Welcome to the World of Proof!
In Further Mathematics, we don't just want to know that something works; we want to be 100% certain why it works. This chapter covers the language of logic. You will learn how to build undeniable arguments and how to spot a "maths lie" using a single example. Don't worry if this seems a bit abstract at first—think of yourself as a mathematical detective building a case that no one can argue with!
1. Proof by Deduction
Proof by deduction is the most common type of proof. It’s like following a trail of breadcrumbs. You start with facts we already know are true (definitions) and use algebraic steps to reach a final conclusion.
How to do it:
- Define your terms: If the question is about even numbers, start by writing an even number as \(2n\). If it's an odd number, use \(2n+1\).
- Set up the expression: Follow the instructions in the question (e.g., "square it" or "add them together").
- Simplify: Use your algebra skills to expand and rearrange.
- Factorise: Usually, you want to show the result is a multiple of something. For example, if you want to prove a result is even, try to factor out a \(2\).
Example: Prove that the product of any two odd numbers is always odd.
Let the two odd numbers be \(2m+1\) and \(2n+1\).
Product: \((2m+1)(2n+1) = 4mn + 2m + 2n + 1\).
Factorise: \(2(2mn + m + n) + 1\).
Since \(2(something) + 1\) is the definition of an odd number, we have proved it!
Quick Review Box:
- Even number: \(2n\)
- Odd number: \(2n + 1\)
- Consecutive integers: \(n, n+1, n+2...\)
Key Takeaway: Deduction is all about logical steps. If your algebra is correct and your starting point is solid, your conclusion must be true!
2. Proof by Exhaustion
This sounds tiring, but it just means "checking every single possibility." You use proof by exhaustion when there are only a limited number of cases to check.
Analogy: Imagine you have a box of 5 lightbulbs and you want to prove they all work. You simply plug each one in, one by one, and test it. Once you've checked all 5, you've proved it "by exhaustion."
When to use it:
- When the question gives you a specific range (e.g., "For \(1 \le n \le 5\)...").
- When you can split a problem into categories (e.g., "When \(n\) is even" and "When \(n\) is odd").
Common Mistake: Forgetting to check one of the cases. If you miss even one, the proof isn't complete!
Key Takeaway: If you can test every possible scenario and they all work, the statement is proved.
3. Disproof by Counter-example
In mathematics, for a statement to be true, it must be true all of the time. To disprove a statement (a conjecture), you only need to find one single case where it doesn't work. This is called a counter-example.
Did you know? You can spend years trying to prove something is true, but someone can come along in one second and destroy your theory just by finding one exception!
Example: Disprove the statement "All prime numbers are odd."
Counter-example: The number \(2\) is prime but it is even. Therefore, the statement is false.
Key Takeaway: Proof requires a general argument; disproof only requires one specific "oops" example.
4. Proof by Mathematical Induction
This is the "heavy hitter" of Further Maths. Proof by induction is used to prove that a formula works for all positive integers (\(n = 1, 2, 3...\)).
The Domino Analogy:
1. The First Domino: You prove the first domino falls (\(n=1\)).
2. The Link: You prove that if any domino falls, the next one must fall (\(k \to k+1\)).
If both are true, every domino in the infinite line will fall!
The Four Essential Steps (B-A-I-C):
- Basis: Show the statement is true for \(n=1\). (Calculate the Left Hand Side and Right Hand Side separately).
- Assumption: Assume the statement is true for \(n=k\). Write down the formula replacing \(n\) with \(k\).
- Inductive Step: Show that it must then be true for \(n=k+1\). This is where the hard algebra happens! Use your "Assumption" to help you.
- Conclusion: Write the standard "closing sentence" (see below).
Application A: Summation of Series
You might be asked to prove a formula like \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\).
Trick: To find the sum for \(k+1\), take the sum for \(k\) (your assumption) and add the \((k+1)^{th}\) term:
\(Sum_{k+1} = Sum_k + Term_{k+1}\).
Application B: Sequences
If a sequence is defined as \(u_{n+1} = u_n + 2n\), you might have to prove a general formula for \(u_n\).
Step-by-step: Use the recurrence relation given in the question to substitute your \(n=k\) assumption into the \(n=k+1\) case.
Application C: Matrices
You can prove powers of matrices, such as \(\mathbf{M}^n\).
Key Rule: Remember that \(\mathbf{M}^{k+1} = \mathbf{M}^k \times \mathbf{M}\). Use your assumption for the \(\mathbf{M}^k\) part and multiply it by the original matrix \(\mathbf{M}\).
The "Standard" Conclusion (Learn this!):
"Since the result is true for \(n=1\), and if it is true for \(n=k\) it is shown to be true for \(n=k+1\), then by induction it is true for all positive integers \(n\)."
Encouragement: Induction algebra can look scary with all the fractions and brackets. Take it slow, factorise early, and remember: the question usually tells you the answer you are trying to reach!
Key Takeaway: Induction is a 3-step engine: Start it (\(n=1\)), show how it continues (\(k \to k+1\)), and conclude.
Summary Checklist
- Can I define an odd/even number algebraically? (Deduction)
- Is the number of cases small enough to check them all? (Exhaustion)
- Have I found one specific example that fails? (Counter-example)
- Have I followed the 4 steps of Induction and written the final conclusion?