Writing and checking a proof

Three correct calculations still leave a question

Consider the claim: the square of every even integer is even. A first draft says, “2² = 4, 4² = 16 and 6² = 36, so the claim is true.” All three calculations are correct. But the passage has not explained what happens to another even integer, such as −8, or why no even integer can fail.

A complete passage says: “Let n be an arbitrary even integer. By definition, n = 2k for some integer k. Then n² = (2k)² = 4k² = 2(2k²). Since 2k² is an integer, n² is even.” The letters stand for any permitted input and its definition witness; they are not abbreviations for the three tested numbers.

Notice what the second passage adds: an exact starting assumption, a definition, an algebraic link, a reason that the final witness is an integer, and the requested conclusion. These features let a reader check the claim for every even integer, including zero and negative ones.

Use familiar algebra and read the proof in stages

Symbols, definitions and mathematical language helps you read the claim precisely. Variables, expressions and substitution explains what a letter may represent. Expanding, factoring and algebraic identities supplies the equality steps. Rational and radical equations explains why a denominator or other restricted operation needs a guard.

The logical ideas needed here are explained at their first use. Separate lessons on statements, implication, quantifiers, direct proof, contradiction and induction develop them further. Those later topics are useful return paths; you do not need their full machinery to check the opening proof.

On a first reading, follow the two passages, the witness diagram, the cancellation example and the paragraph version of a proof. Then check the answered exercises. Return to the two quantifier orders and the complete finite matrix when you can identify the assumptions and conclusion in your own words.

Say exactly what must hold, and for which inputs

A claim must have a definite meaning before it can be proved. In the opening theorem, the domain is ℤ, the integers; the hypothesis is that n is even; the conclusion is that n² is even. “For every” includes every integer satisfying that hypothesis, not only positive inputs or the values you happened to test.

An integer is even when it equals 2k for some integer k. Both parts matter: the form 2k and the integer requirement on k. The equality 3 = 2(3/2) does not prove that 3 is even, because 3/2 is not an integer. Zero is even because 0 = 2 × 0, and −8 is even because −8 = 2(−4).

Read an implication “if P, then Q” as a claim that every permitted input satisfying P also satisfies Q. P names the hypothesis, Q the conclusion, and the symbol ⇒ records this direction. A counterexample must lie in the stated domain, satisfy P, and fail Q. An input outside the domain cannot refute that claim.

State conditions that make expressions defined. For example, a claim containing the real quotient 1/x needs x ≠ 0 wherever that quotient is evaluated. Do not write a comparison at an undefined input and silently count it as false. Similarly, “positive” means greater than zero; it is stronger than “nonnegative”, which permits zero.

Choose an arbitrary input before its dependent witness

“Let n be an arbitrary even integer” introduces n with exactly the theorem’s restrictions. Oddness, positivity or a convenient numerical value must not be added unless the claim already requires it. The definition then supplies an integer k with n = 2k. This k may depend on n.

The symbol ∀ means “for every”; ∃ means “there exists”. The order matters. For every real x there exists a real y greater than x is true: first take an arbitrary x, then choose y = x + 1. The choice is allowed to depend on x, and y − x = 1 > 0 proves the required inequality.

There exists a real y greater than every real x is a different, false claim. Here y must be fixed before x is considered. For any proposed y, the permitted input x = y makes y > x false. A witness depending on the later x cannot satisfy this order.

The adjacent matrix uses the complete finite domain D = {0, 1} and the predicate P(x, y): x = y. Every row contains a true entry, so ∀x ∈ D, ∃y ∈ D, P(x, y) is true. No column contains only true entries, so ∃y ∈ D, ∀x ∈ D, P(x, y) is false. This proves those finite claims, while the preceding arbitrary-real argument proves the separate inequality claim.

Every link needs a reason the reader can identify

In the opening proof, evenness supplies n = 2k with k ∈ ℤ. Substitution and the laws of multiplication give n² = 2(2k²). Closure of the integers under multiplication gives 2k² ∈ ℤ. The definition of evenness then yields the conclusion. The diagram separates the equality from the witness’s type: both are needed.

A proof may use a definition, an assumption explicitly allowed by its method, or an already established fact. When applying a fact, check its hypotheses. For example, integer sums and products are integers; unrestricted quotients of integers need not be integers. The reader should not have to guess which fact licenses a decisive step.

Working backwards from the target can help discover a proof. A target of the form 2r suggests factoring out 2. That is useful scratchwork. In the finished direct argument, start from what is given and show that the target follows; do not assume that the target representation already exists.

Writing “n² = 2r for an integer r, so n² is even” does not prove the opening theorem unless the existence of that integer r has first been derived. Otherwise the passage has assumed exactly what it must show. Writing “obviously” or “therefore” cannot supply the missing reason.

An equality is not an arrow between claims

An equality A = B says that two expressions have the same value. A chain A = B = C asserts each displayed equality. By contrast, P ⇒ Q says that one claim gives another; P ⇔ Q means “P if and only if Q” and requires both P ⇒ Q and Q ⇒ P on the same domain. Use words when they make the relationship easier to read.

For real x, the equivalence x² = 9 ⇔ (x − 3)(x + 3) = 0 ⇔ (x = −3 or x = 3) is valid. Subtracting 9 and factoring are reversible, and the real zero-product property supplies the two roots. Each root also gives square 9, so both directions have been checked.

Replacing the final condition by x = 3 loses a root. The claimed implication x² = 9 ⇒ x = 3 fails at x = −3. If the domain is explicitly restricted to x ≥ 0, the negative root is unavailable and x² = 9 ⇔ x = 3 becomes valid on that restricted domain. Adding this restriction halfway through a proof changes the original claim.

Squaring an equality gives a valid forward implication: a = b ⇒ a² = b² for real a and b. Its converse need not hold, as a = 1 and b = −1 show. When an operation loses information, record a one-way arrow or supply an additional condition that makes the step reversible.

Cancellation belongs only where its factor is nonzero

For real a, b and c with a ≠ 0, ab = ac implies b = c because division by the nonzero a is defined. Conversely, b = c implies ab = ac by multiplication. Thus ab = ac ⇔ b = c is valid under the stated nonzero guard. When a = 0, the products can be equal with b ≠ c, so the guard is essential.

To solve x(x − 1) = 0 over the reals without losing an input, split into x = 0 and x ≠ 0. In the first case, zero already satisfies the equation. In the second case, divide by x to get x − 1 = 0, hence x = 1. Both candidates satisfy the original equation. Cancelling x without this split would discard the solution zero.

A case proof must cover every permitted input. Its branches may use their own assumptions, but those assumptions remain local. The conclusion “x = 1” from the nonzero branch cannot be asserted for the zero branch. After checking each branch, state the combined conclusion for the original domain.

Before dividing an inequality, establish the divisor’s sign as well as its nonzero value. A positive divisor preserves the direction; a negative divisor reverses it. The guard belongs before the operation, because it determines whether that operation and its claimed direction are valid.

A finite check proves a claim only when it covers that domain

Let D = {0, 1, 2, 3}. The claim “for every n ∈ D, n² ≤ 9” can be proved by checking all four inputs exactly: their squares are 0, 1, 4 and 9. The proof is complete because D has no other elements and every displayed comparison is correct.

The same four checks do not prove “for every integer n, n² ≤ 9”. The integer 4 is a counterexample because 4² = 16 > 9. A universal claim over an infinite domain needs an argument that reaches every permitted input, rather than only a longer finite list of successful examples.

A diagram or computation can suggest a claim, reveal a missing condition or provide a counterexample. For proof, identify its exact scope: a complete finite enumeration, an illustration of an arbitrary-input argument, or an approximation with a stated limitation. A visual pattern alone does not justify changing the quantified domain.

A concise paragraph can retain every logical step

The opening proof can be displayed as separate assumption, definition, calculation and conclusion lines while you learn its structure. It can also be written as the short paragraph shown at the start. These presentations establish the same result because they retain the same arbitrary input, typed witness, equality and final definition.

Use complete sentences around equations, name each new variable and keep names consistent. Words such as “by definition”, “because” and “conversely” tell the reader what a line is doing. Punctuation still belongs at the end of a mathematical sentence. Introduce only notation that helps the argument.

Choose detail for the reader’s preparation. A beginner may need the substitution and integer-closure reason written out. A later proof may combine routine arithmetic into one line, but must retain a domain condition or logical direction that affects validity. Concision removes repetition, not necessary hypotheses.

For an equivalence, separate the two directions or give a chain whose steps are all justified equivalences. For several cases, mark their assumptions and show that they exhaust the domain. End by explaining why the established fact is exactly the original claim, rather than leaving the reader at an unrelated true identity.

Check meaning first, then polish the prose

Read the theorem and the last sentence together. Do they have the same domain, hypotheses and conclusion? Then trace the argument forward. At each step, identify the fact used, verify that it applies, and check any new variable’s type and dependence. The first unsupported step is a useful place to focus a repair.

A failed proof does not necessarily mean the theorem is false: the missing integer-closure sentence in the even-square proof can be supplied. A true-looking calculation also does not mean the theorem is true: the negative root refutes the single-root claim. Decide whether to repair the argument, change the claim explicitly, or reject the claim with a permitted counterexample.

After the logic is sound, revise the writing. Remove repeated calculations that add no reason, replace ambiguous pronouns with the relevant object, and state the conclusion clearly. Do not hide a missing step behind longer prose. A useful final reading asks whether another reader can recover why the conclusion follows without guessing an unstated assumption.

Use small cases to inspect a proof’s reach

In Even-square witnesses, compare −2, 0 and 4. Predict k and the final witness 2k². Then inspect the odd input 1: it does not meet the hypothesis and is not a counterexample to the even-input theorem. These exact instances illustrate the written general proof.

In Square-root direction, compare −3 and 3 before revealing the one-way claim. Which input has a true hypothesis and false conclusion? The two-preimage diagram and the general factorisation explain the corrected equivalence; three sampled inputs alone would not prove it over the reals.

In Witness order, choose one row after the input is fixed, then try fixing one column for both rows. The displayed matrix contains every pair from {0, 1} × {0, 1}, so it establishes the two finite quantified truth values exactly. Explain in words why a separate witness for each input is different from one witness for all inputs.

Worked examples

Try it yourself

1. A draft says, “n is even, so n = 2k. Therefore n² = 2(2k²), so it is even.” Supply the missing declarations and reason.

   Show the answer and reasoning

   Let n be an arbitrary even integer; choose k ∈ ℤ with n = 2k; state that 2k² ∈ ℤ by integer multiplication.

   The evenness definition requires an integer witness. Once k is declared integer, the equality n² = 2(2k²) and the integer type of 2k² together prove evenness for every permitted n.

2. Does the equality 3 = 2(3/2) prove that the integer 3 is even? Explain using the exact definition.

   Show the answer and reasoning

   No: the proposed witness 3/2 is not an integer.

   Evenness requires 3 = 2k for some k ∈ ℤ. Solving this equality gives only k = 3/2, outside the required witness domain. An expression shaped like twice a number is insufficient.

3. To prove the sum of two arbitrary odd integers m and n is even, may a writer assume m = 2a + 1 and n = 2a + 1 with the same a? Give a complete repair.

   Show the answer and reasoning

   Use independent integers a and b: m = 2a + 1, n = 2b + 1, so m + n = 2(a + b + 1).

   Using one witness forces m = n and omits unequal odd pairs. The definitions supply possibly different integer witnesses a and b. Their sum plus one is an integer, so the repaired formula proves evenness for every odd pair.

4. Check the real-number claim x² = 9 ⇒ x = 3. Give a counterexample and a correct equivalence on the same domain.

   Show the answer and reasoning

   The input x = −3 refutes the implication. Over ℝ, x² = 9 ⇔ (x = −3 or x = 3).

   At −3 the hypothesis is true and the claimed conclusion false. Factoring x² − 9 = (x − 3)(x + 3) and applying the real zero-product property gives both roots; substituting either root proves the reverse direction.

5. A solution of x(x − 1) = 0 divides by x and gives only x = 1. Which case was lost, and how should the argument be written?

   Show the answer and reasoning

   Retain x = 0 as one solution; in the separate x ≠ 0 case, division gives x = 1. The full solution set is {0, 1}.

   Division by x is unavailable when x = 0. Splitting into zero and nonzero cases covers every real input. Substitution verifies both candidates in the original equation, so no solution is lost or introduced.

6. A writer uses y = x + 1 to justify “there exists one real y greater than every real x”. Why is the witness invalid for that order?

   Show the answer and reasoning

   The chosen y depends on x; a witness introduced before x must be fixed independently of it. Any fixed y fails at x = y.

   The construction proves ∀x ∈ ℝ, ∃y ∈ ℝ, y > x instead. For the reversed order, x = y is a permitted real input and makes y > x false, so no single fixed witness can work.

7. The exact checks 0², 1², 2² and 3² are all at most 9. Which universal claim do they prove, and which all-integer claim fails?

   Show the answer and reasoning

   They prove n² ≤ 9 for every n ∈ {0, 1, 2, 3}. They do not prove it for every n ∈ ℤ; n = 4 refutes that claim.

   The four checks exhaust the stated finite set but not the integers. The permitted integer 4 has square 16 > 9, giving an exact counterexample to the larger-domain claim.

8. A draft starts the even-square proof with “Let n² = 2r for some integer r.” What has it assumed, and what should supply r instead?

   Show the answer and reasoning

   It has assumed the conclusion that n² is even. Start with n = 2k for an integer k and derive r = 2k² ∈ ℤ.

   The integer witness for the square must be obtained from the given evenness of n, not declared as though its existence were already known. Substitution yields n² = 2r and integer closure gives the required type.

9. Write a concise proof that if an integer m is divisible by 5, then m² is divisible by 5. Use “divisible by 5” to mean m = 5k for some k ∈ ℤ.

   Show the answer and reasoning

   Let m be an arbitrary integer divisible by 5. Write m = 5k with k ∈ ℤ. Then m² = 25k² = 5(5k²), and 5k² ∈ ℤ; hence m² is divisible by 5.

   The paragraph declares the arbitrary input, opens the exact definition, justifies the equality and integer witness, and returns to the required conclusion. No finite list of multiples or extra positivity assumption is needed.

What to carry forward

• A proof starts with an exact claim, domain and permitted assumptions, and finishes at that same claim.
• Introduce every variable and witness with its type; a later witness may depend on earlier arbitrary inputs only when the quantifier order permits it.
• Supply reasons for decisive steps and retain nonzero, sign and case conditions before using an operation.
• Equality compares values, implication has a direction, and equivalence requires both directions on the retained domain.
• Complete finite enumeration proves a finite-domain claim; successful samples alone do not prove an unrestricted universal claim.
• Readable prose exposes the logical structure. Revise for clarity after checking that the reasoning is valid.