Mastering Proof by Contradiction and Counter-example in A-Level Maths
Learn how to master proof by contradiction and counter-example, two essential techniques for A-Level Mathematics. This guide covers logical steps, worked examples, and common pitfalls to help you excel in your exams.
Introduction to Mathematical Proof
In A-Level Mathematics, proof is not just about showing that an answer is correct; it is about constructing a rigorous, logical argument that demonstrates a statement is true for all cases within a defined set. Examiners look for structured, clear, and logical steps that lead to a definitive conclusion. Understanding how to use different proof techniques is a core requirement of the specification.
Two of the most powerful tools in your arsenal are proof by contradiction and disproof by counter-example. While one helps you establish the truth of a universal statement, the other allows you to dismantle a false conjecture with a single, well-chosen value. Mastering these will significantly improve your ability to handle abstract algebraic problems and logical reasoning questions.
Understanding Proof by Contradiction
Proof by contradiction is a technique used to prove that a statement is true by showing that its negation leads to a logical impossibility. If assuming the opposite of what you want to prove results in a mathematical absurdity, then your original assumption must have been false, meaning the original statement is true.
To perform a proof by contradiction, follow these steps:
- Assume the negation of the statement you are trying to prove.
- Use logical deduction and known mathematical theorems to manipulate this assumption.
- Continue until you reach a contradiction (e.g., $1 = 0$, or an integer being both even and odd).
- Conclude that since the assumption led to a contradiction, the original statement must be true.
Worked Example: Proof by Contradiction
Proposition: Prove that there are no positive integers $x$ and $y$ such that $x^2 - y^2 = 10$.
Step 1: Assume the negation. Assume there exist positive integers $x$ and $y$ such that $x^2 - y^2 = 10$.
Step 2: Factorise the expression: $(x - y)(x + y) = 10$.
Step 3: Consider the parity of the factors. Let $A = x - y$ and $B = x + y$. Note that $B - A = (x + y) - (x - y) = 2y$. This means $B - A$ must be an even number. For the difference between two numbers to be even, both numbers must be either even or both odd.
Step 4: Check the factors of 10. The pairs of factors of 10 are $(1, 10), (2, 5), (-1, -10), (-2, -5)$.
- If $A=1, B=10$, their difference is 9 (odd).
- If $A=2, B=5$, their difference is 3 (odd).
Conclusion: Since the product of two numbers whose difference is even must be either both even or both odd, their product must be a multiple of 4 (if both even) or odd (if both odd). 10 is neither a multiple of 4 nor odd. This is a contradiction. Therefore, no such integers $x$ and $y$ exist.
Disproof by Counter-example
Sometimes, a statement is simply false. To prove a statement is false, you do not need to show it is false for all cases; you only need to find one single instance where it fails. This is called a counter-example.
When you are asked to prove a statement is false, look for values that sit on the boundaries of the domain, such as negative numbers, zero, or fractions, as these are often where generalisations fail.
Worked Example: Counter-example
Proposition: For all real numbers $x$, $x^2 > x$.
Step 1: Test the conjecture. If we try $x = 2$, $2^2 = 4$, and $4 > 2$ (True). If we try $x = 1$, $1^2 = 1$, and $1 > 1$ (False).
Step 2: Select the counter-example. Let $x = 1$. Since $1^2 = 1$, the statement $1 > 1$ is false. Alternatively, let $x = 0.5$. Then $0.5^2 = 0.25$, and $0.25 < 0.5$.
Conclusion: Since we have found a value ($x=1$ or $x=0.5$) for which the statement does not hold, the conjecture is false.
Common Mistakes to Avoid
- Insufficient Logic: Simply showing that a statement works for a few numbers is not a proof. You must use algebraic manipulation to cover all cases.
- Incorrect Negation: When starting a proof by contradiction, ensure you have correctly identified the negation. For example, the negation of "all $x$ are even" is "there exists at least one $x$ that is odd."
- Ignoring the Domain: Always check if the variables are defined as integers, real numbers, or positive integers. A counter-example that works for real numbers might not be valid if the question specifies integers.
- Vague Conclusions: Always finish your proof with a clear concluding sentence, such as "This contradicts our initial assumption, therefore the statement is true."
Frequently Asked Questions
Q: Do I need to use proof by contradiction for every question? No. Only use it when the question specifically asks for it or when a direct proof seems impossible. Always check if a direct algebraic deduction is simpler first.
Q: What is the difference between a conjecture and a theorem? A conjecture is a statement that is believed to be true but has not yet been proven. A theorem is a statement that has been rigorously proven to be true.
Q: Can I use a counter-example to prove a statement is true? Never. A counter-example can only be used to prove a statement is false. To prove a statement is true, you must use a general logical argument.
Conclusion
Proof by contradiction and counter-example are essential skills for any A-Level mathematician. By assuming the opposite to find a contradiction or searching for that one value that breaks the rule, you develop the logical rigour required for higher-level study. To see these concepts in action with interactive, narrated animations, visit MathInstructor AI and generate a free lesson on proof today.
Topics
Want this explained out loud?
Turn any question into a narrated, animated lesson in seconds.
Try the Studio free