Number and Algebra

Sets

A set is a collection of distinct elements, written by listing its members inside curly braces or by Specifying a property that its members satisfy.

Notation

• $\emptyset$ — the empty set (contains no elements)
• $x \in X$ — the element $x$ belongs to the set $X$
• $x \notin X$ — the element $x$ does not belong to $X$
• $A \subseteq B$ — $A$ is a subset of $B$: every element of $A$ is also an element of $B$
• $|A|$ — the cardinality (number of elements) of a finite set $A$
• $\mathcal{P}(A)$ — the power set of $A$: the set of all subsets of $A$

Set-Builder Notation

A set can be defined by a property:

The vertical bar is read “such that.”

Set Operations

Let $A$ and $B$ be subsets of a universal set $U$.

• Union: $A \cup B = \{x \in U \mid x \in A \mathrm{ or } x \in B\}$
• Intersection: $A \cap B = \{x \in U \mid x \in A \mathrm{ and } x \in B\}$
• Complement: $A" = \{x \in U \mid x \notin A\}$
• Set difference: $A \setminus B = \{x \in A \mid x \notin B\}$

These operations are conveniently visualised with Venn diagrams. In a Venn diagram, the universal Set $U$ is drawn as a rectangle, and subsets are drawn as overlapping circles. The union $A \cup B$ Is the entire region covered by either circle; the intersection $A \cap B$ is the overlapping Region; the complement $A'$ is everything in the rectangle outside the circle for $A$.

De Morgan’s Laws

Theorem. For any subsets $A$ and $B$ of a universal set $U$:

$(A \cup B)' = A' \cap B'$

$(A \cap B)' = A' \cup B'$

Proof of the first law. We show mutual inclusion.

($\subseteq$) Suppose $x \in (A \cup B)'$. Then $x \notin A \cup B$So $x \notin A$ and $x \notin B$. Hence $x \in A'$ and $x \in B'$Which means $x \in A' \cap B'$.

($\supseteq$) Suppose $x \in A' \cap B'$. Then $x \in A'$ and $x \in B'$So $x \notin A$ and $x \notin B$. Therefore $x \notin A \cup B$Giving $x \in (A \cup B)'$.

The second law follows by symmetry or by applying the first law to $A'$ and $B'$. $\blacksquare$

Power Sets

The power set $\mathcal{P}(A)$ of a set $A$ is the set of all subsets of $A$Including $\emptyset$ and $A$ itself.

If $|A| = n$Then $|\mathcal{P}(A)| = 2^n$.

Cardinality of Finite Sets

For finite sets $A$ and $B$:

$|A \cup B| = |A| + |B| - |A \cap B|$

This is the inclusion-exclusion principle for two sets. It subtracts the overlap that would Otherwise be double-counted.

For three sets $A$, $B$, $C$:

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

Functions

A function $f$ is an assignment from a domain ($X$The set of acceptable inputs) to a codomain ($Y$The set into which all outputs must fall), such that:

• Every element in $X$ is mapped to an element in $Y$: $\forall x \in X,\; f(x) \in Y$
• No element in $X$ is mapped to more than one element in $Y$: $f(x_1) = f(x_2) \implies x_1 = x_2$

Notation

A function $f$ with domain $X$ and codomain $Y$:

$f: X \to Y$

Range

The range ($A$) of a function $f: X \to Y$ is the set of all values actually produced:

$A = \{f(x) \mid x \in X\}, \quad A \subseteq Y$

Classes of Functions

• Surjective (onto): Every element of the codomain is hit: $\forall y \in Y,\; \exists x \in X,\; y = f(x)$
• Injective (one-to-one): Distinct inputs give distinct outputs: $f(x_1) = f(x_2) \implies x_1 = x_2$
• Bijective: Both surjective and injective
• Odd: $f(-x) = -f(x)$ for all $x \in \mathbb{R}$
• Even: $f(-x) = f(x)$ for all $x \in \mathbb{R}$

Injectivity, Surjectivity, and Bijectivity — Worked Examples

Inverse Functions

If $f: X \to Y$ is bijective, the inverse function $f^{-1}: Y \to X$ exists and satisfies:

$f^{-1}(f(x)) = x \quad \mathrm{for all } x \in X, \qquad f(f^{-1}(y)) = y \quad \mathrm{for all } y \in Y$

To find $f^{-1}$: write $y = f(x)$Solve for $x$ in terms of $y$Then interchange $x$ and $y$.

Existence condition: A function has an inverse (on its given domain and codomain) if and only if It is bijective.

Self-Inverse Functions

A function $f$ is self-inverse if $f(f(x)) = x$ for all $x$ in the domain, i.e. $f^{-1} = f$.

Common examples: $f(x) = \frac{1}{x}$ on $\mathbb{R} \setminus \{0\}$; $f(x) = a - x$ on $\mathbb{R}$.

Domain Restriction to Achieve Injectivity

A non-injective function can be made injective by restricting its domain.

Scientific Notation and Approximation

Scientific Notation

A number in scientific notation has the form:

$a = b \times 10^k, \quad b \in \mathbb{R},\; 1 \le |b| \lt 10,\; k \in \mathbb{Z}$

Multiplication of Numbers in Scientific Notation

Let $a_1 = b_1 \times 10^{k_1}$ and $a_2 = b_2 \times 10^{k_2}$. Then:

$a_1 \cdot a_2 = b_1 \cdot b_2 \times 10^{k_1 + k_2}$

If $|b_1 \cdot b_2| \ge 10$ or $|b_1 \cdot b_2| \lt 1$Adjust the mantissa back into $[1, 10)$ by Absorbing a factor of $10$ into the exponent.

Significant Figures

The rules for significant figures:

1. All non-zero digits are significant.
2. Zeros between non-zero digits are significant.
3. Leading zeros are not significant.
4. Trailing zeros after a decimal point are significant.
5. Trailing zeros in a whole number without a decimal point are ambiguous.

Examples: $0.00450$ has 3 s.f.; $1020$ has at least 3 s.f.; $1020.$ has 4 s.f.

Absolute and Relative Error

If a quantity’s true value is $x_0$ and the approximate value is $x_a$:

• Absolute error: $|x_a - x_0|$
• Relative error: $\dfrac{|x_a - x_0|}{|x_0|}$

Upper and Lower Bounds

If a value is given as $x = 12.5$ (correct to 1 decimal place), then:

• Upper bound: $12.55$
• Lower bound: $12.45$

The interval is $[12.45,\; 12.55)$. The maximum possible error (absolute) is $\frac{1}{2} \times 10^{-d}$ where $d$ is the number of decimal places.

Sequences and Series

A sequence is a function $f$ with domain $\mathbb{N}$ (or $\mathbb{N}_0$) and codomain $X$. Writing $u_n = f(n)$Every sequence is ordered by its index.

$f: \mathbb{N} \to X, \quad u_n = f(n), \; n \in \mathbb{N}$

Series and Partial Sums

A series is the sum of the terms of a sequence. A partial sum $S_k$ is the sum of the first $k$ terms:

$S_k = \sum_{n=1}^{k} u_n$

Sigma Notation Properties

Sigma notation obeys the following rules (where $c$ is a constant independent of the index):

Linearity:

$\sum_{n=1}^{k} (au_n + bv_n) = a\sum_{n=1}^{k} u_n + b\sum_{n=1}^{k} v_n$

Proof. Distribute the sum over each term and factor constants out. Each term $au_n$ appears Exactly once in the expansion, so grouping gives $a$ times the sum of $u_n$And similarly for $bv_n$. $\blacksquare$

Index shifting: Replacing $n$ with $n + m$ shifts the bounds:

$\sum_{n=p}^{q} u_n = \sum_{n=p+m}^{q+m} u_{n-m}$

Telescoping: If $u_n = v_n - v_{n-1}$Then $\sum_{n=1}^{k} u_n = v_k - v_0$.

Arithmetic Sequences

An arithmetic sequence has a constant common difference $d$ between consecutive terms:

$a_n = a_1 + (n-1)d$

Arithmetic Series — Proof by Pairing

Theorem. $S_k = \frac{k}{2}(a_1 + a_k)$

Proof. Write the sum forward and backward:

$S_k = a_1 + (a_1 + d) + (a_1 + 2d) + \cdots + (a_1 + (k-1)d)$

$S_k = a_k + (a_k - d) + (a_k - 2d) + \cdots + (a_k - (k-1)d)$

Adding term-by-term, each pair sums to $a_1 + a_k$And there are $k$ such pairs:

$2S_k = k(a_1 + a_k) \implies S_k = \frac{k}{2}(a_1 + a_k)$

Substituting $a_k = a_1 + (k-1)d$ gives the alternative form:

$S_k = \frac{k}{2}\big(2a_1 + (k-1)d\big)$

$\blacksquare$

:::caution In the IB formula booklet this is written as:

$S_n = \frac{n}{2}(2u_1 + (n-1)d) = \frac{n}{2}(u_1 + u_n)$

:::

Geometric Sequences

A geometric sequence has a constant common ratio $r$ between consecutive terms:

$u_n = u_1 \cdot r^{n-1}$

Geometric Series — Proof by Subtraction

Theorem. $S_k = \dfrac{u_1(1 - r^k)}{1 - r}$ for $r \ne 1$.

Proof. Write $S_k$ and $rS_k$:

$S_k = u_1 + u_1 r + u_1 r^2 + \cdots + u_1 r^{k-1}$

$rS_k = u_1 r + u_1 r^2 + u_1 r^3 + \cdots + u_1 r^k$

Subtracting:

$S_k - rS_k = u_1 - u_1 r^k$

$S_k(1 - r) = u_1(1 - r^k)$

$S_k = \frac{u_1(1 - r^k)}{1 - r} \qquad \blacksquare$

Convergence of Geometric Series

If $|r| \lt 1$Then $\lim_{k \to \infty} r^k = 0$So:

$S_{\infty} = \frac{u_1}{1 - r}$

If $|r| \ge 1$The series diverges.

Applications: Compound Interest

If a principal $P$ is invested at rate $r$ per period, compounded each period, the value after $n$ Periods is:

$A = P(1 + r)^n$

This is a geometric sequence with $u_1 = P$ and common ratio $1 + r$.

Applications: Annuities

An annuity pays $d$ per period for $n$ periods, with interest rate $r$ per period. The present value Is:

$PV = \frac{d}{r}\left(1 - \frac{1}{(1+r)^n}\right)$

This follows directly from the geometric series sum with first term $\frac{d}{1+r}$ and ratio $\frac{1}{1+r}$.

Logarithms

A logarithm is the inverse function of the exponential $f(x) = b^x$:

$\log_b(b^x) = x, \quad b \in \{r \in \mathbb{R} \mid r \gt 0,\; r \ne 1\}$

The logarithm $\log_b x$ answers the question: “to what power must $b$ be raised to obtain $x$?”

Logarithm Laws — Proofs

Law 1 (Product rule): $\log_a(xy) = \log_a x + \log_a y$

Proof. Let $m = \log_a x$ and $n = \log_a y$. Then $a^m = x$ and $a^n = y$. So $xy = a^m \cdot a^n = a^{m+n}$. Taking $\log_a$ of both sides: $\log_a(xy) = m + n = \log_a x + \log_a y$. $\blacksquare$

Law 2 (Quotient rule): $\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$

Proof. Let $m = \log_a x$ and $n = \log_a y$. Then $a^m = x$ and $a^n = y$. $\frac{x}{y} = \frac{a^m}{a^n} = a^{m-n}$. Taking $\log_a$: $\log_a(x/y) = m - n = \log_a x - \log_a y$. $\blacksquare$

Law 3 (Power rule): $\log_a(x^m) = m\log_a x$

Proof. Let $n = \log_a x$So $a^n = x$. Then $x^m = (a^n)^m = a^{mn}$. Taking $\log_a$: $\log_a(x^m) = mn = m\log_a x$. $\blacksquare$

Change of Base Formula

Theorem. $\log_a x = \dfrac{\log_b x}{\log_b a}$ for any valid bases $a, b \gt 0$, $a, b \ne 1$.

Proof. Let $y = \log_a x$. Then $a^y = x$. Taking $\log_b$ of both sides: $\log_b(a^y) = \log_b x$. By the power rule, $y\log_b a = \log_b x$So $y = \frac{\log_b x}{\log_b a}$. $\blacksquare$

This is particularly useful for computing logarithms in bases other than $10$ or $e$ using a Calculator.

Solving Exponential Equations

When the variable is in the exponent, take logarithms of both sides to bring it down.

Solving Logarithmic Equations

Use the logarithm laws to combine terms, then exponentiate both sides.

Proof by Mathematical Induction

Mathematical induction is a technique for proving statements that are true for all natural numbers (or all integers greater than or equal to some starting value).

Structure of an Inductive Proof

To prove $P(n)$ for all $n \ge n_0$:

1. Base case: Verify $P(n_0)$ is true directly.
2. Inductive hypothesis: Assume $P(k)$ is true for some arbitrary $k \ge n_0$.
3. Inductive step: Using the hypothesis, prove that $P(k+1)$ is true.
4. Conclusion: By the principle of mathematical induction, $P(n)$ is true for all $n \ge n_0$.

The logic is analogous to an infinite chain of dominoes: the base case knocks over the first domino, And the inductive step ensures each domino knocks over the next.

Sum Formula Proofs

Divisibility Proofs

Inequality Proofs

Common Mistakes in Induction

• Forgetting the base case: The inductive step alone proves only an implication $P(k) \implies P(k+1)$. Without the base case, the chain never starts.
• Using what you need to prove: The inductive hypothesis is $P(k)$Not $P(k+1)$. You must derive $P(k+1)$Not assume it.
• Incorrect algebra: Errors in the inductive step (especially with fractions or factorisation) are the most common source of failed induction …/1-number-and-algebra/3_proof-and-logics.
• Wrong starting value: If the statement is only claimed for $n \ge 3$Verify the base case at $n = 3$Not $n = 1$.

The Binomial Theorem

Statement

For any non-negative integer $n$ and any $a, b \in \mathbb{R}$:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Where the binomial coefficient is:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Pascal’s Triangle

The binomial coefficients can be generated recursively via Pascal’s triangle. Each entry is the Sum of the two entries above it:

$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

Proof of Pascal’s identity. Using the factorial definition:

$\binom{n-1}{k-1} + \binom{n-1}{k} = \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{k!(n-1-k)!}$

$= \frac{(n-1)! \cdot k}{k!(n-k)!} + \frac{(n-1)! \cdot (n-k)}{k!(n-k)!}$

$= \frac{(n-1)!\big(k + n - k\big)}{k!(n-k)!} = \frac{n \cdot (n-1)!}{k!(n-k)!} = \frac{n!}{k!(n-k)!} = \binom{n}{k} \qquad \blacksquare$

Properties of Binomial Coefficients

• Symmetry: $\dbinom{n}{k} = \dbinom{n}{n-k}$
• Sum of row: $\displaystyle\sum_{k=0}^{n} \binom{n}{k} = 2^n$ (set $a = b = 1$ in the theorem)
• Alternating sum: $\displaystyle\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$ (set $a = 1,\; b = -1$)

Finding Specific Terms

The general term (the $(k+1)$-th term, since the sum starts at $k=0$) is:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Common Pitfalls

Confusing Subset and Element

$1 \in \{1, 2, 3\}$ but $1 \notin \{\{1\}, 2, 3\}$. In the second set, $\{1\}$ is an element, not $1$. Meanwhile, $\{1\} \subseteq \{1, 2, 3\}$ is true, but $\{1\} \in \{1, 2, 3\}$ is false.

Domain and Range Errors

When finding inverse functions, always check that the original function is bijective on the given Domain. A common error is to write $f^{-1}(x) = \sqrt{x}$ for $f(x) = x^2$ without restricting the Domain to $ contains the hardest questions within the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Number and Algebra with other IB mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Summary

This topic covers the mathematical techniques and concepts related to number and algebra, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.