Direct proof and contrapositive arguments

Two odd groups leave an even total

Five objects and seven objects make twelve. Each odd group contains pairs and one leftover object; the two leftovers form another pair. The calculation 5 + 7 = 12 checks one arrangement. What makes the same conclusion hold for any two odd integers, including negative ones?

An integer is even if it equals 2k for some integer k, and odd if it equals 2k + 1 for some integer k. These are algebraic definitions, so −3 = 2(−2) + 1 is odd and 0 = 2 × 0 is even. A drawing of physical groups illustrates positive counts; it does not by itself cover negative integers.

Write two arbitrary odd integers as m = 2a + 1 and n = 2b + 1, where a and b are integers. Then m + n = 2a + 2b + 2 = 2(a + b + 1). The bracket a + b + 1 is an integer because integers are closed under addition. The sum therefore has the required form 2k and is even.

The proof does not fix a or b to a particular value, nor require them to be equal. Every pair of odd integers has the displayed representations, and every such representation leads to an even sum. That is what replaces a list of successful calculations.

Use familiar algebra; keep the logical direction visible

Symbols, definitions and mathematical language helps you read a claim precisely. Variables, expressions and substitution supplies the algebra, while Expanding, factoring and algebraic identities explains the rearrangements used below. The reciprocal and magnitude examples also use Inequalities, intervals and absolute value.

Statements, truth values and logical connectives, and Implication, equivalence and negation, provide the logical background for the arrow ⇒ and for negating a claim. Here ⇒ means “if … then …”. We explain the route change again where it is needed; these logical topics are preparation, not an excuse to assume the desired conclusion.

On a first reading, follow the odd-sum proof, the direct-proof structure and the operation guards. Then learn the contrapositive using the magnitude example. Return to the quantified notation and rational-number example once you can say which inputs are allowed and which fact each step uses.

Name the inputs, what is given and what must follow

Our odd-sum theorem concerns m and n in ℤ, the integers. Its hypothesis is “m and n are odd”; its conclusion is “m + n is even”. The complete claim is that every integer pair satisfying the hypothesis satisfies the conclusion.

In symbols, ∀m, n ∈ ℤ, ((m is odd ∧ n is odd) ⇒ (m + n is even)). The symbol ∀ means “for every”, and ∧ means “and”. The variables range independently over the same stated integer domain. Their allowed values include zero and negative integers, although only odd inputs satisfy this particular hypothesis.

To prove a statement ∀x ∈ D, (P(x) ⇒ Q(x)) directly, take an arbitrary x in D and assume P(x). Use definitions and established facts to derive Q(x). Here P and Q must each have a defined classical truth value for every allowed input. An arbitrary input means one with no extra restriction beyond the domain and the stated hypothesis.

If the domain has no input satisfying P, the universal implication is true because it has no permitted counterexample. A proof of an implication does not assert that its hypothesis occurs. Our odd-sum claim does have examples, but existence and the universal implication remain separate questions.

A definition supplies a usable representation

The words “some integer” in a definition allow a witness that depends on the input. Oddness of m gives a witness a; oddness of n gives a possibly different witness b. Writing both numbers as 2a + 1 would quietly assume m = n and omit most permitted pairs.

The final witness is k = a + b + 1. Finding an expression that looks like 2k is not enough if k might fail to be an integer. State why this witness belongs to the required domain. Addition and subtraction of integers, and multiplication of integers, stay within ℤ; division does not have that unrestricted closure property.

For example, 3 = 2 × (3/2) does not prove that 3 is even. Its proposed witness 3/2 is not an integer. By contrast, the witness −4 proves that −8 = 2(−4) is even. Definitions control the permitted kind of witness, not just the visual shape of a formula.

Build a chain from the hypothesis, with a reason at each link

A direct proof can be short while remaining complete. First choose the arbitrary input and record the hypothesis. Next replace relevant words by their definitions. Then calculate or apply an established result, keeping its hypotheses visible. Finish by matching the result to the exact required conclusion.

Equality allows substitution and algebraic expansion. If A = B and B = C, then A = C. For implication chains, if P gives R and R gives Q on the retained domain, P gives Q. A step needs a reason: writing a sequence of plausible-looking formulas is not a substitute for showing why each follows.

Working backwards from the desired form can help find a proof. For the odd sum, the target 2k suggests factoring out 2. This is scratchwork: it identifies a useful destination. The final argument must still start from the actual odd-input assumptions and show that its destination is reached.

Assuming the conclusion and deriving a true identity proves nothing about whether the conclusion follows from the hypothesis. Likewise, citing the theorem being proved as a step is circular. State the earlier definition or established result that actually licenses the step.

An algebraic operation needs its own conditions

Let x be any real number with x > 1. Then x is positive and nonzero, so 1/x exists and is positive. Divide 1 < x by the positive x to get 1/x < 1. Thus x > 1 implies 0 < 1/x < 1. Both the existence and the direction of the inequality have been justified.

A convenient total domain for these original reciprocal predicates is D = ℝ ∖ {0}. The hypothesis x > 1 already forces membership in this nonzero domain, but a finite illustration may also display excluded zero so you can see why it is not a permitted row. It is never assigned a reciprocal or a truth value for the full original comparison.

Dividing an inequality by a positive number preserves its direction; dividing by a negative number reverses it. For x = −2, division of 1 < 2 by x gives −1/2 > −1, not −1/2 < −1. Division by zero is unavailable. A sign conclusion must therefore be established before division, not guessed afterwards.

Cancellation also retains its original domain. From (x² − 1)/(x − 1) = x + 1 you may conclude an identity only when x ≠ 1. Factoring the numerator does not give the original quotient a value at 1. An input excluded by the stated domain cannot refute a theorem about permitted inputs.

Start with the negated conclusion to prove the same implication

For defined classical statements P and Q, the contrapositive of P ⇒ Q is ¬Q ⇒ ¬P, where ¬ means “not”. It reverses the direction and negates both entire parts. A direct proof of the contrapositive is a proof of the original implication because their truth values agree for every possible P, Q pair.

The complete table below establishes that agreement. Both implications are false only when P is true and Q is false. The proof route changes, but the claim and its domain do not. In a domain-wide theorem, choose an arbitrary input from the same D, assume the entire conclusion Q fails and derive that the entire hypothesis P fails.

The converse Q ⇒ P reverses without negating and requires its own proof. The inverse ¬P ⇒ ¬Q negates without reversing and also requires its own proof. Neither is the contrapositive. Write the transformed statement in full before doing any algebra.

For a real condition x > 2, its negation is x ≤ 2, not x < 2. Equality belongs to the negation. If the hypothesis is a conjunction A ∧ B, its negation is ¬A ∨ ¬B, allowing either failure. Keep whole parts, boundaries and domains when changing route.

A large square forces a large magnitude, not a positive sign

For every real x, prove: if x² > 4, then |x| > 2. The absolute value |x| is the nonnegative magnitude of x, and |x|² = x². Directly extracting a sign from x² would be a mistake: both 3 and −3 have square 9.

The contrapositive is: if |x| ≤ 2, then x² ≤ 4. Let x be arbitrary and assume |x| ≤ 2. Since |x| ≥ 0, multiplication by |x| preserves the inequality, giving |x|² ≤ 2|x|. Multiplication of |x| ≤ 2 by the positive 2 gives 2|x| ≤ 4. Combining these gives x² = |x|² ≤ 4.

That is a direct proof of ¬Q ⇒ ¬P with P: x² > 4 and Q: |x| > 2. Its equivalence to P ⇒ Q proves the desired theorem for all real x. The equality boundaries are retained: at x = 2 or −2, the original hypothesis is false and the contrapositive conclusion x² ≤ 4 includes equality.

Compare the different claim “x² > 4 implies x > 2”. It fails at x = −3: its hypothesis holds but its conclusion does not. The true direction x > 2 implies x² > 4 has a separate direct proof: both x − 2 and x + 2 are positive, so x² − 4 = (x − 2)(x + 2) > 0. The adjacent number-line comparison shows exactly the extra negative region that prevents reversing this one-sided claim.

A finite check teaches a case; the argument covers the theorem

The exact table may show every input in its declared finite sample. That establishes the claim on that sample, but not on an infinite domain such as ℤ or ℝ. A universal proof must explain an arbitrary permitted input. The algebraic odd-sum proof and the magnitude argument provide that explanation independently of the displayed samples.

A permitted input with the hypothesis true and conclusion false refutes a universal implication. An input with a false hypothesis is not a counterexample merely because its conclusion is false. For x > 2 ⇒ x² > 4, x = 0 has both parts false and satisfies the implication; x = −3 refutes its converse instead.

In the reciprocal example, zero is outside D = ℝ ∖ {0} and has no real reciprocal. It cannot be counted as a failing permitted input. The correct repair for an overbroad reciprocal statement is to state the intended defined domain, not to assign a made-up value or regard an undefined comparison as false.

End at the stated conclusion and name what you used

A reader should be able to identify the arbitrary input, the retained domain, every assumption and the final conclusion without reconstructing them. Use “therefore” only after providing the definition, calculation or earlier result that makes the step valid. A proof can omit routine arithmetic once its structure is clear, but must retain a condition that changes the conclusion.

Before finishing, compare the last sentence with the theorem. Did you prove the requested arrow or its converse? Does a witness have the required kind? Are expressions defined? Was a division sign checked? Did a numerical example replace a general argument? A correct calculation can still answer the wrong logical question.

For an “if and only if” theorem, one arrow alone is insufficient; prove both directions on the same domain. This lesson focuses on direct and contrapositive arguments. Other proof techniques need their own structures and are developed separately.

Use exact cases to inspect the proof’s conditions

In Odd sums, select two integers from the displayed odd list. Predict the separate definition witnesses and the even-sum witness before revealing them. The complete 25-pair table lets you compare negative, equal and unequal inputs; the written arbitrary-integer proof explains why the conclusion extends beyond it.

In Threshold directions, select −3 and predict the original, converse and contrapositive values separately. Compare ±2 to decide where equality belongs. The number-line bands represent the all-real sets x > 2 and x² > 4, clipped to the labelled viewing window with continuation arrows; the sampled table remains a finite illustration.

In Reciprocal guards, compare −1, 1/2, 1 and 2, then inspect excluded zero. Predict which inputs meet the hypothesis and which have a defined reciprocal. All fractions are exact, and no row at zero receives an invented implication value. Explain why the positive divisor, rather than a pattern in the table, justifies the general inequality proof.

Worked examples

Try it yourself

1. Prove that the cube of every even integer is even. State the final integer witness.

   Show the answer and reasoning

   If n = 2k with k ∈ ℤ, then n³ = 8k³ = 2(4k³), so the integer witness is 4k³.

   Take an arbitrary even integer, rather than fixing a numerical value. The definition gives k ∈ ℤ; closure under multiplication makes 4k³ an integer. The final expression has exactly the form required by evenness, including for negative k and zero.

2. Prove that the product of two odd integers is odd. Explain why two separate definition witnesses are used.

   Show the answer and reasoning

   For m = 2a + 1 and n = 2b + 1 with a, b ∈ ℤ, mn = 2(2ab + a + b) + 1, which is odd.

   Expansion gives 4ab + 2a + 2b + 1. The bracket 2ab + a + b is an integer. Distinct witnesses a and b retain arbitrary, independently chosen odd inputs; using one common witness would restrict the proof to equal inputs.

3. Prove x > 3 ⇒ x² > 9 over the reals. Write its contrapositive exactly and decide whether its converse is true.

   Show the answer and reasoning

   The positive factors x − 3 and x + 3 give x² − 9 > 0. The contrapositive is x² ≤ 9 ⇒ x ≤ 3. The converse fails at x = −4.

   For arbitrary x > 3, both factor signs are justified, so the original direct argument is valid. Negating strict inequalities includes equality. The allowed input −4 has square 16 > 9 but does not satisfy x > 3, which refutes only the converse.

4. Prove x² > 9 ⇒ |x| > 3 for every real x by proving its contrapositive directly.

   Show the answer and reasoning

   Assume |x| ≤ 3. Since |x| ≥ 0, x² = |x|² ≤ 3|x| ≤ 9. Thus the contrapositive, and therefore the original implication, holds.

   The negated conclusion is |x| ≤ 3 and the negated hypothesis is x² ≤ 9. Multiplication first by the nonnegative |x| and then by the positive 3 preserves the weak inequalities. Every real input and the equality boundaries remain covered; no sign of x is assumed.

5. A proposed proof says: “Assume x² > 0. Taking a square root gives x > 0.” Is this a direct proof over ℝ? State the error and a correct conclusion.

   Show the answer and reasoning

   It is invalid: x = −1 is a counterexample to the claimed implication. The nonnegative square root gives √(x²) = |x|, so the correct conclusion is |x| > 0, equivalently x ≠ 0.

   The proposed step replaces a magnitude by a signed number. Both negative and positive nonzero inputs have positive squares. A direct proof must retain the meaning of each operation; it cannot repair a false conclusion by choosing only positive examples.

6. Prove that the difference of two rational numbers r = p/q and s = u/v is rational. Are zero or negative denominators allowed?

   Show the answer and reasoning

   r − s = (pv − uq)/(qv) is rational when p, q, u, v are integers and q, v ≠ 0. Negative denominators are allowed; zero denominators are not.

   The numerator is an integer by multiplication and subtraction closure. The denominator is a nonzero integer because both factors are nonzero. The definition of rationality permits either sign of a denominator, and the common-denominator calculation preserves the original nonzero guards.

7. For integers n, prove that if 3n + 2 is odd, then n is odd, by contraposition. State the integer-parity fact used in negating both parts.

   Show the answer and reasoning

   Assume n is even, so n = 2k for k ∈ ℤ. Then 3n + 2 = 6k + 2 = 2(3k + 1) is even. This proves the contrapositive and hence the original implication.

   Every integer has exactly one of the forms 2k or 2k + 1, from division by 2 with remainder 0 or 1. Thus “not odd” means even within the integer domain; it is not a statement about arbitrary real numbers. The witness 3k + 1 is an integer, and 3n + 2 is an integer for every allowed n.

What to carry forward

• A direct proof takes an arbitrary permitted input, assumes the stated hypothesis and derives the exact conclusion.
• Definitions supply witnesses of a specified kind. Keep independent witnesses independent and justify their required integer or nonzero properties.
• Each step uses a definition, an established result or a valid operation with its own conditions; assuming the conclusion is circular.
• Contraposition proves the same implication through ¬Q ⇒ ¬P on the same defined domain. Negate entire parts and include equality boundaries.
• Magnitude and sign are different: x² > 4 forces |x| > 2 but does not force x > 2. A converse is a separate claim.
• Finite exact models illustrate their declared inputs. Universal claims need an arbitrary-input argument, and excluded or undefined inputs receive no invented truth value.