Before You Begin: From School to University Mathematics

School mathematics is calibrated to an exam. Different boards examine different subsets, in different notation, at different depths; and every one of them leaves the same gaps, because a timed paper cannot test whether you understand what a limit is, only whether you can compute one. University mathematics is the opposite: it assumes you know what the objects are, and examines whether you can prove things about them.

This page closes the gap. It defines every term the first-year tracks use, proves every rule it states, and tells you explicitly which parts of school mathematics to keep, which to unlearn, and which to upgrade. If you are confident, skim it and take the self-test in the last section; if any answer surprises you, read the section it came from. Nothing here reappears as an exam topic; it is the foundation the tracks stand on, and the tracks link back here the first time they use a term.

The order is by dependency, not by difficulty. Section 1 is logic, because every proof technique is a logical move with a name. Sections 2–3 are sets and functions, because the whole of the university course is about maps between structured sets. Section 4 is proof technique. Sections 5–8 are the school topics rebuilt rigorously: algebra, trigonometry, geometry, and calculus.

School mathematics teaches procedures: differentiate this, solve that, complete the square. University mathematics studies objects: functions, spaces, maps between them, and the theorems they obey. The transition is the shift from “what is the answer” to “why is this true”, and the reason it trips people up is not difficulty but vocabulary; university mathematics names the moves school mathematics performs silently.

A limit is the first example. At school, $\lim_{x\to 0}\frac{\sin x}{x} = 1$ is a fact to use. At university it is a definition - $\forall\varepsilon > 0\ \exists\delta > 0 : 0 < |x| < \delta \Rightarrow \left|\frac{\sin x}{x} - 1\right| < \varepsilon$; and once the definition is available, “prove that the limit is 1” becomes a different kind of task, one that needs the squeeze theorem and the geometry of the unit circle. This page gives you both layers for every object below: the definition, and the school result it recovers.

Definition (statement). A statement (or proposition) is a declarative sentence that is either true or false. “7 is prime” is a statement; “is 7 prime?” and "$x + 1 = 5$" are not; the last is an open statement, true for some $x$ and false for others.

Connectives. Given statements $P$ and $Q$:

The two rows that surprise people are the last two: a false hypothesis makes an implication vacuously true. “If $2 + 2 = 5$ then I am the Pope” is a true statement. This is not a trick; it is what makes proofs by contradiction and empty-set arguments work.

Definition (converse, contrapositive). For $P \Rightarrow Q$: the converse is $Q \Rightarrow P$; the contrapositive is $\neg Q \Rightarrow \neg P$.

Proposition. An implication and its contrapositive are logically equivalent. An implication and its converse are not.

Proof. Build the truth table. $\neg Q \Rightarrow \neg P$ is false exactly when $\neg Q$ is true and $\neg P$ false, i.e. $Q$ false and $P$ true; the single row where $P \Rightarrow Q$ is false. For the converse, $Q \Rightarrow P$ is false when $Q$ true, $P$ false; a different row, so the converse has a different truth table from the original. $\blacksquare$

Why this matters. Mixing up a statement with its converse is the most common logical error in mathematics. “If $x = 2$ then $x^2 = 4$” is true; its converse “if $x^2 = 4$ then $x = 2$” is false ($x = -2$). Convergent sequences have bounded terms; bounded terms do not give convergence.

Quantifiers. $\forall$ (“for all”) and $\exists$ (“there exists”).

$$\forall\varepsilon > 0\ \exists\delta > 0 : P(\delta) \quad\text{means}\quad \text{for each }\varepsilon,\text{ a }\delta\text{ may be chosen (possibly depending on it).}$$

Proposition (negation rules). $\neg\forall x\,P(x) \equiv \exists x\,\neg P(x)$ and $\neg\exists x\,P(x) \equiv \forall x\,\neg P(x)$. Quantifiers negate in order, and $\forall$ becomes $\exists$.

Why this matters. The negation of “the sequence converges” - $\forall\varepsilon > 0\ \exists N\ \forall n \geq N : |a_n - L| < \varepsilon$ - is “the sequence does not converge”: $\exists\varepsilon > 0\ \forall N\ \exists n \geq N : |a_n - L| \geq \varepsilon$. Every “prove this does not converge” exercise is that string, applied.

Definition. A set is a collection of distinct objects, its elements. $x \in A$ means $x$ is an element; $A \subseteq B$ means every element of $A$ is an element of $B$.

Definition. For sets $A, B$ in a universe $U$:

$$A \cup B = \{x : x \in A \text{ or } x \in B\}, \quad A \cap B = \{x : x \in A \text{ and } x \in B\}, \quad A^c = \{x \in U : x \notin A\},$$ $$A \times B = \{(a, b) : a \in A, b \in B\}, \quad A \setminus B = \{x \in A : x \notin B\}.$$

Proposition (De Morgan). $(A \cup B)^c = A^c \cap B^c$.

Proof. $x \in (A\cup B)^c$ iff $x \notin A \cup B$ iff $x \notin A$ and $x \notin B$ iff $x \in A^c$ and $x \in B^c$. $\blacksquare$

Definition (function). See Section 3; but note the notation for images and preimages: $f(A) = \{f(a) : a \in A\}$ and $f^{-1}(B) = \{x : f(x) \in B\}$. The preimage always exists for any $B$; the inverse function may not.

Definition (cardinality, finite and countable). Two sets are equipotent if there is a bijection between them. A set is countable if it is equipotent to a subset of $\mathbb{N}$; countably infinite if to $\mathbb{N}$ itself; uncountable otherwise.

Theorem (Cantor). $\mathbb{Q}$ is countable; $\mathbb{R}$ is uncountable.

Proof for $\mathbb{Q}$. List fractions by increasing $p + q$ for $p/q$ in lowest terms with $q > 0$; each positive rational appears exactly once, and the list is infinite, so $\mathbb{Q}^+ \sim \mathbb{N}$; reflect to include negatives and zero.

Proof for $\mathbb{R}$ (Cantor’s diagonal). Suppose $\mathbb{R} \sim \mathbb{N}$ via a list $r_1, r_2, \dots$ of reals in $(0,1)$, each with decimal expansion $0.d_{n1}d_{n2}d_{n3}\dots$. Construct $d = 0.e_1e_2e_3\dots$ with $e_n = 5$ if $d_{nn} \neq 5$ and $e_n = 6$ if $d_{nn} = 5$. Then $d \in (0,1)$ and $d \neq r_n$ for every $n$; it differs from $r_n$ in the $n$-th decimal; contradicting the list’s completeness. (Avoiding dual expansions like $0.4999\ldots = 0.5$: the digit choice $5/6$ never produces a trailing $9$.) $\blacksquare$

The theorem to internalise: infinity has sizes. The rationals are dense but countable; the reals are uncountable. Measure theory; the foundation of probability; is built on exactly this distinction.

Definition. A function $f : A \to B$ is a rule assigning to each $a \in A$ exactly one $f(a) \in B$. $A$ is the domain, $B$ the codomain, and $f(A)$ the image.

Definition. $f$ is injective ($f(a_1) = f(a_2) \Rightarrow a_1 = a_2$), surjective ($f(A) = B$), and bijective if both.

Proposition. $f : A \to B$ is bijective iff it has an inverse $f^{-1} : B \to A$ with $f^{-1}(f(a)) = a$ and $f(f^{-1}(b)) = b$.

Proof. If $f$ is bijective, define $f^{-1}(b) = a$ where $a$ is the unique element with $f(a) = b$; unique by injectivity, existing by surjectivity. The two compositions are identities by construction. Conversely, if $f^{-1}$ exists, injectivity and surjectivity follow from the two composition identities. $\blacksquare$

Definition (composition). $(g \circ f)(x) = g(f(x))$, defined where $f(x) \in$ domain of $g$. Composition is associative - $(h \circ g) \circ f = h \circ (g \circ f)$; and not commutative.

Proposition. $(g \circ f)$ injective $\Rightarrow$ $f$ injective.

Proof. $f(a_1) = f(a_2) \Rightarrow g(f(a_1)) = g(f(a_2)) \Rightarrow (g\circ f)(a_1) = (g \circ f)(a_2) \Rightarrow a_1 = a_2$ by injectivity of $g \circ f$. Note $g$ itself need not be injective. $\blacksquare$

Functions are the objects of university mathematics: every structure; a group, a metric space, a measure; comes with the class of functions that preserve it. The vocabulary here (injective, surjective, bijective, inverse, composition) is the vocabulary of every definition that follows.

Direct proof. Assume $P$; derive $Q$ by a chain of implications.

Proof by contrapositive. To prove $P \Rightarrow Q$, prove $\neg Q \Rightarrow \neg P$. Used when the contrapositive gives more to work with: proving ”$n^2$ even $\Rightarrow n$ even” is easiest as ”$n$ odd $\Rightarrow n^2$ odd”, since “odd” has a usable form ($n = 2k+1$).

Proof by contradiction. Assume $P$ and $\neg Q$; derive any falsehood. The irrationality of $\sqrt{2}$ is the classic:

Proof. Suppose $\sqrt{2} = p/q$ in lowest terms. Then $p^2 = 2q^2$, so $p^2$ is even, so $p$ is even; write $p = 2k$. Then $4k^2 = 2q^2$, so $q^2 = 2k^2$, so $q$ is even too. Both even contradicts lowest terms. $\blacksquare$

Note the lemma it used; ”$p^2$ even $\Rightarrow p$ even”; which itself needs the contrapositive form above. Proofs are layered.

Proof by cases. Partition and prove each part: “even $n$” and “odd $n$”.

Proof by induction. To prove $S(n)$ for all $n \geq n_0$: prove $S(n_0)$ (base) and $S(n) \Rightarrow S(n+1)$ (inductive step).

Proof that it works (well-ordering principle). Suppose some $S(n)$ fails; let $n^*$ be the least such; it exists because $\mathbb{N}$ is well-ordered. Then $n^* > n_0$, so $S(n^* - 1)$ holds, so $S(n^*)$ holds by the inductive step - contradiction. $\blacksquare$

Worked example. Prove $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.

Base. $n = 1$: $1 = \frac{1 \cdot 2}{2}$. Step. Assume $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. Then

$$\sum_{k=1}^{n+1} k = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2},$$

which is the formula at $n+1$. $\blacksquare$

Proof by counterexample. To disprove $\forall x\,P(x)$, exhibit one $x$ with $\neg P(x)$. This is the only way to disprove a universal statement, and one counterexample ends the matter; which is why finding them is so valuable.

These are used silently in every proof. Fluency matters: a proof can fail on algebra.

Expansion and factorisation.

$$(a+b)^2 = a^2 + 2ab + b^2, \qquad (a-b)(a+b) = a^2 - b^2, \qquad a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2).$$

The quadratic formula and completing the square.

$$ax^2 + bx + c = 0 \iff x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$

obtained by completing the square; and completing the square is also how you find the vertex of a parabola and derive the canonical form of any quadratic in $x$ and $y$.

Proposition (AM–GM). For $a, b \geq 0$: $\frac{a+b}{2} \geq \sqrt{ab}$, with equality iff $a = b$.

Proof. $(a - b)^2 \geq 0 \iff a^2 - 2ab + b^2 \geq 0 \iff a^2 + b^2 \geq 2ab \iff \frac{a+b}{2} \geq \sqrt{ab}$ (adding $2ab$ and halving; taking positive square roots). Equality iff $(a-b)^2 = 0$. $\blacksquare$

Inequalities you must be able to manipulate.

• $x < y$ and $c > 0 \Rightarrow cx < cy$; but for $c < 0$ the inequality reverses. Dividing by a negative without flipping is the most common school-level error, and in analysis it proves false theorems.
• Triangle inequality: $|x + y| \leq |x| + |y|$, with equality iff $x, y$ have the same sign. Proof: $|x+y|^2 = (x+y)^2 = x^2 + 2xy + y^2 \leq x^2 + 2|xy| + y^2 = (|x|+|y|)^2$ using $xy \leq |xy|$.
• Reverse triangle: $\big||x| - |y|\big| \leq |x - y|$. Proof: apply the triangle inequality to $x = (x - y) + y$ and to $y = (y - x) + x$, then combine.

Proposition (Bernoulli). For $x > -1$ and $r \in \mathbb{N}$: $(1+x)^r \geq 1 + rx$.

Proof. Induction. Base $r = 1$: equality. Step: assume $(1+x)^r \geq 1 + rx$. Multiply both sides by the positive $(1+x)$: $(1+x)^{r+1} \geq (1+rx)(1+x) = 1 + (r+1)x + rx^2 \geq 1 + (r+1)x$. $\blacksquare$

Exponentials and logarithms. $a^x$ and $\log_a x$ are inverse functions; $\log(xy) = \log x + \log y$; $\log(x^r) = r \log x$. The natural logarithm uses $e = \lim_{n\to\infty}(1 + 1/n)^n$; a limit, justified rigorously in the real-analysis track. $\log$ of a negative number is undefined over the reals.

Definition (radian). An angle of $\theta$ radians subtends an arc of length $\theta$ on the unit circle. $\pi$ radians $= 180°$. All calculus assumes radians; $\frac{d}{dx}\sin x = \cos x$ is false in degrees, because the derivative of $\sin$ at $0$ is $\pi/180$-ish in degrees and exactly $1$ in radians.

Identities, proved not listed.

Pythagoras. $\sin^2\theta + \cos^2\theta = 1$: it is the equation of the unit circle, since $\cos\theta, \sin\theta$ are its coordinates.

Angle addition.

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta, \qquad \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.$$

Proof of the first. Take points $P = (\cos\alpha, \sin\alpha)$ and apply the rotation through $\beta$, whose matrix is $\begin{pmatrix}\cos\beta & -\sin\beta\\ \sin\beta & \cos\beta\end{pmatrix}$; the rotated coordinates are exactly the two formulas, since rotation by $\beta$ maps the unit circle to itself and sends the angle $\alpha$ to $\alpha + \beta$. $\blacksquare$

Even–odd. $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$: cosine is even, sine is odd; visible from the unit circle.

Double angle and the half-angle substitution. Setting $\alpha = \beta$: $\sin 2\theta = 2\sin\theta\cos\theta$, $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$. Rearranged: $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ - the identity every Fourier series computation uses.

Proposition (the fundamental limits). $\lim_{x\to 0}\frac{\sin x}{x} = 1$ and $\lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac12$.

Proof of the first (squeeze theorem). For $0 < x < \pi/2$, compare areas: $\triangle OAB \leq \text{sector} \leq \triangle OAC$ gives $\frac{1}{2}\sin x \leq \frac{x}{2} \leq \frac{1}{2}\tan x$, so

$$\cos x \leq \frac{\sin x}{x} \leq 1,$$

and since $\cos x \to 1$, the squeeze theorem (proved in the real-analysis track from the $\varepsilon$–$\delta$ definition) gives the limit. The second follows: $\frac{1 - \cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \to \frac{1}{2}$. $\blacksquare$

Definition (distance, midpoint, gradient). In $\mathbb{R}^2$, $d(P, Q) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ (Pythagoras); the gradient of the line through $P, Q$ is $m = \frac{y_1 - y_2}{x_1 - x_2}$.

Proposition. Lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are perpendicular iff $m_1 m_2 = -1$.

Proof. A line of gradient $m$ has direction vector $(1, m)$. Two direction vectors are perpendicular iff their dot product vanishes: $(1, m_1)\cdot(1, m_2) = 1 + m_1 m_2 = 0$. $\blacksquare$

Conic sections. The circle $x^2 + y^2 = r^2$; the parabola $y = ax^2 + bx + c$; the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$; the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Each is the slice of a cone, and each is a level set of a quadratic; which is why completing the square classifies them all.

Vectors and the dot product. $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$, so $\mathbf{u}\cdot\mathbf{v} = 0$ means perpendicular; the algebraic form of orthogonality that every later course generalises.

Definition (derivative). $f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$ when the limit exists.

The standard derivatives, with one proved.

Proof for $\sin$, the one needing trigonometry. From the definition and the angle-addition formula,

$$\frac{\sin(x+h) - \sin x}{h} = \sin x\,\frac{\cos h - 1}{h} + \cos x\,\frac{\sin h}{h} \to \sin x \cdot 0 + \cos x \cdot 1 = \cos x,$$

using the two fundamental limits of §6. The factor $h$ in the denominator and the radians convention are what make this come out clean; which is why the course uses radians. $\blacksquare$

Rules. Product: $(fg)' = f'g + fg'$. Quotient: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$. Chain: $(g \circ f)'(x) = g'(f(x))\,f'(x)$.

Proof of the chain rule (with the hypothesis stated). If $g$ is differentiable at $f(x)$ and $f$ at $x$, and the difference quotients of $f$ stay bounded away from zero where needed, then

$$\frac{g(f(x+h)) - g(f(x))}{h} = \frac{g(f(x+h)) - g(f(x))}{f(x+h) - f(x)} \cdot \frac{f(x+h) - f(x)}{h}.$$

The subtlety; the denominator $f(x+h) - f(x)$ can be zero for arbitrarily small $h$; is handled by writing the difference quotient of $g$ as a continuous function of the increment, which is why the honest proof defines $\psi(k) = \frac{g(f(x)+k) - g(f(x))}{k}$ for $k \neq 0$ and $\psi(0) = g'(f(x))$, making $\psi$ continuous. Then $g(f(x+h)) - g(f(x)) = \psi(f(x+h) - f(x))$ holds even when the increment is zero, and the limit factors. $\blacksquare$

Definition (integral). $\int_a^b f(x)\,dx$ is the limit of Riemann sums $\sum f(\xi_i)\,\Delta x_i$ as the mesh goes to zero.

The fundamental theorem of calculus. If $f$ is continuous and $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$; and $\int_a^b f = F(b) - F(a)$ for any antiderivative $F$.

Proof of the first part. $\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h}f$, and by continuity $f$ is within $\varepsilon$ of $f(x)$ on $[x, x+h]$ for small $h$, so the average is within $\varepsilon$ of $f(x)$. $\blacksquare$

The school results recovered. Integration by parts from the product rule - $(fg)' = f'g + fg'$ integrates to $\int fg' = fg - \int f'g$. Substitution from the chain rule; $\frac{d}{dx}g(f(x)) = g'(f(x))f'(x)$ read backwards. Both are theorems about antiderivatives, and both fail without the hypotheses, which is why the derivative table above was proved rather than listed.

Some school notation is discarded, and using it flags your work:

Proof hygiene. Never write $\Rightarrow$ between two statements you have not proved connect. Never use the theorem you are proving. State where every hypothesis is used. If a step says “clearly” and it is not, it is not clear.

Attempt these before track 1. Answers below; each links back to its section.

1. Write the negation of: “for every $\varepsilon > 0$ there is an $N$ such that for all $n \geq N$, $|a_n| < \varepsilon$”. (§1)
2. Prove that $f(x) = x^3$ is injective on $\mathbb{R}$. (§3)
3. Prove that $n^3 - n$ is divisible by $6$ for all $n \in \mathbb{N}$. (§4, cases + induction)
4. Show $x^2 > x$ for all $x > 1$, and find every $x$ with $x^2 < x$. (§5)
5. Prove $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$, stating where the formula fails. (§6)
6. Show the lines $y = 2x + 1$ and $x + 2y = 3$ are perpendicular. (§7)
7. Differentiate $x^x$ for $x > 0$, stating the rule used. (§8, chain + log)
8. Prove $\frac{d}{dx}\cos x = -\sin x$ using the fundamental limits. (§8)

Answers.

1. $\exists\varepsilon > 0\;\forall N\;\exists n \geq N : |a_n| \geq \varepsilon$.
2. $f(a) = f(b) \Rightarrow a^3 = b^3 \Rightarrow (a - b)(a^2 + ab + b^2) = 0$; $a^2 + ab + b^2 = (a + b/2)^2 + \frac{3b^2}{4} \geq \frac{3b^2}{4} > 0$ unless $a = b = 0$, so $a = b$.
3. $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers, so one is even and one is a multiple of 3.
4. $x^2 > x \iff x(x-1) > 0 \iff x > 1$ or $x < 0$; $x^2 < x \iff 0 < x < 1$.
5. Divide $\sin(\alpha+\beta)$ by $\cos(\alpha+\beta)$, then divide numerator and denominator by $\cos\alpha\cos\beta$; it fails when $\tan\alpha\tan\beta = 1$, i.e. when $\alpha + \beta = \pi/2$ and the tangent is undefined.
6. Gradients $2$ and $-1/2$: product $-1$.
7. $x^x = e^{x\ln x}$, so $\frac{d}{dx} = e^{x\ln x}(\ln x + 1) = x^x(\ln x + 1)$; chain rule with $f = \ln$ and $g = e^{(\cdot)}$.
8. $\frac{\cos(x+h) - \cos x}{h} = \cos x\,\frac{\cos h - 1}{h} - \sin x\,\frac{\sin h}{h} \to -\sin x$.

• Complex Numbers Review; the first page of the complex-analysis track, which extends this page’s algebra.
• Introduction to Topology; the $\varepsilon$–$\delta$ definitions of §1 made into a subject.
• Smooth Manifolds; where the coordinate geometry of §7 becomes a theory.
• Measure-Theoretic Probability - where §2’s countability distinction decides which theorems hold.

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