Proof by Introducing a Definition

Definitions Turn Words into Proof Conditions

In Combining Forward and Backward Reasoning, we used a conclusion to identify a useful target and then established that target from the hypotheses. A definition often tells us exactly what that target must look like. If a conclusion uses a term such as odd, the definition gives a form that can be written down and checked. If a conclusion concerns a midpoint or a distance, its definition likewise supplies a precise condition to work with.

A definition assigns a precise meaning to a term or symbol. In a proof, introducing or invoking a definition means replacing that term with the condition it names. This is not a new assumption. It is a way to make explicit what a statement already means. For example, the definition of oddness says that an integer $n$ is odd exactly when there is an integer $k$ such that $n=2k+1$. If a proof assumes $n$ is odd, it may use such an integer $k$; if it must prove $n$ is odd, it must produce an integer of the required kind.

A definition can also introduce useful vocabulary for a recurring mathematical idea. We will define the midpoint of two real numbers and the distance between two real numbers. In each case, the proof method is the same: state the definition with its domain, translate the claim into the defining condition, and verify that condition using facts already established.

A definition is a precise translation, not a shortcut around proof. If the goal uses a defined term, unpack the term into its exact condition. Then prove that condition, including any requirements on the objects involved.

A Reliable Definition-Based Method

Definitions are most helpful when used deliberately. First identify the term in the conclusion or hypothesis that carries the main mathematical content. Then write its definition in a form suited to the direction of the argument. A hypothesis involving a defined property may provide a representation or inequality. A conclusion involving that property may specify what must be established.

Identify the defined term.
Notice whether the hypothesis or conclusion contains a property, object, or relation that has a precise definition.

Write the definition explicitly.
Include the relevant domain: an integer witness must be an integer, and a real variable must be a real number.

Choose the direction that helps.
Expand a defined hypothesis to obtain usable information. For a defined conclusion, identify the exact condition that must be shown.

Use established facts to verify the condition.
Carry out the algebra or order reasoning, and check that any proposed witness belongs to the required domain.

Return to the original wording.
State that the defining condition has been met, so the original defined property follows.

This method makes the logical role of a definition visible. When proving an implication, we assume the stated hypothesis and establish the condition required by the conclusion. We do not treat the conclusion as if it were already known. In particular, writing down a formula that resembles the definition is not enough if a required existence claim or domain condition has been omitted.

Definitions Can Name Familiar Objects

A useful definition need not introduce an unfamiliar kind of number. It can give a concise name to a familiar construction. For real numbers $a$ and $b$, we define their midpoint by $$ \operatorname{mid}(a,b)=\frac{a+b}{2}. $$ The formula defines a real number because sums of real numbers are real and division by the nonzero real number $2$ is defined. The definition gives us a compact way to state and prove facts about the number halfway between $a$ and $b$.

Worked Example: The Midpoint Lies Between Its Endpoints

Proposition. If $a,b\in\mathbb R$ and $a\leq b$, then $$ a\leq \operatorname{mid}(a,b)\leq b. $$

Proof. Let $a,b\in\mathbb R$ and suppose $a\leq b$. Adding $a$ to both sides gives $2a\leq a+b$. Since $2>0$, division by $2$ preserves the inequality, so $$ a\leq \frac{a+b}{2}. $$ Adding $b$ to both sides of $a\leq b$ gives $a+b\leq 2b$. Dividing by the positive number $2$ gives $$ \frac{a+b}{2}\leq b. $$ By the definition of $\operatorname{mid}(a,b)$, these inequalities say $$ a\leq \operatorname{mid}(a,b)\leq b. $$ This proves the proposition.

The hypothesis allows equality, and the proof includes that case: when $a=b$, the defined midpoint is $(a+a)/2=a$. No strict inequality was needed. The definition supplied the expression to compare with the endpoints, while the order properties of the real numbers justified each comparison.

This example illustrates a useful proof habit. A defined symbol should not remain mysterious in the argument. Once the midpoint appears in the goal, substitute its defining expression and prove the resulting inequalities. At the end, naming the symbol again connects the calculation to the original statement.

Unpacking a Definition Can Prove a New Property

The same approach works for relations between two objects. Define the distance between real numbers $a$ and $b$ by $$ d(a,b)=|a-b|. $$ We take the standard case definition of absolute value and its elementary properties as familiar here; they are developed formally later in the course. Here $d$ is notation for a real-valued function of two real inputs. The definition does not say that distance is always positive: it can be zero when the inputs are equal. It also does not make a claim about every possible notion of distance in every setting. It defines this particular quantity on the real numbers.

Worked Example: Distance Is Symmetric and Vanishes Exactly at Equality

Proposition. For real numbers $a$ and $b$, $d(a,b)=d(b,a)$, and $d(a,b)=0$ if and only if $a=b$.

Proof. By the definition of distance, $$ d(b,a)=|b-a|=|-(a-b)|=|a-b|=d(a,b), $$ using the absolute-value property $|-x|=|x|$. This proves symmetry.

For the zero condition, again use the definition: $$ d(a,b)=0 \quad\Longleftrightarrow\quad |a-b|=0. $$ The absolute value of a real number is zero exactly when that number is zero, so this is equivalent to $a-b=0$, which is equivalent to $a=b$. Thus $d(a,b)=0$ if and only if $a=b$.

Both conclusions follow by expanding the newly defined notation. Symmetry comes from reversing the order of subtraction and using $|-x|=|x|$; the zero condition reduces to equality of the two inputs. The phrase “if and only if” requires both directions, and the equivalences above verify each direction rather than just one.

A definition-based proof can also clarify the meaning of a phrase defined by an inequality. For a real number $c$ and a nonnegative real number $r$, we will say that $x$ is within $r$ units of $c$ when $|x-c|\leq r$. This definition is convenient when discussing values near a center $c$. It is often useful to express the same condition as an interval, since ordinary order inequalities can be easier to inspect.

Worked Example: Rewriting a Distance Condition as an Interval

Proposition. Let $c,x\in\mathbb R$ and $r\in\mathbb R$ with $r\geq0$. Then $$ |x-c|\leq r \quad\Longleftrightarrow\quad c-r\leq x\leq c+r. $$

Proof. We use the definition of absolute value by cases. First suppose $x-c\geq0$. Then $|x-c|=x-c$, so $|x-c|\leq r$ is equivalent to $x-c\leq r$, or $x\leq c+r$. In this case $x\geq c$, and $r\geq0$ gives $c-r\leq c\leq x$. Thus both interval inequalities hold.

Now suppose $x-c<0$. Then $|x-c|=-(x-c)=c-x$, so $|x-c|\leq r$ is equivalent to $c-x\leq r$, or $x\geq c-r$. In this case $x<c$, and $r\geq0$ gives $x<c\leq c+r$, hence $x\leq c+r$. Again both interval inequalities hold. These two cases prove that $|x-c|\leq r$ implies $c-r\leq x\leq c+r$.

For the reverse implication, suppose $c-r\leq x\leq c+r$. If $x-c\geq0$, then $|x-c|=x-c$, and $x\leq c+r$ gives $x-c\leq r$. If $x-c<0$, then $|x-c|=c-x$, and $c-r\leq x$ gives $c-x\leq r$. In either case $|x-c|\leq r$. This proves the reverse implication, and therefore the equivalence.

What Definitions Do—and Do Not—Permit

When a term is defined, using it correctly means respecting its full definition. If the definition says that a witness is an integer, it is not enough to produce a real number. If a definition has a nonnegative parameter, that hypothesis must be stated before using it. If a condition is defined by an equation, the equation must be established rather than merely suggested by a diagram or an informal description.

A definition can be used in either direction, but the direction matters. For example, from $n$ odd, we may use the existence of an integer $k$ with $n=2k+1$. To prove that a particular integer is odd, we must exhibit such a $k$ and verify it is an integer. The first use unpacks the definition; the second checks the definition. Both are legitimate, but they have different proof tasks.

Where the defined term appears	Definition-based move	What must be checked
Hypothesis	Expand it into the condition it guarantees.	Use only the conclusion actually supplied by the definition.
Conclusion	Identify the condition that would establish it.	Verify every equation, inequality, or witness requirement.
New notation	Replace the notation by its defining formula when needed.	Confirm the inputs lie in the stated domain.

A definition cannot be used circularly. You may use the definition of the property you assumed. But when that property is the conclusion, you must establish its defining condition from other justified facts rather than assume it.

Introducing a definition can make a proof shorter, but clarity is more important than brevity. State the definition before relying on the new term, distinguish the term from the objects it describes, and return to the desired claim after verifying the defining condition. If the definition includes a quantifier such as “there exists,” the proof must provide an object satisfying it. If the definition includes “for every,” a proof must address an arbitrary object in the specified domain.

Definitions also help separate familiar language from mathematical precision. “Halfway between” suggests a midpoint, but the formula $(a+b)/2$ specifies exactly which real number is meant. “Close to” can be informal, whereas $|x-c|\leq r$ specifies a definite condition once $c$ and $r$ are given. A proof can then use algebra and order properties rather than relying on an intuitive picture.

The essential strategy is to move between two equivalent descriptions: the compact defined term and the explicit condition that gives it meaning. Use whichever description makes the next justified step visible. When the condition is established, translate back to the defined language to finish the proof.

Check Your Understanding

For each question, focus on the exact condition supplied by the relevant definition and on what the proof still needs to verify.

If $n$ is an odd integer, what does the definition guarantee? What must be shown if the goal is to prove that a particular integer $n$ is odd?
In the midpoint proposition, where does the expression $(a+b)/2$ come from, and why does the hypothesis $a\leq b$ include the case $a=b$?
Why is the domain of the witness important when a proof uses a definition of the form “there exists an integer $k$”?
For $r\geq0$, what role does the sign condition on $r$ play when $|x-c|\leq r$ is rewritten as an interval?
In the distance proposition, which absolute-value properties connect $d(a,b)$ to $d(b,a)$ and to the condition $a=b$?
Explain why assuming a defined property in order to prove that same property would be circular reasoning.