Calculus rests on one quiet assumption: that the number line has
no holes. The
idea of a limit
only makes sense if the value a sequence is "closing in on" actually
exists. The rationals \mathbb{Q} are not enough — they
are riddled with gaps. The real numbers \mathbb{R} are exactly
\mathbb{Q} with every gap filled.
First, what \mathbb{R} shares with \mathbb{Q}:
both are ordered fields. You can add, subtract, multiply and divide (a
field), and there is a consistent order \le
compatible with that arithmetic (an ordered field). What sets
\mathbb{R} apart is one extra property — completeness
— and the whole point of this page is to see, concretely, why \mathbb{Q}
fails it.
A gap you can name: \sqrt{2}
The diagonal of a unit square has length \sqrt{2}, a number whose
square is 2. It sits somewhere on the line — between
1.41 and 1.42, between
1.414 and 1.415, and so on forever. Yet
no fraction lands on it. We prove this by contradiction.
Step 1 — assume the opposite. Suppose
\sqrt{2} were rational. Then we could write it as a
fraction in lowest terms,
\sqrt{2} = \frac{p}{q}, \qquad p, q \in \mathbb{Z},\ q \ne 0,
where p and q share no common factor
(any fraction can be reduced to this form, so this costs us nothing).
Step 2 — clear the square root. Square both sides and multiply through by
q^2:
2 = \frac{p^2}{q^2} \quad\Longrightarrow\quad p^2 = 2q^2.
Step 3 — read off that p is even. The right-hand
side 2q^2 is 2 times an integer, so
p^2 is even. But the square of an odd number is
odd ((2k+1)^2 = 4k^2+4k+1), so for
p^2 to be even, p itself must be even.
Write p = 2k.
Step 4 — substitute and simplify. Put
p = 2k back into p^2 = 2q^2:
(2k)^2 = 2q^2 \quad\Longrightarrow\quad 4k^2 = 2q^2 \quad\Longrightarrow\quad q^2 = 2k^2.
Step 5 — read off that q is even too. By exactly
the same reasoning as Step 3, q^2 = 2k^2 is even, so
q is even.
Step 6 — the contradiction. We have shown both
p and q are even — they share the
factor 2. But Step 1 fixed the fraction in lowest terms,
with no common factor. This is impossible. The only faulty link in the chain was the
assumption that \sqrt{2} is rational, so that assumption is false:
\sqrt{2} \notin \mathbb{Q}.
-
There is no rational number r with
r^2 = 2. Equivalently,
\sqrt{2} \notin \mathbb{Q}. So the equation
x^2 = 2 has a solution on the number line but
not in \mathbb{Q} — a genuine gap.
-
\mathbb{R} is a complete ordered field: an
ordered field in which every nonempty set bounded above has a least upper bound
(its supremum)
lying in \mathbb{R}. This single extra axiom — completeness —
is what fills every gap, including the one at \sqrt{2}, and it
characterises \mathbb{R} uniquely.
Watch the gap survive any zoom
The rationals are dense: between any two of them lies another, so they pack
the line with no visible holes. Step through the figure to scatter rationals just below
\sqrt{2} (where (p/q)^2 < 2) and just
above it (where (p/q)^2 > 2). They crowd in from both sides — yet
the point \sqrt{2} itself, marked in the centre, is claimed by
none of them. Refresh to redraw at a fresh random zoom level: however far you
magnify, the gap is always there.
Calling \mathbb{R} a field is shorthand for a
short list of axioms governing + and
\times: both are commutative and associative, multiplication
distributes over addition, there are identities 0 and
1, every element has an additive inverse, and every
nonzero element has a multiplicative inverse. \mathbb{Q}
satisfies all of these too — as do the complex numbers.
Ordered adds a relation \le that is total and
plays nicely with arithmetic: if a \le b then
a + c \le b + c, and if also
0 \le c then ac \le bc. This is why
\mathbb{C} is not an ordered field — you cannot order it
compatibly with multiplication. Both \mathbb{Q} and
\mathbb{R} are ordered fields; the watershed between them is the
next property.
Density and completeness are different things, and \mathbb{Q}
shows they come apart. Dense means: between any two rationals lies a third
— concretely, \tfrac{a+b}{2}. So the rationals leave no
interval-sized hole; you can approximate any real to any precision with a fraction.
But the set
S = \{\, x \in \mathbb{Q} : x > 0,\ x^2 < 2 \,\}
is nonempty and bounded above (by 2, say), yet has
no least upper bound in \mathbb{Q}: any rational
upper bound can be nudged a little smaller and still bound S,
because the would-be boundary \sqrt{2} is missing. In
\mathbb{R} that least upper bound exists and equals
\sqrt{2}. Filling such gaps — guaranteeing every bounded set its
boundary point — is exactly the
completeness axiom,
first made precise by
Dedekind and
Cauchy.