Direct Proof

A direct proof begins with the stated hypotheses and develops the conclusion through valid deductions. State the definitions and let the variables represent arbitrary objects satisfying the hypotheses.

The same structure works for integers, sets and inequalities. Start with an arbitrary permitted case, translate the hypotheses into useful mathematical form, and finish by showing that the result has the form required by the conclusion.

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Translate parity into integer form

An integer is even when it can be written as $2k$ for an integer k. It is odd when it can be written as $2k+1$ . These representations handle positive, negative and zero cases where appropriate.

To prove a statement about two odd integers m and n, introduce separate integer witnesses:

m=2r+1,\qquad n=2s+1.

The witnesses r and s may differ. Giving them the same name would restrict the problem to equal odd integers, which is not what a general two-input claim allows.

Prove a sum is even

Claim: the sum of two odd integers is even.

Let m and n be arbitrary odd integers. Write $m=2r+1$ and $n=2s+1$ for integers r and s. Then

m+n=(2r+1)+(2s+1)=2(r+s+1).

Because integers are closed under addition, $r+s+1$ is an integer. The sum is therefore twice an integer and is even.

The final sentence is part of the proof. It connects the algebraic expression to the definition of evenness. For a concrete check, $7+11=18$ , but those particular values were not used to justify the general claim.

Prove a product is odd

Claim: the product of two odd integers is odd.

Use the same arbitrary representations and expand:

mn=(2r+1)(2s+1)=4rs+2r+2s+1=2(2rs+r+s)+1.

The expression $2rs+r+s$ is an integer, so the product has the form twice an integer plus one. It is odd.

Taking $r=s$ gives the special case that the square of an odd integer is odd. A special case can follow from a general theorem; proving a special case alone would not establish the full theorem.

Divisibility is an existence statement

For integers d and n with $d\ne0$ , the notation $d\mid n$ means that $n=dk$ for some integer k. It does not mean the decimal quotient was rounded to an integer, and it is not the same as the fraction $d/n$ .

Claim: if $d\mid a$ and $d\mid b$ , then $d\mid(a+b)$ .

The hypotheses provide integers r and s with $a=dr$ and $b=ds$ . Therefore

a+b=dr+ds=d(r+s).

Since $r+s$ is an integer, d divides the sum. The same reasoning proves that d divides any integer linear combination $ua+vb$ :

ua+vb=d(ur+vs),\qquad u,v\in\mathbb Z.

The integer restriction on u and v matters. Multiplying a multiple of d by an arbitrary real number need not preserve divisibility.

Divisibility through a chain

Suppose $a\mid b$ and $b\mid c$ , with a and b non-zero integers. There are integers r and s with $b=ar$ and $c=bs$ . Substitution gives $c=a(rs)$ , so $a\mid c$ . This proves transitivity for the stated divisibility relation.

For example, $3\mid12$ and $12\mid60$ imply $3\mid60$ . The proof works because the two integer multipliers combine, not because these particular numbers are small.

Prove a set inclusion using an arbitrary element

Claim: $A\cap B\subseteq A$ .

Take any $x\in A\cap B$ . By the definition of intersection, x belongs to A and to B. In particular, $x\in A$ . Every member of the intersection therefore belongs to A, which is exactly the required inclusion.

The argument does not assume that the intersection is non-empty. It shows that any member, if present, must have the required property. If there are no members, the inclusion still holds.

The same method proves that if $A\subseteq B$ and $B\subseteq C$ , then $A\subseteq C$ . Start with $x\in A$ , use the first inclusion to obtain $x\in B$ , then the second to obtain $x\in C$ .

Prove a set equality in both directions

Claim: $A\setminus(B\cup C)=(A\setminus B)\cap(A\setminus C)$ .

For the first inclusion, let $x\in A\setminus(B\cup C)$ . Then x is in A and not in $B\cup C$ . Hence x is in neither B nor C. It follows that x belongs to both $A\setminus B$ and $A\setminus C$ , so x is in their intersection.

For the reverse inclusion, suppose x belongs to both $A\setminus B$ and $A\setminus C$ . Then x is in A, outside B and outside C. It is therefore outside $B\cup C$ , so $x\in A\setminus(B\cup C)$ .

The two inclusions establish equality. Each direction uses the definitions; no assumed picture is needed. A Venn diagram can illustrate the result, but a diagram of one particular collection would not replace this arbitrary-member argument.

Use definitions to establish inequalities

Claim: if $x>y$ and $z>0$ for real numbers, then $xz>yz$ .

From $x>y$ , the difference $x-y$ is positive. Multiplying it by the positive z gives $z(x-y)>0$ . Expanding yields $xz-yz>0$ , so $xz>yz$ .

The hypothesis $z>0$ cannot be dropped. If $z=-1$ , order reverses; if $z=0$ , the products are equal. This proof explains the condition beside the rule rather than treating it as an isolated warning.

Another standard inequality follows from a square:

(x-y)^2\ge0\quad\Rightarrow\quad x^2+y^2\ge2xy.

Equality occurs exactly when $x=y$ . The argument uses the non-negativity of real squares, so its domain must be real numbers.

Make the conclusion explicit

A direct proof is complete when the reader can see why the final expression satisfies the target definition. “This looks even” is weaker than “it equals 2 times an integer”. “The sets have the same size” is weaker than proving that they have the same members.

Do not work from the desired conclusion as though it were already true. It is legitimate to explore backwards while planning, but the written proof must present a justified route from established assumptions. Mathematical Proof explains the difference between discovering an argument and presenting one.

Practice questions

Prove that the sum of an even integer and an odd integer is odd.
Prove that if $5\mid a$ and $5\mid b$ , then $5\mid(2a-3b)$ .
Prove that $A\subseteq A\cup B$ .

Check your answers

Write the integers as $2r$ and $2s+1$ . Their sum is $2(r+s)+1$ , which is odd because $r+s$ is an integer.
Write $a=5r$ and $b=5s$ . Then $2a-3b=5(2r-3s)$ ; the bracket is an integer, so 5 divides the result.
Take an arbitrary $x\in A$ . Membership in A is enough for membership in $A\cup B$ , by the definition of inclusive union. Hence every element of A belongs to the union.