1.2 Mathematical Reasoning, Logic, and Proof
Key Takeaways
- Inductive reasoning proposes a pattern from examples; deductive reasoning proves a conclusion from definitions, accepted facts, and valid inference.
- A universal claim can be disproved by one valid counterexample, but repeated confirming examples do not prove it.
- The contrapositive of a conditional is logically equivalent to the original statement; the converse and inverse are not automatically equivalent.
- Truth tables, proof by cases, direct proof, contradiction, and contrapositive proof are practical tools for multiple-choice reasoning items.
Reasoning as tested mathematics
The AEPA Mathematics profile explicitly includes mathematical communication, connections, reasoning, and logic. That matters because teacher-certification mathematics expects more than reaching a numeric answer. A candidate must recognize why a conclusion follows, whether a student explanation is valid, and which representation or counterexample clarifies a claim.
Inductive reasoning moves from observed cases to a conjecture. It is useful for noticing patterns: 1 + 3 = 4, 1 + 3 + 5 = 9, and 1 + 3 + 5 + 7 = 16 suggest that the sum of the first n odd positive integers is n squared. The pattern is compelling, but the examples alone do not prove the statement for every positive integer.
Deductive reasoning starts from definitions, axioms, theorems, or previously established facts and reaches a conclusion by valid steps. A proof that the sum of two odd integers is even is deductive: write the odd integers as 2a + 1 and 2b + 1, add them, and get 2(a + b + 1), which is divisible by 2.
Conditional logic essentials
Most formal logic items can be translated into if-then form. For a conditional p -> q, p is the hypothesis and q is the conclusion. The conditional is false only when p is true and q is false. The related statements matter because distractors often swap them.
| Statement | Form | Relationship to p -> q |
|---|---|---|
| Original conditional | If p, then q | Baseline claim |
| Converse | If q, then p | Not automatically equivalent |
| Inverse | If not p, then not q | Not automatically equivalent |
| Contrapositive | If not q, then not p | Logically equivalent |
| Biconditional | p if and only if q | Requires both directions |
For example, let p be "an integer is divisible by 6" and q be "the integer is divisible by 3." The original conditional is true. The contrapositive, "If an integer is not divisible by 3, then it is not divisible by 6," is also true. The converse, "If an integer is divisible by 3, then it is divisible by 6," is false because 9 is divisible by 3 but not by 6.
Proof and disproof tools
Use a direct proof when definitions lead straight to the conclusion. Use a contrapositive proof when the negated conclusion creates an easier route. A classic example is proving "If n squared is even, then n is even" by proving the contrapositive: if n is odd, then n squared is odd. If n = 2k + 1, then n squared = 4k squared + 4k + 1 = 2(2k squared + 2k) + 1, which is odd.
Use proof by contradiction when assuming the opposite creates an impossibility. Use proof by cases when the domain naturally splits, such as even and odd integers or positive, zero, and negative values. Use a counterexample when a universal claim says "all," "every," or "for any" and you can find one permitted case where the conclusion fails.
Worked example: judging a student claim
A student says, "The square of every integer is greater than the original integer because 2 squared is greater than 2, 3 squared is greater than 3, and 4 squared is greater than 4." The examples are true but the universal claim is false. The integer 0 gives 0 squared = 0, not greater than 0, and the integer 1 gives 1 squared = 1. The reasoning is inductive, and the counterexamples disprove the claim.
Common traps
Do not treat a diagram or several numerical cases as a proof unless the argument covers all possible cases. Do not assume a biconditional from a one-way conditional. Do not forget vacuous truth in formal logic: if the hypothesis p is false, p -> q is true in a truth table. On the exam, that fact usually appears when a row has p false and the distractor incorrectly marks the conditional false.
For constructed reasoning in your scratchwork, write claims in a compact structure: given, need to show, key definition, conclusion. This keeps proof items from turning into word puzzles. When answer choices contain explanations, prefer the one that identifies the mathematical reason, not merely the one that repeats a correct answer.
Reading explanation choices
Many reasoning items do not ask, "Which answer is true?" They ask which explanation best justifies a result. In those items, two choices may contain true statements, but only one addresses the claim being proved. For example, if the claim concerns all integers, an explanation about several positive examples is too weak. If the claim concerns divisibility, an explanation using decimal approximation may be irrelevant even if the final number is correct.
A compact scratchwork routine helps. First, mark the quantifier: all, some, none, exactly one, at least one, or if and only if. Second, identify the object set: integers, real numbers, triangles, functions, or sample outcomes. Third, decide whether the answer must prove, disprove, classify, or translate. Fourth, test the logical direction. This routine catches the most common proof traps on a multiple-choice exam: proving a converse, giving examples instead of a proof, using a hidden assumption, or treating a special case as if it were general.
When a truth table appears, build intermediate columns. For p <-> (q or not p), compute not p first, then q or not p, then the biconditional. Rushing straight to the final column is where most row errors happen.
For the conditional statement p -> q, under which truth values is the statement false?
A claim says, "Every integer greater than 1 that is odd is prime." What is the most efficient way to disprove it?
Which strategy best proves the statement "If n squared is even, then n is even"?