Proof by Cases - Forms of Argument

Mathematical Writing - Vivaldi Franco 2014

Proof by Cases
Forms of Argument

Some mathematical arguments are made tidier by breaking them into a number of cases, of which precisely one holds, while all lead to the desired conclusion.

EXAMPLE. Consider the following statement:

The solution set of the inequality $$2|x| \geqslant |x-1|$$ is the complement of the open interval $$(-1,1/3)$$.

PROOF. Let $$x$$ be a real number. There are three cases.

(1)

(2)

(3)

So the required solution set is the collection of the values of $$x$$ such that $$x \leqslant -1$$, or $$1/3\leqslant x < 1$$, or $$x\geqslant 1$$, which is the union of two rays:

$$ (-\infty ,-1]\cup [1/3,1)\cup [1,\infty )= (-\infty ,-1]\cup [1/3,\infty ). $$

This set is obtained from the real line by removing the open interval $$(-1,1/3)$$, as claimed. $$\square $$

The opening sentence ’Let $$x$$ be a real number’ acknowledges that all real values of $$x$$ will have to be considered. The second sentence announces that the proof will branch into three cases, determined by the sign of the expressions within absolute value. Note the careful distinction between strict and non-strict inequalities, to avoid missed or repeated values of $$x$$.

The structure of this proof is shaped by the structure of the absolute value function, which is piecewise-defined—see Sect. 5.6. The presence of piecewise-defined functions usually leads to proofs by cases.

EXAMPLE. A proof by cases on divisibility.

Let $$n$$ be an integer; then $$n^5-n$$ is divisible by 30.

PROOF. We factor the integer 30 and the polynomial $$n^5-n$$:

$$ 30=2\times 3\times 5 \qquad \quad n^5-n=n(n-1)(n+1)(n^2+1). $$

For each prime $$p=2,3,5$$ we will show that, for any integer $$n$$, at least one factor of $$n^5-n$$ is divisible by $$p$$.

1.

2.

3.

The proof is complete. $$\square $$

After the factorisations we declare our intentions. There are three cases, one for each prime divisor $$p$$ of 30. Each prime in turn leads to $$p$$ cases, corresponding to the possible values of the remainder $$j$$ of division by $$p$$. To simplify book-keeping, we consider also negative remainders, and then pair together the remainders which differ by a sign. Note the presence of the symbols $$\pm $$ and $$\mp $$ within the same sentence. The positive sign in the first expression matches the negative sign in the second expression, and vice-versa.