Checking Is Valuable—but What Has It Established?
Consider the claim:
The sum of two odd integers is even.
We can check examples:
\[ 3+5=8, \qquad 7+11=18, \qquad -9+13=4. \]These examples support the claim, but infinitely many pairs of odd integers remain. A proof explains all of them at once.
Every odd integer can be written as \(2k+1\). If two odd integers are \(2a+1\) and \(2b+1\), then
\[ (2a+1)+(2b+1) =2a+2b+2 =2(a+b+1). \]Because \(a+b+1\) is an integer, the sum has the form \(2n\) and is therefore even.
The examples showed that the claim worked in selected cases. The argument from the definition of odd integers showed why it works in every case.
\[ \boxed{\text{examples test; proof establishes}} \]Proof, Verification, and Evidence
| Method | What It Does | What It Can Establish |
|---|---|---|
| Example | Tests one case | That the claim holds in that case |
| Several examples | Tests a pattern repeatedly | Evidence, but not a universal proof over an infinite domain |
| Counterexample | Finds a case where the hypothesis holds and conclusion fails | Disproves a universal claim |
| Direct verification | Substitutes or checks a proposed object against defining conditions | Whether that object satisfies those conditions |
| Exhaustive verification | Checks every case in a finite domain | A proof for that finite domain |
| Proof | Connects assumptions to a conclusion through justified reasoning | That the claim holds throughout its stated domain |
The strength of a method depends on the claim. One substitution is enough to verify that \(x=4\) solves a particular equation. It is not enough to prove an identity for every real \(x\).
What Makes a Proof Informal?
An informal proof uses ordinary mathematical language and familiar reasoning instead of a fully formal symbolic system. Informal does not mean careless, approximate, or optional.
A sound informal proof still:
- states the claim and domain clearly,
- identifies assumptions and definitions,
- justifies every material step,
- covers all required cases,
- avoids using the conclusion as an assumption, and
- ends with the claimed result.
For example:
The sum of two multiples of \(5\) is a multiple of \(5\). If the numbers are \(5a\) and \(5b\), their sum is \(5a+5b=5(a+b)\). Since \(a+b\) is an integer, the sum is a multiple of \(5\).
The argument is concise and conversational, but its logical structure is complete.
Anatomy of a Sound Argument
| Part | Question | Example |
|---|---|---|
| Claim | What must be shown? | The sum is even |
| Domain | What objects are under discussion? | Integers |
| Hypothesis | What is given or assumed? | Both integers are odd |
| Definition | How can the hypothesis be represented? | Odd integers have form \(2k+1\) |
| Reasoning chain | How does the hypothesis lead to the conclusion? | Factor the sum as \(2\) times an integer |
| Conclusion | Why has the claim now been established? | The sum has the definition of an even integer |
When an argument feels incomplete, identify which part of this structure is missing.
A Reliable Direct-Proof Method
- State the claim precisely. Include its domain and conditions.
- Assume the hypothesis. For \(p\rightarrow q\), begin with an arbitrary object satisfying \(p\).
- Unpack definitions. Rewrite terms such as even, divisible, rational, subset, or factor into usable forms.
- Apply established facts. Use definitions, laws, axioms, and previously proved results.
- Maintain equivalence or valid implication. Explain why each step follows.
- Reach the required form. Connect the final expression or statement to the definition of the conclusion.
- Close the argument. State that the desired conclusion follows for the arbitrary object.
The word arbitrary matters. If the proof uses no special feature beyond the stated hypothesis, the reasoning applies to every object in the domain.
Starting from Definitions
Definitions often provide the most direct route from a hypothesis to a conclusion.
| Concept | Useful Algebraic Form |
|---|---|
| \(n\) is even | \(n=2k\) for some integer \(k\) |
| \(n\) is odd | \(n=2k+1\) for some integer \(k\) |
| \(a\) divides \(b\) | \(b=ak\) for some integer \(k\) |
| \(r\) is rational | \(r=\frac{a}{b}\), where \(a,b\in\mathbb{Z}\) and \(b\ne0\) |
| \(A\subseteq B\) | Every element of \(A\) is an element of \(B\) |
| \(x\) is a root of \(f\) | \(f(x)=0\) |
Definitions do two jobs. They explain what a term means and provide a form that can be manipulated or checked.
Direct Verification by Substitution
To verify a proposed solution, substitute it into the original relationship and check every condition.
Verify that \(x=4\) solves
\[ 3x-5=7. \]Substitute \(4\) for \(x\):
\[ 3(4)-5=12-5=7. \]The left side equals the right side, so \(x=4\) is a solution.
Verification does not require reproducing the method that found the candidate. It asks a narrower question: does the candidate satisfy the original requirements?
For an inequality, check the truth of the resulting statement. For a function inverse, composition may be the required test. For a geometric construction, lengths, angles, incidence, or congruence conditions may need to be checked.
Verifying and Proving Identities
An identity claims equality for every value in a stated domain. Checking a few values gives evidence but does not prove the identity.
Consider
\[ (a+b)^2=a^2+2ab+b^2. \]Testing \(a=2\) and \(b=3\) verifies one case:
\[ (2+3)^2=25 \]and
\[ 2^2+2(2)(3)+3^2=4+12+9=25. \]To prove the identity, use the definitions of exponentiation and the distributive law:
\[ \begin{aligned} (a+b)^2 &=(a+b)(a+b)\\ &=a(a+b)+b(a+b)\\ &=a^2+ab+ab+b^2\\ &=a^2+2ab+b^2. \end{aligned} \]The symbolic derivation applies to every permitted \(a\) and \(b\), so it proves the identity.
Exhaustive Verification of Finite Cases
Checking cases becomes a proof when the domain is finite and every case has been covered correctly.
Claim: For every integer \(n\) with \(1\le n\le4\),
\[ n^2+n \]is even.
| \(n\) | \(n^2+n\) | Even? |
|---|---|---|
| \(1\) | \(2\) | Yes |
| \(2\) | \(6\) | Yes |
| \(3\) | \(12\) | Yes |
| \(4\) | \(20\) | Yes |
Because the stated domain contains exactly these four cases, the table proves the claim over that domain.
If the claim concerned every integer, the same four checks would not be exhaustive. A general proof would be needed:
\[ n^2+n=n(n+1), \]and one of two consecutive integers is always even, so their product is even.
Informal Geometric and Diagrammatic Proof
A diagram can explain why an algebraic relationship holds, provided the reasoning uses general structure rather than an accidental feature of one drawing.
Imagine a rectangle with side lengths \(a+b\) and \(c\). Split it into two rectangles with widths \(a\) and \(b\), both with height \(c\).
Computing the whole area gives
\[ (a+b)c. \]Computing the two pieces and adding gives
\[ ac+bc. \]Both calculations measure the same region, so
\[ (a+b)c=ac+bc. \]The picture supports an informal proof of the distributive law for positive lengths. Algebra extends the law beyond this geometric interpretation to broader number systems where the operations are defined.
A diagram is not automatically a proof. Labels, construction rules, and stated properties—not visual appearance alone—must justify the conclusion.
Direct Proof of a Conditional
To prove
\[ p\rightarrow q \]directly, assume an arbitrary object satisfies \(p\) and show that it must satisfy \(q\).
Claim:
If an integer is divisible by \(6\), then it is divisible by \(3\).
Assume \(n\) is an integer divisible by \(6\). By definition,
\[ n=6k \]for some integer \(k\). Since \(6k=3(2k)\),
\[ n=3(2k). \]Because \(2k\) is an integer, \(n\) is divisible by \(3\). Therefore, every integer divisible by \(6\) is divisible by \(3\).
The proof did not test a special integer. It used the general form of every integer divisible by \(6\).
Context, Units, and Assumptions
Mathematical verification must include the conditions that gave the symbols meaning.
Suppose a cost model is
\[ C=45+70h, \qquad h\ge0, \]where \(h\) is labor time in hours and \(C\) is cost in dollars.
The pair \(h=2\), \(C=185\) satisfies the formula:
\[ 45+70(2)=185. \]It also satisfies \(h\ge0\), and the units are consistent. The pair is therefore verified within the stated model.
However, this does not prove that the company actually charged \(\$185\). That practical conclusion also depends on assumptions such as:
- the fee and rate were applied as stated,
- no tax, discount, or overtime rate was added, and
- the recorded labor time was accurate.
Verification confirms the relationship under its assumptions; it does not independently verify the assumptions themselves.
Worked Example: Sum of Consecutive Integers
Prove informally that the sum of two consecutive integers is odd.
Represent an Arbitrary Pair
Let the first integer be \(n\). The next consecutive integer is \(n+1\).
Add Them
\[ n+(n+1)=2n+1. \]Use the Definition
Since \(n\) is an integer, \(2n+1\) has the form of an odd integer.
Conclude
Therefore, the sum of any two consecutive integers is odd.
The word “any” is justified because \(n\) represented an arbitrary integer.
Worked Example: Verifying a Proposed Solution
Verify whether \(x=3\) solves
\[ \frac{x+1}{x-1}=2. \]Check the Restriction
The denominator requires
\[ x\ne1. \]The candidate \(x=3\) is permitted.
Substitute into the Original Equation
\[ \frac{3+1}{3-1} =\frac{4}{2} =2. \]Compare Both Sides
The left side equals the right side, so \(x=3\) is a valid solution.
If a candidate had made the denominator zero, it would fail before any equality check.
Worked Example: A Divisibility Claim
Prove: If integers \(a\) and \(b\) are both divisible by \(4\), then \(a+b\) is divisible by \(4\).
Use the Hypotheses
Because \(a\) is divisible by \(4\),
\[ a=4m \]for some integer \(m\). Because \(b\) is divisible by \(4\),
\[ b=4n \]for some integer \(n\).
Combine the Quantities
\[ a+b=4m+4n=4(m+n). \]Use Closure
Since integers are closed under addition, \(m+n\) is an integer.
Conclude from the Definition
The sum \(a+b\) is \(4\) times an integer, so it is divisible by \(4\).
Proof and Verification Audit
| Audit Question | Reason |
|---|---|
| Is the claim stated precisely? | Prevents proving a different statement |
| Is the domain explicit? | Definitions and operations may depend on the number system |
| Are hypotheses used but not enlarged? | Prevents hidden assumptions |
| Does each material step have a reason? | Closes logical gaps |
| Were restrictions preserved? | Prevents invalid operations or candidates |
| Does the argument cover every required case? | Distinguishes proof from sampling |
| Was the conclusion assumed anywhere? | Detects circular reasoning |
| For verification, was the original relation checked? | A transformed relation may admit invalid candidates |
| Are units and assumptions consistent? | Preserves contextual validity |
| Does the final sentence match the claim? | Confirms completion |
Common Reasoning Errors
Treating Examples as a Universal Proof
No finite sample proves an unrestricted claim over an infinite domain.
Assuming the Conclusion
A direct proof of \(p\rightarrow q\) begins with \(p\), not with \(q\).
Circular Reasoning
Restating the claim in different words does not establish it.
Using an Unjustified Step
Familiar-looking algebra still needs valid laws, definitions, or previously established facts.
Dividing by a Quantity That Might Be Zero
Confirm that a divisor is nonzero before dividing.
Checking a Transformed Equation but Not the Original
Squaring, multiplying by variable expressions, and other steps can introduce invalid candidates.
Ignoring a Missing Case
An argument covering positive inputs may not cover zero or negative inputs.
Relying on a Diagram's Appearance
A figure that looks perpendicular, equal, or to scale does not prove those properties.
Confusing Algebraic Verification with Factual Confirmation
A value can satisfy a model even when the model's measurements or assumptions are inaccurate.
Practice
- Verify whether \(x=5\) solves \(4x-7=13\).
- Verify whether \(x=-2\) solves \(\frac{x+4}{x}= -1\), including the domain restriction.
- Prove directly that the sum of two even integers is even.
- Prove directly that the product of an even integer and any integer is even.
- Prove that the difference of two multiples of \(7\) is a multiple of \(7\).
- Explain why checking \(n=1,2,3,4\) does not prove \(n^2+n\) is even for every integer.
- Explain why checking every day of a seven-day schedule can prove that at least one employee is assigned each day, provided the schedule is fixed and complete.
- Use algebra to prove \(x(x+3)=x^2+3x\).
- Find the flaw: “Assume \(a=b\). Divide both sides of \(a^2=ab\) by \(a-b\).”
- A model gives \(V=100-4t\) for \(0\le t\le25\). Verify the pair \((t,V)=(15,40)\), including units if \(t\) is minutes and \(V\) is liters.
Selected Answers
- \(4(5)-7=13\), so \(x=5\) is a solution.
- The restriction is \(x\ne0\), so \(-2\) is permitted. Then \(\frac{-2+4}{-2}=-1\), so it is a solution.
- Let the integers be \(2a\) and \(2b\). Their sum is \(2a+2b=2(a+b)\), which is even.
- Let the even integer be \(2a\) and the other integer be \(b\). Their product is \(2ab=2(ab)\), which is even.
- Write the numbers as \(7a\) and \(7b\). Their difference is \(7a-7b=7(a-b)\), a multiple of \(7\).
- Those checks cover only four integers; the stated domain is infinite. A general argument is required.
- The domain has exactly seven identified cases. Correctly checking all seven is exhaustive.
- By the distributive law, \(x(x+3)=x\cdot x+x\cdot3=x^2+3x\).
- If \(a=b\), then \(a-b=0\). Dividing by \(a-b\) divides by zero and is invalid.
- \(100-4(15)=40\), \(15\) lies in the domain, and the verified output is \(40\) liters after \(15\) minutes.
Summary
- Examples provide evidence and verify individual cases; they do not generally prove universal claims over infinite domains.
- A counterexample disproves a universal claim by satisfying its hypothesis and failing its conclusion.
- An informal proof uses ordinary mathematical language but still requires complete, justified reasoning.
- A direct proof assumes an arbitrary object satisfies the hypothesis and derives the conclusion.
- Definitions often convert verbal properties into useful algebraic forms.
- Direct verification substitutes a candidate into the original relationship and checks every condition.
- Checking every case proves a claim when the domain is finite and the enumeration is complete.
- An identity requires a general derivation, not merely matching numerical examples.
- Restrictions, units, cases, and assumptions remain part of proof and verification.
- A final audit helps detect circular reasoning, hidden assumptions, missing cases, and invalid operations.