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.

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.

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.

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.

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. There are ten odd numbers, so the sum is \(10^2=100\). Ten successive L-shaped borders form a ten-by-ten square.
2. Take \(x=\tfrac12\). Then \(x^2=\tfrac14\lt\tfrac12\). One allowed counterexample disproves the universal claim.
3. Write the integers as \(2a\) and \(2b\). Their sum is \(2(a+b)\), twice an integer, so it is even.
4. \((2a+1)(2b+1)=4ab+2a+2b+1=2(2ab+a+b)+1\). This is one more than an even integer.
5. No. The converse allows \(x=-3\), because \((-3)^2=9\). Squaring hides a sign distinction.
6. A proof for all positive integers is missing. At \(n=41\), the value is \(41(41+1+1)=41\cdot43\), which is composite.
7. Consecutive integers have opposite parity. One of \(n\) and \(n+1\) is even, so their product has a factor of two.
8. Dividing by \(x\) assumes \(x\ne0\) and loses a solution. Factor \(x(x-1)=0\) to obtain \(x=0\) or \(x=1\).

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.