Direct proof
Learn direct proof for Year 11 Specialist Mathematics in Queensland (QCAA). This is the first and most important proof method: you assume what you are given and reason forward, step by justified step, to the conclusion.
You will learn to represent an even integer as 2k and an odd integer as 2k+1, then use algebra to prove statements about integers — that sums and products behave as claimed and that divisibility follows — the proof foundation for the rest of the course.
Theory
A direct proof argues straight from what is given to what must be shown, one justified step at a time. In Year 11 Specialist Mathematics (QCAA, Queensland) it is the first proof method: represent an even integer as 2k and an odd integer as 2k+1, then use algebra to prove statements about integers.
An implication "if P then Q" claims that whenever the hypothesis \(P\) holds, the conclusion \(Q\) must follow. A direct proof establishes it by assuming \(P\) and reasoning forward, using algebra and known facts, until \(Q\) is reached.
The key move is to replace words by algebra. An even integer is a multiple of \(2\), so it is written \(2k\); an odd integer is one more than an even one, so it is written \(2k+1\). Here \(k\) is any integer, so the argument covers every case at once.
More generally, an integer divisible by \(d\) is written \(dk\). A result of the form \(2\times(\text{integer})\) is even; a result of the form \(2\times(\text{integer})+1\) is odd. Reaching one of these forms is what finishes a parity proof.
When two integers are independent, give them different letters, \(2m\) and \(2n\). Using \(2m\) twice would force the two integers to be equal and prove nothing general.
The two general representations that start almost every integer proof (\(k\) an integer):
Sum of two even integers — factor out the \(2\):
Product of two odd integers — expand, then group to the form \(2k+1\):
How to write a direct proof about integers
- State what is given and what is to be shown, and represent each integer in general form: even \(=2m\), odd \(=2n+1\), divisible by \(d\) \(=dk\) — different letters for independent integers.
- Combine the expressions as the statement requires: add, multiply or square them.
- Simplify by expanding and then factoring to reveal \(2(\ldots)\) or \(2(\ldots)+1\).
- Conclude: name the resulting integer and state that the form makes the number even, odd, or divisible as claimed.
Represent each even integer with its own multiple of \(2\), then add and factor:
| \(\text{integers}\) | \(=\) | \(2m \text{ and } 2n\) |
| \(2m + 2n\) | \(=\) | \(2(m+n)\) |
Since \(m+n\) is an integer, \(2(m+n)\) is \(2\times(\text{an integer})\), which is even.
The sum equals \(2(m+n)\), so it is even.
Write each odd integer as \(2\times(\text{integer})+1\), expand, then group to the form \(2k+1\):
| \(\text{integers}\) | \(=\) | \(2a+1 \text{ and } 2b+1\) |
| \((2a+1)(2b+1)\) | \(=\) | \(4ab + 2a + 2b + 1\) |
| \(=\) | \(2(2ab+a+b) + 1\) |
With \(k = 2ab+a+b\) an integer, the product has the form \(2k+1\).
The product equals \(2(2ab+a+b)+1\), so it is odd.
Represent the even integer as \(2m\) and the odd integer as \(2n+1\), then add:
| \(2m + (2n+1)\) | \(=\) | \(2m + 2n + 1\) |
| \(=\) | \(2(m+n) + 1\) |
As \(m+n\) is an integer, the sum has the form \(2\times(\text{integer})+1\).
The sum equals \(2(m+n)+1\), so it is odd.
Use divisibility by \(3\) to write \(n=3m\), then square it:
| \(n\) | \(=\) | \(3m\) |
| \(n^2\) | \(=\) | \((3m)^2\) |
| \(=\) | \(9m^2\) |
Since \(m^2\) is an integer, \(9m^2\) is \(9\times(\text{an integer})\).
\(n^2 = 9m^2\), so it is divisible by \(9\).
Common pitfalls
Frequently asked questions
What is a direct proof?
A direct proof assumes the hypothesis is true and reasons forward, using algebra and known facts one step at a time, until the conclusion is reached.
How do you write an even number and an odd number in a proof?
An even integer is written \(2k\) and an odd integer is written \(2k+1\), where \(k\) is any integer, so a single argument covers every case.
Why use different letters for the two integers?
Independent integers get different letters, \(2m\) and \(2n\). Using \(2m\) for both would force the integers to be equal and would not prove the general statement.
How do you prove the sum of two even integers is even?
Let the integers be \(2m\) and \(2n\). Then \(2m+2n=2(m+n)\), which is \(2\times(\text{an integer})\), so the sum is even.
Is checking a few examples a valid proof?
No. Examples can suggest a result but cannot prove it. A direct proof must use the general forms so that it holds for every integer.
How do you know when a parity proof is finished?
When the algebra reaches \(2\times(\text{integer})\) the number is even, and when it reaches \(2\times(\text{integer})+1\) it is odd. Naming that integer completes the proof.