Claims, Conditions, and Counterexamples

Find the conditions that limit a mathematical claim, follow an if–then statement in the correct direction, and use counterexamples and numerical checks without extending them beyond the cases they cover.

Mathematical claims often apply only under stated conditions. Learn to find those conditions, follow an if–then claim in the correct direction, and use a counterexample to expose a claim that is too broad.

Consider the equation

ax=b.

Dividing both sides by $a$ gives

x=\frac{b}{a},

but only when $a\ne0$ . That condition matters. If $a=0$ and $b\ne0$ , the original equation has no solution. If $a=0$ and $b=0$ , every value of $x$ is a solution. A correct conclusion must remain within the cases covered by its conditions.

Exercise: Identify a necessary condition

The equation $x^2=y$ is rearranged as $x=\sqrt{y}$ . What condition makes this conclusion correct when $x$ and $y$ are real numbers?

Choose the missing condition

x is nonnegative.

y is strictly positive.

x is an integer.

Select one choice, then check.

HintTry a negative value

If $x=-2$ , then $x^2=4$ , but $\sqrt{4}=2$ .

SolutionAccount for both square roots

From $x^2=y$ , the general real conclusion is

x=\pm\sqrt{y},

with $y\ge0$ . The narrower conclusion $x=\sqrt{y}$ also requires $x\ge0$ .

Follow an If–Then Claim in One Direction

The statement

P\Rightarrow Q

means that whenever condition $P$ holds, conclusion $Q$ must also hold. For example:

If an integer is divisible by $4$ , then it is even.

The claim works from divisibility by $4$ to evenness. It does not say that every even integer is divisible by $4$ ; the even integer $6$ shows that the reverse claim is false.

Exercise: Keep the stated direction

Suppose:

If an integer is divisible by $6$ , then it is divisible by $3$ .

Which reverse claim is not justified?

Choose the unjustified claim

If an integer is divisible by 3, then it is divisible by 6.

Every integer divisible by 6 is also divisible by 3.

Select one choice, then check.

HintTest a small multiple of 3

The integer $9$ is divisible by $3$ . Check whether it is divisible by $6$ .

SolutionTest the reverse

The integer $9$ is divisible by $3$ , but it is not divisible by $6$ . Therefore the original implication is true, but its reverse is not.

When both directions are true, writers may say “if and only if” and use the symbol $\Leftrightarrow$ . Later chapters will introduce this notation when a result genuinely requires both directions.

Test a Broad Claim with a Counterexample

Words such as “every,” “all,” and “for any” make a claim about every allowed case. Several successful examples can make such a claim plausible, but they do not cover all possible values.

For example, the values $1$ , $2$ , and $3$ satisfy $x^2\ge x$ . They do not prove that the inequality holds for every positive real number. At $x=1/2$ ,

\left(\frac12\right)^2=\frac14<\frac12.

This one counterexample disproves the broad claim. It also reveals the missing interval: the inequality fails when $0<x<1$ .

Exercise: Repair a claim using its counterexample

The claim “if $x^2=9$ , then $x=3$ ” is false over the real numbers. Which revision is complete?

Choose the repaired claim

If x² = 9, then x = 3 or x = -3.

If x² = 9, then x is positive.

If x² = 9, then x > 3.

Select one choice, then check.

HintUse the counterexample

The value $-3$ also has square $9$ .

SolutionInclude every valid case

Factoring gives

x^2-9=(x-3)(x+3)=0.

A product is zero when at least one of its factors is zero. Therefore $x-3=0$ or $x+3=0$ , which gives $x=3$ or $x=-3$ .

State What a Calculation Establishes

A calculation can verify a particular case. An experiment can report what happened for particular data and settings. Both are useful, but neither alone establishes a claim about every possible case.

If a program tests an identity on 100 inputs and every result agrees, then the identity passed those 100 checks. The test may reveal implementation errors and increase confidence, but a general mathematical conclusion still needs reasoning that covers the full stated domain.

Exercise: Limit a conclusion to the checked case

Computing $2+5=5+2$ provides what kind of support for the claim $a+b=b+a$ for all real numbers $a$ and $b$ ?

Choose the supported conclusion

A numerical check for one pair of numbers.

A proof for every pair of real numbers.

A definition of addition.

Select one choice, then check.

HintCount the covered cases

A claim about all pairs of real numbers covers more than the selected pair $2$ and $5$ .

SolutionDescribe only what was checked

The computation checks the claim for the pair $a=2$ and $b=5$ . It does not cover every pair of real numbers. A general argument or an accepted property of real-number addition is needed to justify the claim for all allowed values.

The practical questions are short: What is being claimed? Under which conditions? Does the reasoning run in the stated direction? Can one allowed case disprove the claim? What cases did the calculation or experiment actually check? These questions will recur whenever later chapters introduce a formula or result.