Number and Algebra | IB - Wyatt's Notes

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.

• $\emptyset$ — 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 $A$

A set can be defined by a property:

The vertical bar is read “such that.”

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 $A$.

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 $B'$. $\blacksquare$

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

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 $A$, $B$, $C$:

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

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 $Y$: $f(x_1) = f(x_2) \implies x_1 = x_2$

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

$f: X \to Y$

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

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

• 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 $x \in \mathbb{R}$

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.

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 $\mathbb{R}$.

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

A number in scientific notation has the form:

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

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.

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.

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

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

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.

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.

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

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 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 $\sum_{n=1}^{k} u_n = v_k - v_0$.

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

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

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:

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

$\blacksquare$

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

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

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:

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

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

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

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.

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 $1 + r$.

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 $\frac{1}{1+r}$.

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 $x$?”

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 $\log_a$: $\log_a(x^m) = mn = m\log_a x$. $\blacksquare$

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.

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

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

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).

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.

• 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 $n = 1$.

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:

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

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:

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

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

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

• 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 $a = 1,\; b = -1$)

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

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

Sets are collections that define membership — like a guest list for a party where you must be on the list to enter. Venn diagrams visualise how sets overlap, making abstract relationships concrete. Functions are the workhorses of mathematics: they take inputs and produce outputs with machine-like consistency. Logarithms are the undo button for exponentiation — they answer the question “what power must I raise the base to in order to get this number.” Induction is a chain of dominoes: prove the first falls, prove each one knocks over the next, and the whole chain must fall. The binomial theorem distributes the work of expanding powers across combinations.

$1 \in \{1, 2, 3\}$ 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.

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.

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    A[1_Number And Algebra] --> B[Key Concepts]
    A --> C[Core Principles]
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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 demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

• Number And Algebra
• Proof and Logic
• Complex Numbers