The Real Number System

You have used these numbers for years. This topic asks a different question: which rules are we assuming, and what can be proved from them. That shift — from calculating to justifying — is what CAPE is testing.

Sets of numbers

Each set contains the one above it:

$$\mathbb{N} \subset \mathbb{W} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$$

The irrational numbers are the reals that are not rational — $\sqrt2$, $\pi$, $e$. They fill in the gaps that $\mathbb{Q}$ leaves.

The field axioms

For all $a, b, c \in \mathbb{R}$:

and the one that links them, the distributive law:

$$a(b+c) = ab + ac$$

Exam questions name the axiom being used. In $3(x + 4) = 3x + 12$ the step is distributive; in $2 + x = x + 2$ it is commutative; in $x \times 1 = x$ it is the multiplicative identity.

Closure

A set is closed under an operation if performing it on members of the set always lands you back inside the set.

$\mathbb{N}$ fails under subtraction because $3 - 5 = -2$, which is not a natural number. $\mathbb{Z}$ fails under division because $3 \div 2 = 1.5$, which is not an integer.

The irrationals are not closed under any of the four operations:

$$\sqrt2 + (-\sqrt2) = 0 \qquad \sqrt2 \times \sqrt2 = 2$$

Both results are rational, from irrational inputs.

Rational and irrational numbers

A number is rational exactly when its decimal expansion terminates or recurs.

$$0.75 = \frac34 \qquad 0.\dot3 = \frac13 \qquad 0.\dot4 = \frac49$$

To convert a recurring decimal, let $x$ be the number, multiply by a power of 10 that shifts one full period, and subtract.

Express $0.\dot1\dot8$ as a fraction.

$$x = 0.181818\ldots$$ $$100x = 18.181818\ldots$$ $$99x = 18$$ $$x = \frac{18}{99} = \frac{2}{11}$$

An irrational number has a decimal expansion that never terminates and never settles into a repeating block.

Proving irrationality

The standard argument is by contradiction.

Prove that $\sqrt2$ is irrational.

Suppose, for contradiction, that $\sqrt2$ is rational. Then

$$\sqrt2 = \frac{p}{q}$$

for integers $p$ and $q$ with no common factor — the fraction is in lowest terms.

Squaring and rearranging:

$$2 = \frac{p^2}{q^2} \quad \Rightarrow \quad p^2 = 2q^2$$

So $p^2$ is even, and therefore $p$ is even (an odd number squared is odd). Write $p = 2m$:

$$(2m)^2 = 2q^2 \quad \Rightarrow \quad 4m^2 = 2q^2 \quad \Rightarrow \quad q^2 = 2m^2$$

So $q^2$ is even, and therefore $q$ is even.

But now $p$ and $q$ are both even, so they share the factor 2 — contradicting the assumption that the fraction was in lowest terms.

The assumption must be false, so $\sqrt2$ is irrational. $\blacksquare$

Working with the axioms

Two results worth knowing, both proved the same way.

The sum of a rational and an irrational is always irrational.

Suppose $r$ is rational and $i$ is irrational, and suppose for contradiction that $r + i = s$ is rational. Then

$$i = s - r$$

But $\mathbb{Q}$ is closed under subtraction, so $s - r$ is rational — making $i$ rational, which contradicts the assumption. Therefore $r + i$ is irrational. $\blacksquare$

The sum of two irrationals can be either.

$$\sqrt2 + \sqrt3 \text{ is irrational} \qquad \sqrt2 + (-\sqrt2) = 0 \text{ is rational}$$

So no general rule exists, and a question asking for one is asking you to notice that.