Hey everyone, and welcome to the ultimate guide to playing devil’s advocate in your IB Math exams. Sounds fun, right? Sometimes in math, the most powerful way to prove you’re right is to start by assuming you’re wrong. It feels a bit like a logic puzzle or a detective story, where you follow a trail of clues to uncover a hidden truth. We’re going to dive into two of the most elegant and powerful techniques in your mathematical toolkit: the swift takedown of a counter-example and the mind-bending art of proof by contradiction. These aren’t just abstract concepts; they are problem-solving strategies that show up frequently, especially in IB Math AA HL. Let’s get ready to dismantle some bad assumptions and build some rock-solid proofs!
Table of Contents
The Power of a Single ‘Nope’: Disproving with Counter-Examples
Imagine someone makes a bold, universal statement like, “All birds can fly.” It sounds pretty reasonable, right? But then you think of a penguin or an ostrich. Boom. That single example—that one case where the rule fails—is called a counter-example. It’s all you need to prove the universal statement is false. In mathematics, this is an incredibly efficient way to debunk a claim.
Rule: Disproving a Universal Statement
A universal statement is a claim that is supposed to be true for all elements in a set (e.g., “for any real number x…”). To disprove it, you only need to find one single counter-example for which the statement is false.
Example: Finding a Counter-Example
Statement: For any real number $x$, if $x^2 > 25$, then $x > 5$.
Let’s analyze this. Our brains immediately jump to numbers like 6 or 10. For $x=6$, we get $6^2 = 36 > 25$, and $6 > 5$. It works! But the statement claims this is true for any real number.
What about negative numbers? Let’s test $x = -10$.
Step 1: Check the ‘if’ condition ($x^2 > 25$).
$(-10)^2 = 100$. Since $100 > 25$, the condition is true.
Step 2: Check the ‘then’ conclusion ($x > 5$).
Is $-10 > 5$? Absolutely not. The conclusion is false.
Conclusion: We found a case where the ‘if’ part is true but the ‘then’ part is false. Therefore, $x = -10$ is a counter-example, and the original statement is proven to be false.
Hint: Hunt for Edge Cases
When you see a universal statement, always test the tricky numbers! Think about negatives, zero, fractions, and irrational numbers. These ‘edge cases’ are often where counter-examples are hiding.
The Mind-Bending Art of Proof by Contradiction
Okay, now for the main event. What if you can’t find a counter-example? What if you need to prove something is true, but a direct approach is messy? Enter proof by contradiction. This is a beautiful, indirect method of proof that is a cornerstone of higher mathematics.
The Strategy: Proof by Contradiction
The core idea is to prove a statement is true by showing that if it were false, it would lead to a logical impossibility (a contradiction).
- Assume the Opposite: Start by assuming that the statement you want to prove is false.
- Follow the Logic: Use logical steps and mathematical rules to see where this assumption leads.
- Find the Contradiction: Show that your assumption leads to something nonsensical—like $1=2$, or a number being both even and odd, or a violation of your initial conditions.
- Conclude: Since your assumption led to a contradiction, it must be wrong. Therefore, the original statement must be true.
The Classic Case: Proving $\sqrt{5}$ is Irrational
Let’s walk through a classic IB Math AA HL problem: proving a number is irrational. We’ll tackle $\sqrt{5}$. A direct proof is tough, but a proof by contradiction is beautifully clean.
Example: Prove that $\sqrt{5}$ is irrational.
Step 1: Assume the opposite.
Let’s assume $\sqrt{5}$ is rational. By definition, this means we can write it as a fraction in its simplest form: $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ share no common factors.
Step 2: Manipulate the equation.
Square both sides: $(\sqrt{5})^2 = (\frac{p}{q})^2$, which gives us $5 = \frac{p^2}{q^2}$.
Rearrange the equation: $5q^2 = p^2$.
Step 3: Make deductions.
This tells us that $p^2$ must be a multiple of 5. A key property of prime numbers is that if $p^2$ is a multiple of 5, then $p$ itself must be a multiple of 5. So, we can write $p = 5k$ for some integer $k$.
Step 4: Substitute and deduce again.
Let’s substitute $p=5k$ back into our equation $5q^2 = p^2$:
$5q^2 = (5k)^2$
$5q^2 = 25k^2$
Divide both sides by 5: $q^2 = 5k^2$.
Look familiar? This means $q^2$ is a multiple of 5, which in turn means $q$ must also be a multiple of 5.
Step 5: Find the contradiction.
We have deduced that $p$ is a multiple of 5, AND $q$ is a multiple of 5. This means they share a common factor of 5. But this directly CONTRADICTS our initial assumption that the fraction $\frac{p}{q}$ was in its simplest form!
Conclusion: Our assumption that $\sqrt{5}$ is rational led to a logical impossibility. Therefore, the assumption must be false. We conclude that $\sqrt{5}$ is irrational.
The Equation Puzzle: Proving No Integer Solutions
This technique isn’t just for number theory. It’s great for proving that certain equations don’t have the types of solutions we’re looking for, like integers.
Example: Show that $3x^2 – 6x + 1 = 0$ has no integer solutions.
Step 1: Assume an integer solution exists.
Let’s assume there is an integer solution, $x=k$, where $k$ is an integer. If so, it must satisfy the equation:
$3k^2 – 6k + 1 = 0$
Step 2: Rearrange to find a contradiction.
Let’s isolate the constant term:
$1 = 6k – 3k^2$
Now, look for a common factor on the right-hand side. We can factor out a 3:
$1 = 3(2k – k^2)$
Step 3: Analyze the result.
Since $k$ is an integer, $k^2$ is an integer, and $2k$ is an integer. Their difference, $(2k – k^2)$, must also be an integer. Let’s call this integer $m$.
Our equation becomes $1 = 3m$.
This equation states that 1 is a multiple of 3. But this is clearly false! This is our CONTRADICTION.
Conclusion: Our assumption that an integer solution exists led to the absurd statement that 1 is a multiple of 3. Therefore, the assumption was wrong, and the equation has no integer solutions.
Thinking Big: Proving There’s No Largest Odd Integer
Proof by contradiction can also be used to prove broad, conceptual ideas about entire sets of numbers. Let’s prove there’s no such thing as a “biggest” odd number.
Example: Prove that there is no largest odd integer.
Step 1: Assume the opposite.
Assume there *is* a largest odd integer. Let’s call it $N$.
Step 2: Use the properties of the assumption.
By definition, an odd integer can be written in the form $2k+1$ for some integer $k$. So, $N = 2k+1$.
Step 3: Construct a number that breaks the assumption.
Let’s create a new number, $M$, by taking our supposed largest odd integer and adding 2 to it: $M = N + 2$.
Let’s analyze $M$. Substitute the expression for $N$:
$M = (2k+1) + 2 = 2k + 3$
We can rewrite this as $M = 2(k+1) + 1$. Since $k+1$ is an integer, $M$ is in the form $2 \times (\text{integer}) + 1$. This means $M$ is also an odd integer.
Now, let’s compare the size of $M$ and $N$. By our definition, $M = N + 2$, which clearly means $M > N$.
Step 4: State the contradiction.
We started by assuming $N$ was the largest odd integer. But we just constructed another integer, $M$, which is also odd and is larger than $N$. This is a CONTRADICTION.
Conclusion: Our assumption that a largest odd integer exists must be false. Therefore, there is no largest odd integer.
Key Takeaways
Mastering these two proof techniques is a huge step up in your mathematical maturity. They require you to think critically, creatively, and with precision.
- Counter-Examples: Your go-to weapon against ‘for all’ or ‘for any’ statements. One hit is all it takes. Always check negative numbers, zero, and fractions.
- Proof by Contradiction: Your secret weapon for trickier proofs. Embrace the idea of assuming the opposite and hunting for the logical flaw. This method is especially powerful for proving irrationality, non-existence of solutions, and general number theory properties.
Practice is key! The more you use these methods, the more natural they will become. You’ll start to recognize the patterns and know exactly when to deploy each strategy.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
Question 1: Disprove the statement: “For any integer $n$, the expression $n^2 + n + 41$ produces a prime number.”
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Solution: We need to find a counter-example. Let’s test some values of $n$. For $n=1, 1+1+41=43$ (prime). For $n=2, 4+2+41=47$ (prime). This seems to work for small numbers. Let’s try a trickier one, like $n=40$.
$n^2 + n + 41 = 40^2 + 40 + 41 = 1600 + 40 + 41 = 1681$.
Is 1681 prime? Notice that $41^2 = 1681$. So $1681 = 41 \times 41$, which is not prime. Another easy counter-example is $n=41$: $41^2 + 41 + 41 = 41(41+1+1) = 41 \times 43$.
Thus, $n=40$ (or $n=41$) is a valid counter-example. The statement is false.
Question 2: Disprove the statement: “If a quadrilateral has four equal sides, then it must be a square.”
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Solution: A square has four equal sides and four right angles. To find a counter-example, we need a quadrilateral with four equal sides but without four right angles. A rhombus (that is not a square) fits this description perfectly. A rhombus has four equal sides, but its angles are not necessarily 90 degrees. Therefore, a rhombus is a counter-example, and the statement is false.
Question 3: Prove by contradiction that the sum of a rational number and an irrational number is irrational.
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Solution: Let $r$ be a rational number and $i$ be an irrational number. Let their sum be $S = r+i$.
Assume the opposite: Assume $S$ is rational.
If $r$ is rational, we can write $r = \frac{a}{b}$ for integers $a, b$ ($b \neq 0$). If $S$ is rational, we can write $S = \frac{c}{d}$ for integers $c, d$ ($d \neq 0$).
Our equation is $\frac{c}{d} = \frac{a}{b} + i$.
Rearranging for $i$: $i = \frac{c}{d} – \frac{a}{b} = \frac{bc-ad}{bd}$.
Since $a,b,c,d$ are all integers, $bc-ad$ is an integer and $bd$ is a non-zero integer. This means $i$ can be expressed as a fraction of two integers, which means $i$ is rational.
This is a contradiction, as we were given that $i$ is irrational. Therefore, our assumption that the sum $S$ is rational must be false. The sum of a rational and an irrational number must be irrational.
Question 4: Prove by contradiction that $\sqrt{7}$ is irrational.
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Solution: Assume $\sqrt{7}$ is rational. Then $\sqrt{7} = \frac{p}{q}$ where $p, q$ are integers with no common factors and $q \neq 0$.
Squaring both sides gives $7 = \frac{p^2}{q^2}$, which means $7q^2 = p^2$.
This shows $p^2$ is a multiple of 7. Since 7 is prime, $p$ must also be a multiple of 7. So, let $p = 7k$ for some integer $k$.
Substitute this back: $7q^2 = (7k)^2 = 49k^2$.
Divide by 7: $q^2 = 7k^2$.
This shows $q^2$ is a multiple of 7, which means $q$ is also a multiple of 7.
We have shown that both $p$ and $q$ are multiples of 7. This contradicts our initial assumption that $p$ and $q$ had no common factors. Therefore, $\sqrt{7}$ must be irrational.
Question 5: Prove that the equation $5x^2 – 10x + 2 = 0$ has no integer solutions.
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Solution: Assume there is an integer solution $x=k$.
Substitute into the equation: $5k^2 – 10k + 2 = 0$.
Rearrange the equation: $5k^2 – 10k = -2$.
Factor out the common factor on the left side: $5(k^2 – 2k) = -2$.
Since $k$ is an integer, $(k^2 – 2k)$ is also an integer. Let’s call it $m$.
The equation becomes $5m = -2$. This implies that -2 is a multiple of 5. This is clearly false, giving us our contradiction.
Therefore, our initial assumption was wrong, and there are no integer solutions.
Question 6: Disprove the statement: “The difference between any two prime numbers is always an even number.”
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Solution: We need a counter-example. The statement claims the difference is *always* even. Most prime numbers are odd (e.g., 3, 5, 7, 11…). The difference between any two odd numbers is even (e.g., $11-7=4$). However, there is one prime number that is even: the number 2. Let’s test a difference involving 2.
Consider the prime numbers 5 and 2. Their difference is $5 – 2 = 3$. The number 3 is odd.
This is a counter-example, so the statement is false.
Question 7: Prove by contradiction that there is no smallest positive rational number.
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Solution: Assume there *is* a smallest positive rational number. Let’s call it $r$.
Since $r$ is a positive rational number, $r > 0$.
Now, let’s construct a new number, $m$, by dividing $r$ by 2: $m = \frac{r}{2}$.
Since $r$ is rational, $r = \frac{a}{b}$ for integers $a, b$. Then $m = \frac{a}{2b}$, which is also a rational number. Since $r > 0$, we know $m > 0$. So $m$ is a positive rational number.
Let’s compare $m$ and $r$. We have $m = \frac{r}{2}$. Since $r$ is positive, this means $m < r$.
We have found a positive rational number $m$ that is smaller than $r$. This contradicts our initial assumption that $r$ was the smallest positive rational number. Therefore, our assumption was false, and no such number exists.
Question 8: Prove that if $n^2$ is an even integer, then $n$ must also be an even integer. Use proof by contradiction.
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Solution: We want to prove: If $n^2$ is even, then $n$ is even.
Assume the opposite: Assume that $n^2$ is even, but $n$ is odd.
If $n$ is odd, we can write it in the form $n = 2k+1$ for some integer $k$.
Let’s see what $n^2$ is: $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.
Since $k$ is an integer, $(2k^2 + 2k)$ is also an integer. Let’s call it $m$. So, $n^2 = 2m+1$.
By definition, this means $n^2$ is an odd number.
This contradicts our given information that $n^2$ is an even integer. Therefore, our assumption that $n$ is odd must be false. If $n^2$ is even, $n$ must be even.
Question 9: Disprove the statement: For all real numbers $a$ and $b$, $\sqrt{a^2 + b^2} = a + b$.
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Solution: We need a counter-example. Let’s pick some simple non-zero values for $a$ and $b$. Let $a=3$ and $b=4$.
Left-Hand Side (LHS): $\sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Right-Hand Side (RHS): $a + b = 3 + 4 = 7$.
Since $5 \neq 7$, the statement is false for this case. Therefore, $a=3, b=4$ is a counter-example.
IB Exam Style Question
Question 10: (a) Prove by contradiction that for any prime number $p > 2$, $p$ must be an odd number. (b) Hence, prove that the product of any two prime numbers greater than 2 is always an odd number.
Reveal Solution
Part (a) Solution:
We are given that $p$ is a prime number and $p > 2$. We want to prove that $p$ is odd.
Assume the opposite: Assume that $p$ is an even number.
By definition, an even number has a factor of 2. So, if $p$ is even, it must be divisible by 2.
However, by definition, a prime number has exactly two distinct positive divisors: 1 and itself. Since $p$ is divisible by 2, and we are given $p>2$, $p$ must have at least three divisors: 1, 2, and $p$.
This contradicts the definition of a prime number. Therefore, our assumption that $p$ is even must be false. A prime number $p>2$ must be odd.
Part (b) Solution:
Let $p_1$ and $p_2$ be two prime numbers greater than 2. From part (a), we know that both $p_1$ and $p_2$ must be odd numbers.
An odd number can be written in the form $2k+1$ for some integer $k$.
Let $p_1 = 2k_1 + 1$ and $p_2 = 2k_2 + 1$ for some integers $k_1$ and $k_2$.
Now, let’s find their product:
$p_1 \times p_2 = (2k_1 + 1)(2k_2 + 1)$
$= 4k_1k_2 + 2k_1 + 2k_2 + 1$
$= 2(2k_1k_2 + k_1 + k_2) + 1$
Since $k_1$ and $k_2$ are integers, the expression $(2k_1k_2 + k_1 + k_2)$ is also an integer. Let’s call it $m$.
So, the product $p_1 \times p_2 = 2m + 1$.
By definition, any number in the form $2m+1$ is an odd number. Therefore, the product of any two prime numbers greater than 2 is always odd.