0.1 From patterns to mathematical arguments - Courses of Computing4All
Precalculus for High School Students 0.1 From patterns to mathematical arguments
Precalculus for High School Students
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Learning goals: Distinguish an example from a proof; state a precise conjecture; use counterexamples and algebraic reasoning.

Begin with a question

The sums \(1\), \(1+3\), \(1+3+5\), and \(1+3+5+7\) equal \(1,4,9,16\). What might the sum of the first ten positive odd numbers be? Predict before calculating. A mathematician notices structure, proposes a claim, tests its boundaries, and then asks why it must hold. Speed is useful, but a convincing explanation is the goal.

Examples, conjectures, and proof

A conjecture is a claim suggested by evidence. Checking five examples can support a conjecture but cannot establish a statement about every integer. A counterexample is one allowed case in which the claim fails. It disproves a universal claim immediately. Always specify the universe: integers, positive integers, real numbers, or some restricted interval. The claim that every number has a larger square is false for \(x=\tfrac12\), even though it works for integers greater than one.

One counterexample defeats an “every number” claimOne counterexample defeats an “every number” claim0½11.5012xyClaim: x² ≥ x for every real x.Test x = ½:x² = ¼, which is less than ½.Blue: y = x²Green: y = xFor 0 < x < 1, the square is smaller.At x = ½, the blue point lies below the green point.A single allowed counterexample is enough to disprove a universal claim.
The graph makes the counterexample visible: \(\tfrac14<\tfrac12\). Checking several inputs where the claim works cannot remove this failure. Always check which numbers the claim includes.

For the odd-number pattern, arrange square tiles in an \(n\)-by-\(n\) square. Enlarging an \((n-1)\)-by-\((n-1)\) square adds an L-shaped border containing \(2n-1\) tiles. This explains why the successive totals are squares. Algebra records the same structure: \[n^2-(n-1)^2=2n-1.\] Beginning with one tile and adding these borders gives \(1+3+\cdots+(2n-1)=n^2\). The picture and algebra explain all positive integer cases together.

Border 1: 1 tileBorder 2: 3 tilesBorder 3: 5 tilesBorder 4: 7 tilesBorder 5: 9 tiles1 + 3 + 5 + 7 + 9 = 25
Reading the picture. Each new color adds an L-shaped border to the previous square. The fifth border has nine tiles, and all five borders together make a \(5\times5\) square. In general, the \(n\)th border has \(2n-1\) tiles. The diagram explains why the sum of the first \(n\) positive odd numbers is \(n^2\); it is not just a numerical coincidence.

A proof using definitions

An even integer has the form \(2k\), and an odd integer has the form \(2k+1\), with \(k\) an integer. To prove that the sum of two odd integers is even, use different integer variables: \[(2a+1)+(2b+1)=2(a+b+1).\] Since \(a+b+1\) is an integer, the result is twice an integer and therefore even. Writing both odd integers as \(2k+1\) would address only equal odd numbers; independent variables make the argument general.

Two odd collections combine into complete pairsTwo odd collections combine into complete pairs5 = 2(2) + 1one extra7 = 2(3) + 1one extraThe two extrasmake one more pair.Together: 6 pairs = 12 objectsIn general: a pairs + b pairs + 1 extra pair = a + b + 1 pairs.(2a + 1) + (2b + 1) = 2(a + b + 1), so the sum is even.
The picture illustrates 5 + 7 using positive collections. The algebraic proof in the lesson covers all odd integers, including negative ones, because a and b may be any integers.

Conditional statements and their converses

“If a number is divisible by four, then it is even” is true: \(4k=2(2k)\). Its converse, “If a number is even, then it is divisible by four,” is false; six is a counterexample. A true implication does not automatically work backward. This distinction matters when solving equations: squaring both sides preserves a true equality, but the squared equation can have extra solutions.

A true “if–then” statement may fail in reverseA true “if–then” statement may fail in reverseEven integersMultiples of 4… −8, −4, 0, 4, 8 …4k = 2(2k)always even6Forward: trueMultiple of 4 → evenConverse: falseEven does not guaranteea multiple of 4.6 = 2 × 3, but 6/4 is not an integer.The inner set is entirely inside the set of even integers.The number 6 belongs to the larger set but lies outside the smaller set.
A containment diagram separates an implication from its converse. Every multiple of four is even, but the counterexample 6 shows that an even number need not be a multiple of four. The listed integers are examples; both sets continue indefinitely.

Communicating and improving an argument

Write the claim, identify assumptions, justify each step, and state the conclusion in the language of the question. In a group, ask “Which step explains why this is always true?” rather than only comparing answers. An error journal should contain the incorrect idea, a counterexample or correction, and a new problem that checks the repaired understanding. These habits make unfamiliar problems manageable.

Practice: attempt before reading the solutions

Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.

  1. Predict \(1+3+5+\cdots+19\) and justify it.
  2. Disprove: every real number satisfies \(x^2\ge x\).
  3. Prove that the sum of two even integers is even.
  4. Prove that the product of two odd integers is odd.
  5. Is the converse of “If \(x=3\), then \(x^2=9\)” true over the real numbers?
  6. A student tests \(n^2+n+41\) for five positive integers and declares it always prime. What is missing?
  7. Explain why \(n(n+1)\) is even for every integer \(n\).
  8. Find the flaw in dividing \(x^2=x\) by \(x\) and reporting only \(x=1\).
Worked solutions and reasoning

  1. First count the terms: the nth positive odd number is \(2n-1\). Setting \(2n-1=19\) gives \(2n=20\), so there are ten terms, not nineteen.

    The sum of the first n odd numbers is \(n^2\), giving \(10^2=100\). The square-tile diagram explains the rule: each successive odd number adds the next L-shaped border. As an arithmetic check, pair the first and last terms: \((1+19)+(3+17)+(5+15)+(7+13)+(9+11)=5(20)=100\).

  2. To disprove a claim about every real number, we need one real input for which the claimed inequality is false. Numbers strictly between zero and one are useful because multiplying such a number by itself makes it smaller.

    Choose \(x=\tfrac12\). Then \(x^2=(\tfrac12)^2=\tfrac14\), and \(\tfrac14\lt\tfrac12\). This allowed input contradicts \(x^2\ge x\), so the universal claim is false. We do not need to show that it fails at every input; for example, it does hold at x = 2.

  3. Start with the definition of even: an even integer is twice an integer. Write the two numbers as \(2a\) and \(2b\), where a and b are independent integers. Different variables allow the numbers to be different.

    Their sum is \(2a+2b=2(a+b)\). Integers are closed under addition, so \(a+b\) is an integer. The sum therefore has the form twice an integer and is even. This argument includes positive, negative, and zero values; checking a few numerical examples alone would not establish all cases.

  4. Write arbitrary odd integers as \(2a+1\) and \(2b+1\), with integer a and b. Expand every term of the product: \[(2a+1)(2b+1)=4ab+2a+2b+1=2(2ab+a+b)+1.\]

    The expression \(2ab+a+b\) is an integer, so the result is one more than twice an integer. That is exactly the definition of odd. The proof applies to all integer choices, including negative ones. For a numerical check, \(3\cdot5=15=2(7)+1\), but the algebra is what makes the argument general.

  5. The original statement goes from \(x=3\) to \(x^2=9\). Its converse reverses the direction: “If \(x^2=9\), then \(x=3\).” To test it, solve the hypothesis completely.

    Factoring gives \(x^2-9=(x-3)(x+3)=0\), so \(x=3\) or \(x=-3\). The value −3 satisfies the hypothesis but not the conclusion. Thus the converse is false over the real numbers. Squaring removes the sign information; it cannot distinguish 3 from −3.

  6. Five successful tests provide evidence, not a proof about infinitely many positive integers. A proof must explain why every allowed input works, or a single counterexample can show that the claim is false.

    Here a counterexample is available: at \(n=41\), \[n^2+n+41=41^2+41+41=41(41+1+1)=41\cdot43=1763.\] Both factors exceed one, so the value is composite. The missing step is therefore not simply checking more small cases: the proposed “always prime” statement is actually false.

  7. Consecutive integers have opposite parity: if one is even, the other is odd. We can make that observation into a proof by considering the two possibilities for n.

    If \(n=2k\), then \(n(n+1)=2k(2k+1)=2[k(2k+1)]\). If \(n=2k+1\), then \(n+1=2(k+1)\), so \(n(n+1)=2[(2k+1)(k+1)]\). In either case the bracket is an integer, making the product even. These cases cover every integer, including negative integers and zero.

  8. Division by x is permitted only after assuming \(x\ne0\). That assumption removes a possible solution before we have checked it. In fact, x = 0 satisfies the original equation because \(0^2=0\).

    Keep all possibilities by moving terms to one side and factoring: \[x^2-x=0\quad\Longrightarrow\quad x(x-1)=0.\] The zero-product property gives x = 0 or x = 1. Substitution confirms both. Dividing by an expression containing the unknown can lose solutions when that expression might be zero.

Reflection

Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.

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