6.2 Deductive Proof
SL 1.6
- three ways to lay out the proof of an identity: LHS to RHS, both sides separately, and by equivalence
- how to write a deductive (direct) proof of a statement “if \(P\), then \(Q\)”
- how to represent even, odd and consecutive numbers with algebra
Proving an identity
To prove \(A \equiv B\) there are three common layouts.
Layouts for proving \(A \equiv B\)
Method A — LHS to RHS. Start from one side and change it, step by step, into the other: \(\;A = \cdots = \cdots = B\).
Method B — both sides. Simplify each side separately until they reach the same expression \(C\): \(\;A = \cdots = C\) and \(B = \cdots = C\), so \(A = B\).
Method C — equivalence. Turn the statement into one that is clearly true, using only reversible steps: \(\;A = B \Leftrightarrow \cdots \Leftrightarrow (\text{something true})\).
Method A is the one the IB expects most often. Start from the more complicated side. Never write the result you want to prove at the start and work on both sides together (unless every step is reversible, as in Method C).
Example 1 Prove the identity \((n + 1)^3 - (n - 1)^3 \equiv 6n^2 + 2\).
Start from the left-hand side:
\[ \begin{aligned} \text{LHS} &= \left(n^3 + 3n^2 + 3n + 1\right) - \left(n^3 - 3n^2 + 3n - 1\right) \\ &= 6n^2 + 2 \\ &= \text{RHS} \end{aligned} \]
Example 2 Prove the identity \(\dfrac{1}{k} - \dfrac{1}{k + 1} \equiv \dfrac{1}{k^2 + k}\), for \(k \neq 0, -1\).
\[ \text{LHS} = \frac{(k + 1) - k}{k(k + 1)} = \frac{1}{k(k + 1)} = \frac{1}{k^2 + k} = \text{RHS} \]
Example 3 Use Method B to prove that \((x + 3)^2 - (x - 1)^2 \equiv 8(x + 1)\).
Simplify each side separately:
\[ \text{LHS} = (x^2 + 6x + 9) - (x^2 - 2x + 1) = 8x + 8, \qquad \text{RHS} = 8(x + 1) = 8x + 8 . \]
Both sides equal \(8x + 8\), so LHS \(\equiv\) RHS.
Deductive proof
Many statements have the form “if \(P\) is true, then \(Q\) is true”, written \(P \Rightarrow Q\). In a deductive proof (or direct proof) we assume \(P\), and use facts and logical steps to reach \(Q\).
For whole numbers, these algebraic forms are very useful (with \(n, m \in \mathbb{Z}\)):
| Number | Written as |
|---|---|
| an even number | \(2n\) |
| an odd number | \(2n + 1\) |
| two consecutive integers | \(n,\; n + 1\) |
| a multiple of 3 | \(3n\) |
| two different odd numbers | \(2n + 1,\; 2m + 1\) (use two letters!) |
Example 4 Prove that the sum of two consecutive odd numbers is always a multiple of 4.
Let the consecutive odd numbers be \(2n + 1\) and \(2n + 3\), where \(n \in \mathbb{Z}\).
\[ (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1) . \]
Since \(n + 1 \in \mathbb{Z}\), the sum is a multiple of 4.
Example 5 Prove that if \(n\) is odd, then \(n^2 - 1\) is a multiple of 8.
Assume \(n\) is odd, so \(n = 2k + 1\) for some \(k \in \mathbb{Z}\).
\[ n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k = 4k(k + 1) . \]
\(k\) and \(k + 1\) are consecutive integers, so one of them is even, and \(k(k + 1)\) is a multiple of 2. Therefore \(4k(k + 1)\) is a multiple of \(4 \times 2 = 8\).
Example 6 Prove that if \(0 < x < y\), then \(x^2 < y^2\).
Assume \(0 < x < y\). Then \(y - x > 0\) and \(y + x > 0\). So
\[ y^2 - x^2 = (y - x)(y + x) > 0, \]
because it is the product of two positive numbers. Hence \(x^2 < y^2\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Starting with the result and working on both sides | Start from one side (Method A) | Otherwise you assume what you want to prove. |
| “It works for \(n = 1, 2, 3\), so it is true” | Use a general \(n\) | Examples are not a proof (see 6.3). |
| Two odd numbers written \(2n + 1\) and \(2n + 1\) | \(2n + 1\) and \(2m + 1\) | The same letter makes them the same number. |
| Forgetting to state \(n \in \mathbb{Z}\) | Say what your letters are | Proofs must be complete. |
Exercises
1 Prove that \((x + 4)^2 - (x - 4)^2 \equiv 16x\).
LHS \(= (x^2 + 8x + 16) - (x^2 - 8x + 16) = 16x =\) RHS.
2 Prove that \(\dfrac{2}{x - 1} - \dfrac{2}{x + 1} \equiv \dfrac{4}{x^2 - 1}\), for \(x \neq \pm 1\).
LHS \(= \dfrac{2(x + 1) - 2(x - 1)}{(x - 1)(x + 1)} = \dfrac{4}{x^2 - 1} =\) RHS.
3 Prove that the sum of any three consecutive integers is a multiple of 3.
\(n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)\), with \(n + 1 \in \mathbb{Z}\).
4 Prove that the product of two odd numbers is odd.
\((2n + 1)(2m + 1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1\), which is odd since \(2nm + n + m \in \mathbb{Z}\).
5 Prove that the difference between the squares of two consecutive integers is always odd.
\((n + 1)^2 - n^2 = 2n + 1\), which is odd.
6 Prove that \(n^3 - n\) is a multiple of 6 for every integer \(n\).
\(n^3 - n = (n - 1)n(n + 1)\), a product of three consecutive integers. Among them at least one is even and exactly one is a multiple of 3, so the product is a multiple of \(2 \times 3 = 6\).
7 Prove that for all real \(a\) and \(b\), \(a^2 + b^2 \ge 2ab\).
\(a^2 + b^2 - 2ab = (a - b)^2 \ge 0\), since a square is never negative. So \(a^2 + b^2 \ge 2ab\).
8 Prove that the sum of two consecutive odd numbers is a multiple of 4.
Let the numbers be \(2n + 1\) and \(2n + 3\), \(n \in \mathbb{Z}\). Their sum is \(4n + 4 = 4(n + 1)\), a multiple of 4.
9 Prove that \((n + 3)^2 - (n - 3)^2\) is a multiple of 12 for every integer \(n\).
\((n^2 + 6n + 9) - (n^2 - 6n + 9) = 12n\), which is a multiple of 12 because \(n \in \mathbb{Z}\).
10 Prove that \(x^2 - 6x + 10 > 0\) for all real \(x\).
Complete the square: \(x^2 - 6x + 10 = (x - 3)^2 + 1\). Since \((x - 3)^2 \ge 0\), the expression is at least \(1 > 0\).