non-negative magnitude whose square is x². When x ≥ 0, the absolute value equals x, so the original equation is valid on that restricted domain. A numerical counterexample disproves the unrestricted claim; the definition of absolute value explains the repaired identity.

Answer H: Let w be a perpendicular side and l the side parallel to the wall. The constraint is 2w + l = 80. The area is A = w(80 − 2w) = 800 − 2(w − 20)², so the unconstrained-wall maximum is 800 m² at w = 20 m and l = 40 m. For the changed problem, l ≤ 30, so that earlier solution is infeasible. Write A = l(80 − l)/2 = 40l − l²/2. Its derivative 40 − l is positive for 0 < l ≤ 30, so the maximum in that interval is at l = 30 m. Then w = 25 m and A = 750 m². Both cases use exactly 80 m of fencing.

Review the pattern of your errors. If A was wrong but B was right, the gap may concern changing percentage bases rather than general word problems. If C produced two answers, work on reversible transformations and substitution. If D had the right dimensions but wrong units for area, repair dimensional interpretation. If H reused the first dimensions after the wall restriction changed, focus on feasible sets. The purpose of the key is to identify a next learning action, not merely to award yourself a score.

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19. Delayed and changed-condition tests

Immediately repeating a solution can feel fluent because the previous steps remain available in memory. A delayed attempt asks a different question: can you retrieve the method after the worked solution is no longer fresh? Choose a later study session, cover the examples and attempt a problem from a mixed set. The exact delay should fit the learner and course; the important feature is that the answer is not being copied from a just-viewed template.

Delayed test one: A container starts with 4 litres of water. One inlet adds 1.5 litres per minute while a constant-rate outlet removes 0.5 litre per minute. Assuming sufficient capacity and both flows operating throughout, how much water is present after 8 minutes? Before calculating, state whether the initial amount must be included and whether the two rates should be added or subtracted.

Answer: The net rate is 1 litre per minute, so the increase is 8 litres. Including the original 4 litres gives 12 litres. Check using separate totals: 4 + 1.5 × 8 − 0.5 × 8 = 4 + 12 − 4 = 12. The number 8 alone would describe the increase, not the final amount. If the outlet opens only after the first four minutes, the changed answer is 4 + 1.5 × 8 − 0.5 × 4 = 14 litres.

Delayed test two: Solve (x² − 16)/(x − 4) = 9 over the real numbers. Then change the right side to eight and solve again. Answer: The domain excludes x = 4. Cancelling on the valid domain gives x + 4 = 9, hence x = 5. Direct substitution gives (25 − 16)/(5 − 4) = 9. In the changed version, the sole candidate is x = 4, which is excluded. Therefore the changed equation has no solution. The same cancellation step can lead to a valid answer in one case and no answer in another.

Delayed test three: A learner claims that because a polynomial takes positive values at x = 0 and x = 2, it cannot have a root between them. Refute the claim with a specific polynomial. One answer is f(x) = (x − 1)². It equals one at both endpoints but equals zero at x = 1. Equal endpoint signs do not rule out a root. The claim confuses a useful sufficient condition for a crossing with a necessary condition for every root.

Add an explanation test: choose one result and teach it aloud without looking at the page. State the meaning of each variable, why the first equation is valid, which transformation could fail and how you checked the final answer. If your explanation stalls, identify that exact point. The repair might be a definition, a diagram or a small calculation, rather than another complete AI-generated solution.

Keep a compact progress record with four entries: the problem, the assistance used, the first unresolved step and the result of the delayed attempt. Compare like with like. A harder question solved with one hint may show more development than an easy question solved without help. Use the record to choose the next practice task, not to manufacture a universal score for mathematical ability.

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20. When to stop, ask a person and close the loop

Stop the AI-assisted attempt when you cannot verify a crucial step and further generated explanations only repeat it. Write the exact line, your interpretation and the reason it remains doubtful. Bring that record to a teacher, tutor or knowledgeable person. This is a productive handoff because it identifies the mathematical obstruction rather than presenting an unfiltered conversation and asking someone else to find the issue.

Seek human help when the question is ambiguous, a diagram is unreadable, different authoritative sources use different conventions, or the proof depends on a theorem beyond your current understanding. Also seek help when repeated practice reveals the same prerequisite difficulty. If distributing a negative sign is still unreliable, a complicated optimisation problem may be the wrong place to repair it. A short targeted lesson can be more useful than another advanced worked solution.

Use stronger review when mathematics affects real consequences. A correct school-style calculation is not sufficient validation for engineering safety, medication decisions, contracts, significant financial commitments or other high-stakes applications. The model may omit constraints that a qualified professional would recognise. This guide teaches mathematical checking; it does not authorise using a generated result as professional advice or bypassing an established review process.

For ordinary practice, finish with a clear acceptance record. State the final answer with its units and conditions. Record the strongest independent check. Explain whether you found all solutions or only one candidate, whether the answer is exact or approximate, and what remains uncertain. Then complete a fresh or changed-condition problem without access to the previous explanation. These steps separate a correct response today from a method you can use tomorrow.

You have met the learning goal when you can formulate a problem, choose a representation, explain the method, reject a plausible error and verify the conclusion with decreasing assistance. You do not need to memorise every example in this article. Keep the habits that transfer: preserve meaning, expose assumptions, justify transformations, and return to the original question before accepting an answer.

Continue through the Super Intelligence Learning Hub for the wider practical curriculum. For broader comparison of genuinely different approaches, use Multiple Super Intelligence Approaches on One Problem. For subject-specific Additional Mathematics routing, the Additional Mathematics AI guide supplies a separate subject context. This article’s central task remains your own mathematical independence.

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