Mathematical Proof

Start by stating the claim and its domain

Consider “the product of consecutive integers is even”. In precise form, the claim is that for every $n\in\mathbb Z$, $n(n+1)$ is divisible by 2. The domain includes negative integers and zero, not only the positive examples that are easiest to picture.

A proof should state what may be assumed and what must be established. If the claim is an implication, begin with its hypothesis. If the claim is a biconditional, identify two directions. If it uses an inequality, preserve the restrictions that make each transformation legal.

There is no need to make a proof artificially symbolic. Clear sentences can carry the logical steps. Symbols are useful where they reduce ambiguity rather than obscure it.

Examples suggest a claim; an argument establishes it

The products $2\cdot3$, $3\cdot4$ and $4\cdot5$ are even, but those three checks do not establish the statement for every integer. A counterexample would settle a universal claim negatively, while a finite collection of successes normally does not settle it positively.

The distinction is sharp in the expression $n^2+n+41$. For $n=0,1,2$, it gives the prime numbers 41, 43 and 47. The same pattern does not continue forever: at $n=40$,

A long run of successes is not a substitute for a reason covering all cases. Computation remains useful for finding patterns, checking algebra and searching for counterexamples. Its role must match the claim being made.

A complete proof by cases

Every integer n is even or odd. Handle both cases.

If n is even, there is an integer k with $n=2k$. Then

The bracketed expression is an integer, so the product is even.

If n is odd, there is an integer k with $n=2k+1$. Then $n+1=2(k+1)$, and

Again the bracketed expression is an integer. The cases are exhaustive, so the result holds for every integer n. The conclusion does not depend on the example values used to motivate it.

Notice the work done by the definition of evenness. The proof does not merely say “there is a 2 somewhere”; it rewrites the entire expression as 2 times an integer.

Choose a method that matches the statement

A direct proof develops the conclusion from the hypotheses. A contrapositive proof establishes $\neg q\Rightarrow\neg p$ instead of $p\Rightarrow q$. A contradiction proof assumes the claim fails and derives an impossibility. Induction combines a starting case with a step that covers the next integer case.

These are not competing styles to choose at random. A claim about an odd square may be easier to prove by considering what happens when the input is even. A claim that an object cannot exist may be naturally suited to contradiction. A formula indexed by all positive integers may be suited to induction.

The dedicated Direct Proof, Proof by Contradiction and Mathematical Induction pages develop these methods with full examples.

Existence and uniqueness are separate tasks

To prove that there exists an integer satisfying $n^2-n=20$, exhibit $n=5$ and verify $25-5=20$. That is a complete existence proof. It does not prove uniqueness: $n=-4$ is another solution.

To prove a unique real solution to $2x+3=11$, existence is verified by $x=4$. For uniqueness, suppose u and v both solve the equation. Then $2u+3=2v+3$, so $2u=2v$ and $u=v$. Any two solutions must agree.

The second argument explains why there cannot be a different solution. Merely finding one answer does not always provide that information.

Equality needs the right kind of argument

For sets A and B, proving $A=B$ usually means proving $A\subseteq B$ and $B\subseteq A$. One inclusion can leave extra elements on the other side. For example, $\{1,2\}\subseteq\{1,2,3\}$ is true, but the sets are not equal.

For algebraic expressions, show a chain of justified equivalent rewrites on the stated domain. Cancelling a factor such as $x-1$ requires $x\ne1$. Equal simplified formulas do not restore excluded inputs.

For geometric objects, the hypotheses must describe the general figure, not a convenient special drawing. The existing Proof in Geometry and Coordinate Proofs pages apply these principles to geometric arguments.

Distinguish an arbitrary object from a chosen witness

In a universal proof, “let n be an arbitrary integer” means that no additional property has been imposed. Replacing n with 5 would abandon that generality. In an existence proof, choosing a particular valid value is exactly what is needed.

Different arbitrary objects may require different auxiliary variables. If m and n are odd, write $m=2r+1$ and $n=2s+1$. Using the same r for both would accidentally assume $m=n$.

A variable introduced under an assumption must also remain within its scope. If a proof chooses a witness depending on x, it cannot later claim that the same witness works for every x unless that stronger fact is established.

Audit the argument, not just the final line

A useful proof review asks four questions. Are the hypotheses clear? Is each inference justified? Are all cases covered? Does the final statement match the original claim?

Common failures include assuming the conclusion, proving the converse instead, dividing by a quantity that might be zero, and turning an observation into a universal rule. For instance, deriving $x=1$ from $x^2=x$ by dividing by x loses the valid solution $x=0$. The correct factorisation is $x(x-1)=0$.

A contradiction also needs to be genuinely impossible under the assumptions. Reaching an unexpected but valid value is not enough. State which assumption or established result the contradiction violates.

Related topics

Practise complete forward arguments in Direct Proof. Then compare Proof by Contradiction with Mathematical Induction. Truth Tables provides a finite setting in which exhaustive verification really can settle a logical equivalence.