Simple deductive proof

Definitions are the safest starting point in elementary proof because they translate words into algebra. An even integer has the form $2k$; an odd integer has the form $2k+1$; and an integer divisible by $d$ has the form $dk$, where the auxiliary variable $k$ is an integer.

The phrase “let $n=2k$ for some $k\in\mathbb Z$” does two jobs: it uses the definition of evenness and records that $k$ is an integer. Without $k\in\mathbb Z$, the expression $2k$ would not by itself prove that a number is even.

Known results may be used when they are established or elementary—for example, closure of the integers under addition and multiplication, the distributive law, and the fact that among consecutive integers one is even. Name the fact when it is central to the argument.

Notation can make a proof concise. The statement $d\mid n$ means “$d$ divides $n$,” so there exists an integer $k$ such that $n=dk$. It does not mean the fraction $\frac{d}{n}$.

A direct proof follows the direction of the implication. Start by assuming the hypothesis, translate it into a useful mathematical form, apply valid algebra or known results, and finish by matching the definition of the conclusion.

A reliable proof frame is: Let an arbitrary object satisfy the hypothesis; then represent it using the relevant definition; therefore transform the expression; since the remaining quantity belongs to the required set, the conclusion follows.

The word “arbitrary” matters. It signals that no special value was chosen, so the reasoning applies to every object satisfying the hypothesis. A direct proof should not assume the conclusion or use the result it is trying to establish.

Not every algebraic line needs a sentence, but every non-obvious logical transition should be justified. A short proof can still be complete.

An algebraic proof uses accepted algebraic laws to show that a claim is true for every permitted value. When proving an identity, begin with one side—usually the more complicated side—and transform it until it becomes the other. Do not write the unproved identity as your first equality and then manipulate both sides simultaneously. This is a reliable introductory method, not the only valid one: you may also use a clearly labelled chain of equivalent statements that reduces the target identity to an established truth, provided every transformation is reversible on the stated domain. Do not assume the conclusion is already true.

The symbol $=$ joins expressions that have the same value in the current context. The symbol $\equiv$ may be used for an identity that holds throughout its domain. An equation such as $x+2=7$ is true only for particular $x$; $(x+1)^2\equiv x^2+2x+1$ is true for all real $x$.

Every operation must preserve truth. Expanding, factoring, collecting like terms, or adding the same expression to both sides is valid. Dividing by a variable requires proof that the variable is non-zero; otherwise a possible case may be lost.

A proof of an identity should be a chain of exact equalities. Numerical approximations cannot establish an exact algebraic statement.

Parity describes whether an integer is even or odd. The definitions $2k$ and $2k+1$ convert parity statements into algebra. After simplifying, aim for $2K$ to prove evenness or $2K+1$ to prove oddness, with $K\in\mathbb Z$.

Use distinct integer parameters for independent numbers. If $a$ and $b$ are arbitrary odd integers, write $a=2k+1$ and $b=2m+1$. This includes unequal odd integers and does not impose an accidental relationship.

Parity arguments often require factoring. An expression such as $4k^2+4k+1$ becomes $2(2k^2+2k)+1$, revealing the required odd form.

A stronger result may come from using an extra known fact. For instance, every odd square is congruent to $1$ modulo $8$, because the product of two consecutive integers is even.

To prove $d\mid N$, finish by writing $N=dK$ for an integer $K$. If the hypothesis says $d\mid a$, begin with $a=dk$, where $k\in\mathbb Z$. This definition turns divisibility into ordinary algebra.

Linear combinations preserve a common divisor. If $d\mid a$ and $d\mid b$, then $d\mid(ra+sb)$ for any integers $r,s$. Substitute $a=dk$ and $b=dm$, factor out $d$, and use integer closure.

Factoring is often the key to divisibility proofs. Rather than expanding a complicated expression, look for a product containing a guaranteed multiple of the required divisor.

The converse of a true divisibility statement need not be true. From $6\mid n$ we may conclude $3\mid n$, but $3\mid n$ does not imply $6\mid n$. The integer $9$ is a counterexample.

Consecutive integers differ by 1, so represent them as $n,n+1,n+2,\ldots$. Consecutive even integers differ by 2 and may be written $2k,2k+2,2k+4$; consecutive odd integers may be written $2k+1,2k+3,2k+5$.

Among two consecutive integers, one is even. Therefore their product is even. Among three consecutive integers, one is divisible by 3 and at least one is even, so their product is divisible by 6.

Symmetric representations can simplify algebra. Three consecutive integers centred at $n$ may be written $n-1,n,n+1$. Their outer terms have a natural difference-of-squares structure.

When a problem asks for a statement that is “always true,” the proof must cover every integer starting value, including negative integers and zero. Algebraic representations do this automatically.

A valid proof preserves truth at every step and covers the entire stated domain. To audit an argument, ask: Were all variables defined? Was the hypothesis used correctly? Is every operation valid? Does the final line match the conclusion? Were any cases silently excluded?

Common invalid arguments include assuming the conclusion, using non-exhaustive examples as proof, automatically reversing an implication, dividing by a quantity that may be zero, and treating an approximate graph as exact proof. When solving $x^2=a$ with $a>0$, consider both $x=\pm\sqrt a$; the principal square root $\sqrt a$ itself is single-valued and nonnegative.

A counterexample is the quickest way to show that a universal claim is false. It must satisfy every hypothesis and violate the conclusion. A value outside the stated domain is not a valid counterexample.

An identity proof shows a statement is always true throughout its domain. An equation-solving calculation finds particular values. Confusing these goals often produces circular or incomplete reasoning.

IB proof marks reward communication as well as algebra: introduce variables, state membership such as $k\in\mathbb Z$, justify the decisive step, and finish with a conclusion.

Attempt each proof by first translating the key words into algebra. Then compare your work with the solution structure. More than one valid proof may exist, but every solution must cover the full domain and justify its decisive step.

Keep proofs concise: introduce variables, perform the necessary reasoning, state why the resulting quantity has the required form, and conclude. Avoid adding unrelated calculations that make the logical chain harder to see.