4.1 Disproving statements

Before we discuss ways of proving statements, it is important that you understand what it means to disprove a statement. Remember that mathematical statements are either true or false, but not both (See Section 2.1).

Consider the following statement.

Claim 4.1.

For any two real numbers $x$ and $y$ , if $x^{2}=y^{2}$ then $x=y$ .

This statement is not true: take $x=-3$ and $y=3$ . Then $x^{2}=9$ and $y^{2}=9$ , therefore $x^{2}=y^{2}$ . But this does not mean $x=y$ because we know $-3\neq 3$ .

Statements like this (with a universal quantifier) can be disproved by finding a single example for which the statement is false. Such an example is called a counterexample to the statement.

When presented with a statement that is false, it can be tempting to try to fix it in some way or another. Consider the following modified versions of Claim 4.1.

Claim 4.2.

For any two positive real numbers $a$ and $b$ , if $a^{2}=b^{2}$ then $a=b$ .

Claim 4.3.

For any two non-negative real numbers $a$ and $b$ , if $a^{2}=b^{2}$ then $a=b$ .

Claim 4.4.

For any two real numbers $a$ and $b$ , if $a^{2}=b^{2}$ then $a=b$ or $a=-b$ .

Claim 4.5.

For any two negative real numbers $a$ and $b$ , if $a^{2}=b^{2}$ then $a=b$ .

Claims 4.2, 4.3, 4.4 and 4.5 are all true, but they are not counterexamples to Claim 4.1; they do not disprove Claim 4.1. Rather, in each claim we have either modified the hypothesis to restrict the collection of objects so that the statement is true, or we have modified the conclusion.

Exercise 4.6.

Disprove the statement ‘all odd integers between $2$ and 14 are prime’.

Solution.Consider the integer $9$ which is between $2$ and $14$ but is not prime since it can be written as $3\times 3$ . Hence not all integers between $2$ and $14$ are prime.

Exercise 4.7.

Disprove the statement ‘for all real numbers $x$ , $x^{2}+x>0$ ’.

Solution.Let $x=0$ then $x^{2}+x=0$ . Hence $x^{2}+x$ is not greater than $0$ for all real $x$ .

[Of course, there are other counterexamples that you could use.]

Exercise 4.8.

Disprove the statement ‘for all positive integers $x$ , $x^{2}+x+41$ is prime’.

Solution.If you evaluate $x^{2}+x+41$ for each of $x=1$ , $x=2$ , and so on, it appears this quadratic always gives prime numbers.

However, for $x=41$ we have

$(41)^{2}+41+41=41(41+1+1)=41\times 43$

which is not prime.

[This quadratic is a famous example.]