Use definitions to make a general argument, or show that the opposite claim cannot hold.
AA HL
Papers 1 and 2
No GDC needed
About 20–30 min
By the end of this lesson
distinguish proof from evidence and counterexample
write a direct proof using arbitrary integers
build a contradiction with a precise assumption
Understand the idea
Testing several even integers may suggest that their squares are even. A proof must explain every allowed case at once. Writing an arbitrary even integer as $2k$, with k an integer, makes that possible without listing any cases.
Sometimes the direct route is awkward. Assume the opposite of the claim, keep the assumptions precise, and derive an impossibility. The contradiction tells us that the opposite assumption cannot hold.
Checkpoint
Correct. A later permitted case may still fail.
Not quite. Finite checking does not establish a universal implication. Successful examples do not disprove a statement either.
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A later permitted case may still fail.
Put the idea into notation
A universal claim has the form “for every x in a stated domain, P(x)”. A single valid counterexample disproves it; a list of successful examples does not prove it.
For integers k, even numbers have form $2k$ and odd numbers have form $2k+1$. For a nonzero integer d, divisibility by d means the number equals $dm$ for an integer m.
In contradiction, assume the negation of the desired conclusion. For “there is no smallest positive rational”, its negation is “there is a smallest positive rational”. Do not assume the conclusion you are trying to establish.
Example 1
Prove a statement for arbitrary inputs
Prove that the sum of two odd integers is even.
Let the integers be $2m+1$ and $2n+1$, where m,n are integers.
The sum is $2(m+n+1)$.
Since $m+n+1$ is an integer, the sum is even.
Why use two different integer parameters?
Using different integer parameters allows the two numbers to be different. The final integer condition connects the algebra back to the definition.
Example 2
Refute rather than keep testing
Is $n^2+n+11$ prime for every nonnegative integer n?
Try $n=11$: $121+11+11=143$.
$143=11\times13$, so it is composite.
The universal claim is false.
Why choose n = 11?
Choose n = 11 because every term in n² + n + 11 then has a factor 11. The resulting second factor is 13, so the value is composite. A counterexample need not be the first input where the claim fails.
Example 3
Create an impossible smaller value
Prove that there is no smallest positive rational number.
Assume a smallest positive rational q exists.
Then q/2 is rational and $0\lt q/2\lt q$.
This contradicts the assumed minimality of q, so no such smallest number exists.
Why must the smaller value stay in the set?
The proposed smaller value must remain in the same set. This argument does not work for positive integers because half an integer need not be an integer.
In the exam
Quick check
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The integer condition makes the parity claim valid.
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The integer condition makes the parity claim valid.
Divisibility is exactly an integer-multiple statement.
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Divisibility is exactly an integer-multiple statement.
A counterexample needs an allowed input and a false conclusion.
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A counterexample needs an allowed input and a false conclusion.
The proposed counterexample must stay in the set and exceed the assumed largest member.
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The proposed counterexample must stay in the set and exceed the assumed largest member.
The interval has largest member 1. The contradiction fails because 2 is outside the set.
Show explanation
The interval has largest member 1. The contradiction fails because 2 is outside the set.
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