What you'll learn
- What a counterexample is and why a single counterexample can disprove a universal statement.
- How to check the precise conditions and domain of a claim.
- How to find effective counterexamples involving numbers, algebra and functions.
- How to write a short but logically complete disproof.
Mathematical statements
A mathematical statement is a sentence that is either true or false. For example, “Every even integer is divisible by 2” is true, while “Every prime number is odd” is false.
Many statements make a claim about every possible object of a particular kind. Words such as all, every, any and for all signal this kind of claim.
Universal statement
A universal statement claims that a property is true for every value or object in a stated domain. It can often be written in the form “For all xxx in the domain, property P(x)P(x)P(x) is true.”
The domain is the set of values or objects being considered. Examples include the integers, positive integers, real numbers or triangles. A value can only be used as a counterexample if it belongs to the stated domain.
Recognising a universal statement
Consider the claim: “For every integer nnn, the expression n2+n+1n^2+n+1n2+n+1 is odd.”
- The phrase “for every integer” shows that this is a universal statement.
- Its domain is the set of integers, so negative integers, zero and positive integers are all allowed.
- To disprove the statement, it would be enough to find one integer nnn for which n2+n+1n^2+n+1n2+n+1 is even.
Words to notice
When you see all, every, any or for all, immediately consider whether a single carefully chosen case could make the claim fail.
What is a counterexample?
Counterexample
A counterexample is a specific value or object that satisfies the conditions of a universal statement but does not satisfy its conclusion.
A universal statement says that there are no exceptions. Therefore, finding just one valid exception proves that the statement is false.
In symbols, suppose a claim has the form
For every x in D,P(x) is true.\text{For every }x\text{ in }D,\quad P(x)\text{ is true}.For every x in D,P(x) is true.To disprove it, you need to find some x∈Dx\in Dx∈D for which P(x)P(x)P(x) is false.
One valid exception is enough
You do not need to identify every value for which the claim fails. One counterexample is sufficient to disprove a universal statement.
Disproving a claim about prime numbers
Disprove the statement: “Every prime number is odd.”
- The claim concerns all prime numbers, so choose a value that is genuinely prime.
- Take the prime number 2. It has exactly two positive factors, 1 and 2, so it meets the condition of being prime.
- However, 2 is even, not odd. Therefore, 2 is a counterexample, and the statement is false.
Simply writing “2” is usually not a complete disproof. You should explain both why it satisfies the conditions and why it contradicts the conclusion.
Choosing a useful counterexample
Random testing can find a counterexample, but a more reliable approach is to think about cases where mathematical behaviour changes.
Useful values often include:
- zero;
- 1 or −1-1−1;
- negative numbers;
- fractions or non-integers;
- boundary values;
- equal values when a claim assumes only an ordering;
- small even, odd, prime or composite integers.
The best choice depends on the domain and the structure of the claim.
Testing a claim involving squares
Disprove the statement: “For every real number xxx, x2>xx^2>xx2>x.”
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Because the domain is the real numbers, fractions are allowed. Values between 0 and 1 are promising because squaring them makes them smaller.
-
Choose x=12x=\frac12x=21. Then
x2=(12)2=14.x^2=\left(\frac12\right)^2=\frac14.x2=(21)2=41. -
Since 14<12\frac14<\frac1241<21, the required inequality x2>xx^2>xx2>x is false.
-
Therefore, x=12x=\frac12x=21 is a counterexample, so the original statement is false.
Ignoring the domain
A proposed counterexample must satisfy the original conditions. For a statement about positive integers, zero, negative numbers and fractions cannot be used as counterexamples.
Disproving implications
Some universal statements have the form:
If condition A is true, then conclusion B is true.
This is called an implication. To disprove it, you must find a case where condition A is true but conclusion B is false.
A case where condition A is false does not disprove the implication, because the statement makes no claim about such a case.
Disproving an implication about divisibility
Disprove the statement: “For every integer nnn, if n2n^2n2 is divisible by 4, then nnn is divisible by 4.”
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A counterexample must have n2n^2n2 divisible by 4 while nnn itself is not divisible by 4.
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Choose n=2n=2n=2. Then
n2=22=4,n^2=2^2=4,n2=22=4,which is divisible by 4.
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However, 2 is not divisible by 4.
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Thus n=2n=2n=2 satisfies the condition but not the conclusion, so it is a counterexample and the statement is false.
Countering an if–then statement
For “if A, then B”, look specifically for A true and B false. No other combination disproves the statement.
Counterexamples involving algebraic identities
A claimed identity is an equation said to be true for every permitted value of its variables. If substituting one permitted value gives unequal sides, the claimed identity has been disproved.
Disproving a claimed identity
Disprove the claim that, for all real numbers aaa and bbb,
a2+b2=a+b.\sqrt{a^2+b^2}=a+b.a2+b2=a+b.-
Choose values that satisfy the domain, such as a=1a=1a=1 and b=1b=1b=1.
-
The left-hand side becomes
a2+b2=12+12=2.\sqrt{a^2+b^2} =\sqrt{1^2+1^2} =\sqrt2.a2+b2=12+12=2. -
The right-hand side becomes
a+b=1+1=2.a+b=1+1=2.a+b=1+1=2. -
Since 2≠2\sqrt2\neq22=2, the two sides are not equal. Therefore, a=1a=1a=1 and b=1b=1b=1 form a counterexample, so the claimed identity is false.
A numerical counterexample disproves a claimed identity, but it does not explain the full algebraic reason for the failure. That is acceptable when the task asks you to show disproof by counterexample.
Disproof is not the same as proof
Checking several examples can suggest that a statement is true, but it does not prove a universal claim. There may be an exception that you have not tested.
For instance, the expression
n2−n+41n^2-n+41n2−n+41produces prime numbers for many small positive integer values of nnn. However, taking n=41n=41n=41 gives
412−41+41=412,41^2-41+41=41^2,412−41+41=412,which is composite. Earlier successful cases could not establish that the expression was always prime.
Examples have different logical roles
One counterexample can disprove a universal statement, but any finite collection of successful examples usually cannot prove that the statement is true for every possible value.
Treating repeated success as proof
Testing 5, 10 or even 100 values does not prove a claim about all integers or all real numbers. A proof must cover the entire domain.
Writing a complete disproof
A clear counterexample argument normally has three parts:
- State your chosen value or object.
- Show that it meets the conditions in the claim.
- Show that the claimed conclusion fails, then state that the original claim is false.
Keep the argument focused. Once a valid counterexample has been fully verified, no further examples are required.
Not every claim can be disproved this way
Counterexamples directly disprove universal statements. To disprove an existential statement such as “There exists an integer with property PPP”, checking a few unsuccessful cases is not enough; you would need to prove that no such integer exists.
In the exam
- Identify the domain and every condition that your counterexample must satisfy.
- Choose a simple value, checking zero, 1, negative values, fractions or boundary cases where permitted.
- Substitute or calculate explicitly, showing that the condition holds but the conclusion fails.
- Finish with a direct sentence: “Therefore, this is a counterexample, so the statement is false.”
- Stop after one fully justified counterexample unless the question asks for more.
Check yourself
- How could you disprove the claim that the sum of any two prime numbers is even?
- Find a counterexample to “For every real number xxx, x2=x\sqrt{x^2}=xx2=x.”
- Why does testing the first 20 positive integers not prove a statement about every positive integer?