Further Algebra — IB Math AA (HL)

1. Partial Fractions

Partial fraction decomposition rewrites a rational function $\dfrac{P(x)}{Q(x)}$ (where degree of $P$ is less than degree of $Q$) as a sum of simpler fractions. This is essential for integration and series work at HL.

Types of Partial Fractions

Denominator factor	Partial fraction form
$(ax+b)$ — linear	$\dfrac{A}{ax+b}$
$(ax+b)^2$ — repeated linear	$\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}$
$(ax^2+bx+c)$ — irreducible quadratic	$\dfrac{Ax+B}{ax^2+bx+c}$

An irreducible quadratic is one that cannot be factored over the reals (discriminant $b^2 - 4ac < 0$). The key technique for finding constants is the cover-up method (substitute the root of each linear factor) or comparing coefficients.

Worked Example 1

Decompose $\dfrac{3x+1}{(x-1)(x+2)}$ into partial fractions

Set up form: $\dfrac{3x+1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$

Multiply both sides by $(x-1)(x+2)$: $3x+1 = A(x+2) + B(x-1)$

Cover-up (substitute roots):
$x=1$: $3(1)+1 = A(3) \Rightarrow A = \dfrac{4}{3}$
$x=-2$: $3(-2)+1 = B(-3) \Rightarrow B = \dfrac{5}{3}$

Result: $\dfrac{3x+1}{(x-1)(x+2)} = \dfrac{4/3}{x-1} + \dfrac{5/3}{x+2}$

2. Proof Techniques

IB HL requires proficiency in four main proof methods. Each serves different types of mathematical claims.

The Four Proof Methods

Direct proof: Assume the hypothesis and derive the conclusion through valid logical steps.
Proof by contradiction: Assume the negation of what you want to prove, then derive a contradiction.
Proof by contrapositive: To prove $P \Rightarrow Q$, prove instead $\neg Q \Rightarrow \neg P$ (logically equivalent).
Mathematical induction: Prove a base case, then prove the inductive step (if true for $n=k$, true for $n=k+1$).

Worked Example 2

Prove by mathematical induction: $\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Base case ($n=1$): LHS $= 1^2 = 1$. RHS $= \dfrac{1 \cdot 2 \cdot 3}{6} = 1$. ✓

Inductive hypothesis: Assume true for $n=k$, i.e., $\displaystyle\sum_{j=1}^{k} j^2 = \dfrac{k(k+1)(2k+1)}{6}$.

Inductive step ($n=k+1$): $$\sum_{j=1}^{k+1} j^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}$$

Factor: $= \dfrac{(k+1)[k(2k+1)+6(k+1)]}{6} = \dfrac{(k+1)(2k^2+7k+6)}{6} = \dfrac{(k+1)(k+2)(2k+3)}{6}$

Conclusion: This is the formula with $n=k+1$. By the principle of mathematical induction, the formula holds for all $n \in \mathbb{Z}^+$. $\square$

Induction Template: Always explicitly state: (1) the base case is verified; (2) the inductive hypothesis; (3) what you are proving in the inductive step; (4) the conclusion invoking the principle of mathematical induction. Missing any of these costs marks.

3. Systems of Three Linear Equations

A system of three equations in three unknowns can be solved using Gaussian elimination (row reduction). The augmented matrix is reduced to row echelon form (REF).

Types of Solutions

Unique solution: One point of intersection (consistent, independent). REF has three pivots.
Infinitely many solutions: The planes share a line or coincide (consistent, dependent). A free variable appears.
No solution: Inconsistent system. A row $[0\;0\;0\;|\;c]$ with $c \ne 0$ appears in REF.

Worked Example 3

Solve the system: $x+y+z=6$, $2x-y+z=3$, $x+2y-z=2$

Write augmented matrix: $$\left[\begin{array}{ccc|c} 1&1&1&6\\ 2&-1&1&3\\ 1&2&-1&2 \end{array}\right]$$

Row reduce ($R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - R_1$): $$\left[\begin{array}{ccc|c} 1&1&1&6\\ 0&-3&-1&-9\\ 0&1&-2&-4 \end{array}\right]$$

$R_3 \leftarrow 3R_3 + R_2$: $$\left[\begin{array}{ccc|c} 1&1&1&6\\ 0&-3&-1&-9\\ 0&0&-7&-21 \end{array}\right]$$

Back-substitute: $-7z = -21 \Rightarrow z = 3$. Then $-3y - 3 = -9 \Rightarrow y = 2$. Then $x + 2 + 3 = 6 \Rightarrow x = 1$.

Solution: $(x, y, z) = (1, 2, 3)$.

Practice Problems

Q1. Decompose $\dfrac{2x^2+x-1}{x(x+1)(x-1)}$ into partial fractions.

Show Solution

$\dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{C}{x-1}$. Multiply through: $2x^2+x-1 = A(x+1)(x-1)+Bx(x-1)+Cx(x+1)$.

$x=0$: $-1 = -A \Rightarrow A=1$. $x=-1$: $2-1-1=B(-1)(-2) \Rightarrow 0=2B \Rightarrow B=0$. $x=1$: $2+1-1=C(1)(2) \Rightarrow C=1$.

Result: $\dfrac{1}{x} + \dfrac{1}{x-1}$

Q2. Prove by induction that $n^3 + 2n$ is divisible by 3 for all positive integers $n$.

Show Solution

Base case ($n=1$): $1+2=3$, divisible by 3. ✓

Assume $k^3+2k = 3m$ for some integer $m$.

Step ($n=k+1$): $(k+1)^3+2(k+1) = k^3+3k^2+3k+1+2k+2 = (k^3+2k) + 3k^2+3k+3 = 3m + 3(k^2+k+1)$, which is divisible by 3. $\square$

Q3. Determine the nature of solutions of: $x+y-z=2$, $2x-y+z=1$, $3x+2y-3z=k$. For what value of $k$ does a solution exist?

Show Solution

Row reduce: after elimination, the third row becomes $[0\;0\;0\;|\;k-5]$. For a solution to exist, we need $k=5$, giving infinitely many solutions (one free variable). For $k \ne 5$, the system is inconsistent.

4. Exam Tips

For partial fractions with a repeated factor $(x-a)^2$, both $\dfrac{A}{x-a}$ and $\dfrac{B}{(x-a)^2}$ are needed.
In induction proofs, never skip the "assume for $n=k$" step — it must be explicitly stated.
When row reducing, keep track of all row operations. A single arithmetic error cascades through the solution.
A zero row in REF with a non-zero augmented entry immediately signals no solution — state this clearly.