Proof by Cases

Proving Every Possibility

In Proof by Contradiction, we proved a claim by assuming its negation and deriving an impossibility. Proof by cases uses a different structure. We divide the possible situations into a collection of cases, show that the cases include every possibility, and prove the desired conclusion separately under each case assumption.

The logical form is simple. If $C_1,C_2,\ldots,C_k$ are cases and at least one of them must hold, then a statement $S$ follows if we prove $C_1\Longrightarrow S$, $C_2\Longrightarrow S$, and so on through $C_k\Longrightarrow S$. It is not necessary for the cases to be mutually exclusive, although nonoverlapping cases often make the proof easier to read.

Proof by cases. To prove $S$, identify cases that cover all situations allowed by the hypotheses. In each case, assume its condition and derive $S$. Since every possible situation belongs to at least one case, $S$ holds without qualification.

For real numbers, the order properties of the real numbers give a particularly useful split: for any real $x$, either $x\geq0$ or $x<0$. These cases are exhaustive and cannot both hold. We will use this division to understand absolute value and then prove an inequality that is useful throughout analysis.

Absolute Value Is Defined by Cases

The absolute value of a real number records its distance from zero without retaining its sign. Its definition is already piecewise:

|x|= \begin{cases} x,&x\geq0,\\ -x,&x<0. \end{cases}

The equality case belongs in the first branch: if $x=0$, then $|x|=x=0$. In the second branch, $x<0$, so $-x>0$. Thus the definition gives a nonnegative value in either situation.

This definition illustrates why cases are useful rather than artificial. Expressions involving $|x|$ have different algebraic forms depending on the sign of $x$. We cannot replace $|x|$ by $x$ for every real $x$, nor can we replace it by $-x$ for every real $x$. A proof must handle both possibilities or use a result that already covers them.

Worked Example: A Square Is Nonnegative

Claim. For every real number $x$, $x^2\geq0$.

Proof by cases. Let $x\in\mathbb R$. Since $x\geq0$ or $x<0$, consider these two cases.

Case 1: $x\geq0$. The product of two nonnegative real numbers is nonnegative. Since $x$ is nonnegative, $x^2=x\cdot x\geq0$.

Case 2: $x<0$. Then $-x>0$, so $(-x)(-x)>0$. But $(-x)(-x)=x^2$, and therefore $x^2>0$, which implies $x^2\geq0$.

Both cases give the required inequality. As $x$ was arbitrary, $x^2\geq0$ for every real $x$.

The strict inequality in the second case is appropriate: if $x<0$, then $x\neq0$, so its square is positive. The first case allows $x=0$, which is why its conclusion is stated as $x^2\geq0$, not $x^2>0$.

A Basic Absolute-Value Bound

The definition yields two inequalities that will let us manage absolute values without repeatedly opening their definition. We prove them together because they express that $|x|$ is at least as large as either $x$ or $-x$.

Proposition. For every real number $x$, $$ |x|\geq x \qquad\text{and}\qquad |x|\geq -x. $$

Proof. Let $x\in\mathbb R$, and divide into the exhaustive cases $x\geq0$ and $x<0$.

If $x\geq0$, then $|x|=x$. Thus $|x|\geq x$ holds with equality. Also, $x\geq0$ implies $x\geq -x$, because adding $x$ to both sides gives $2x\geq0$. Since $|x|=x$, we obtain $|x|\geq -x$.

If $x<0$, then $|x|=-x$. Hence $|x|\geq -x$ holds with equality. Also $-x>x$: adding $-x$ to the inequality $x<0$ gives $0<-x$, and in particular $-x>x$. Therefore $|x|=-x\geq x$.

Both desired inequalities hold in each possible case, proving the proposition for every real $x$.

Worked Example: Solving an Absolute-Value Inequality

Claim. The real solutions of $|2x-1|<3$ are exactly those satisfying $-1<x<2$.

Proof by cases. Set $u=2x-1$. The two exhaustive cases are $u\geq0$ and $u<0$.

Case 1: $2x-1\geq0$. Here $|2x-1|=2x-1$. The inequality becomes $2x-1<3$, so $2x<4$ and $x<2$. The case assumption also gives $2x\geq1$, hence $x\geq\frac12$. Thus this case gives $\frac12\leq x<2$.

Case 2: $2x-1<0$. Here $|2x-1|=-(2x-1)=1-2x$. The inequality becomes $1-2x<3$, so $-2x<2$. Dividing by $-2$ reverses the inequality, giving $x>-1$. The case assumption gives $2x<1$, hence $x<\frac12$. Thus this case gives $-1<x<\frac12$.

The two solution ranges together are $(-1,\frac12)\cup[\frac12,2)=(-1,2)$. The boundary $x=\frac12$ is included: substituting it gives $|2(\frac12)-1|=0<3$. This verifies that the complete solution is $-1<x<2$.

The case assumption is part of the work, not an optional label. Solving the inequality obtained from one branch of the absolute-value definition is not enough: the other branch must also be addressed. Notice, too, that the final union includes every value found in either case and does not create a gap or include an endpoint that fails the original strict inequality.

The Triangle Inequality

A central estimate for absolute values is that the size of a sum cannot exceed the sum of the sizes. For real numbers this is called the triangle inequality. The name comes from geometry, but here we prove the real-number statement directly from the definition of absolute value and the proposition just established.

Theorem (Triangle Inequality). For all real numbers $x$ and $y$, $$ |x+y|\leq |x|+|y|. $$

Proof. Let $x,y\in\mathbb R$. We divide according to the sign of the sum $x+y$; every real sum satisfies either $x+y\geq0$ or $x+y<0$.

Case 1: $x+y\geq0$. By the definition of absolute value, $|x+y|=x+y$. The proposition gives $x\leq|x|$ and $y\leq|y|$. Adding these inequalities yields

|x+y|=x+y\leq |x|+|y|.

Case 2: $x+y<0$. By the definition, $|x+y|=-(x+y)=-x-y$. Apply the proposition to $x$ and $y$, using its inequalities $-x\leq|x|$ and $-y\leq|y|$. Adding gives

|x+y|=-x-y\leq |x|+|y|.

In either case the same conclusion holds. The cases exhaust all real values of $x+y$, so $|x+y|\leq|x|+|y|$ for all real $x,y$.

This proof illustrates a useful design choice. One could split into four cases according to the signs of $x$ and $y$, but that would require more bookkeeping. Instead, the proposition provides bounds that already handle both signs of each input. Splitting on the sign of the expression whose absolute value appears in the conclusion gives just two cases.

Worked Example: A Lower Bound for Two Distances

Claim. For every real $x$, $$ |x-2|+|x+2|\geq4. $$

Proof by cases. The expressions change form at $x=-2$ and $x=2$, so use the exhaustive regions $x\leq-2$, $-2\leq x\leq2$, and $x\geq2$. The endpoints are allowed in adjacent cases; overlap at an endpoint causes no problem.

Case 1: $x\leq-2$. Then $x-2\leq0$ and $x+2\leq0$. Therefore $|x-2|=2-x$ and $|x+2|=-x-2$. Their sum is $$ (2-x)+(-x-2)=-2x\geq4, $$ because $x\leq-2$ implies $-2x\geq4$.

Case 2: $-2\leq x\leq2$. Now $x-2\leq0$ and $x+2\geq0$. Thus $|x-2|=2-x$ and $|x+2|=x+2$, and their sum is $$ (2-x)+(x+2)=4. $$

Case 3: $x\geq2$. Then $x-2\geq0$ and $x+2\geq0$. Hence $|x-2|=x-2$ and $|x+2|=x+2$. Their sum is $$ (x-2)+(x+2)=2x\geq4, $$ because $x\geq2$.

Every real $x$ lies in at least one of these regions, and each case gives a sum at least $4$. This proves the claim. Equality holds throughout the middle region $[-2,2]$, as well as at the two outer boundary points.

The interval endpoints matter. At $x=-2$ and $x=2$, one of the expressions inside the absolute values is zero, and the formulas from adjoining cases agree. Stating the regions with inclusive endpoints makes coverage transparent; the overlap does not invalidate the proof because the conclusion is established in either applicable case.

Planning a Case-Based Proof

A case split should come from the hypotheses, definitions, or the expression being analyzed. Sign is a natural split for absolute value. Parity can be a natural split for an integer expression. A formula involving a maximum may call for comparing its inputs. Before beginning calculations, check that the cases really cover the domain and that each case supplies enough information to prove the same target.

State the target and its domain.
Identify which objects are arbitrary and what conclusion must hold for each one.

Choose a meaningful split.
Use a distinction that changes the relevant definition or supplies a useful inequality.

Check coverage.
Verify that every object allowed by the hypotheses satisfies at least one case condition.

Prove the conclusion within each case.
Use the case assumption explicitly, together with the original hypotheses and established results.

Combine the cases.
State why covering all possibilities means the conclusion holds for the original arbitrary object.

Cases do not need to be equally difficult, and a proof may use more than two. But an unnecessarily large split can hide the main idea. The triangle inequality proof uses two cases based on the sign of the sum, not four cases based on the separate signs of its inputs. Good case selection keeps the proof complete while reducing repetition.

Common Gaps in Case Arguments

A proof that verifies some cases but omits another has not established a universal claim. For example, proving a formula when $x\geq0$ says nothing about negative $x$ unless the desired conclusion is already known there for another reason. The phrase “the other case is similar” is acceptable only when the omitted argument is genuinely parallel and the relevant deductions are clear; it cannot replace a missing logical possibility.

A second common problem is using a case condition in a way that is not justified. From $x<0$, one may conclude $-x>0$, but not $-x<0$. In inequality manipulations, multiplying by a negative number reverses the inequality, as in the absolute-value example. Each branch must preserve the hypotheses and use the applicable algebraic rules.

Proof by cases is not proof by contradiction. In a case proof, the case assumptions describe possible situations and are not denied; the goal is to prove the conclusion in every one. In a contradiction proof, the temporary assumption is the negation of the target and is ultimately ruled out. Either method can be valid, but the logical role of the assumptions is different.

A case proof may also be combined with other proof methods. Within one case, a direct argument may be most convenient; within another, an established theorem may do the work. What matters is that the original domain is fully covered and that each branch reaches the stated conclusion from valid assumptions.

Check Your Understanding

For each question, identify the cases explicitly and verify that they cover every input in the stated domain.

For a real number $t$, write the two cases used by the definition of $|t|$. Which case includes $t=0$, and what is $|0|$?
Prove by cases that $x^2+|x|\geq0$ for every real $x$. In each case, state why both terms are nonnegative.
Solve $|x+1|\leq2$ by splitting into $x+1\geq0$ and $x+1<0$. Include the boundary point where $x+1=0$ in the appropriate case.
Use the proposition $|u|\geq u$ and $|u|\geq-u$ to prove that $|a-b|\leq|a|+|b|$ for all real $a,b$. What are the two cases in the triangle-inequality argument?
A proof of a claim for every real $x$ treats $x>0$ and $x<0$, but does not treat $x=0$. What logical possibility has been omitted? How could the case split be repaired?
Explain why the regions $x\leq-2$, $-2\leq x\leq2$, and $x\geq2$ cover the real line even though their endpoints overlap. Must cases in a proof be mutually exclusive?

What You'll Learn