4.6 Decimal expansions and irrational numbers

Once Dedekind cuts are available, return to the familiar language of decimal expansions and ask a natural question:

How does a decimal such as $10.4352902543...$ determine a real number?

The answer is that a decimal expansion produces a nested family of rational approximations, and those approximations can be turned into a cut.

The structural boundary of the construction

The uniqueness theorem explains why the completed cut model represents the familiar real number system:

Theorem

There is one complete ordered field up to isomorphism

Any two complete ordered fields are isomorphic by an order-preserving field isomorphism. Thus a construction that satisfies the complete ordered-field axioms produces the real number system in the structural sense.

This theorem is a stated course boundary. The present notes use it to explain why different rigorous constructions describe the same real system, but do not prove the uniqueness theorem in this course.

Decimal expansions give lower and upper fences

Consider the decimal

$$x=10.4352902543\ldots$$

Form the sequence of lower bounds

$$S=\{10,\ 10.4,\ 10.43,\ 10.435,\ \ldots\}$$

and the sequence of upper bounds

$$T=\{11,\ 10.5,\ 10.44,\ 10.436,\ \ldots\}.$$

The idea is simple:

• the lower list truncates the decimal and stays below the target,
• the upper list moves one step above the truncation and stays above the target.

So the real number is trapped inside narrower and narrower rational intervals.

Worked example

Nested intervals for $\sqrt{2}$

Even without committing to a particular decimal such as $10.435...$, the same method can be seen clearly with $\sqrt{2}$:

$$1\lt \sqrt{2}\lt 2,$$$$1.4\lt \sqrt{2}\lt 1.5,$$$$1.41\lt \sqrt{2}\lt 1.42,$$$$1.414\lt \sqrt{2}\lt 1.415.$$

Each extra digit shrinks the interval and gives a better rational fence.

Figure. A decimal expansion does not merely list digits. It creates a chain of smaller and smaller rational intervals containing the target number.

Turning a decimal expansion into a cut

For an integer part $k\in Z$ and digits $d_j\in\{0,\ldots,9\}$, put

$$t_n=k+\sum_{j=1}^n d_j10^{-j},\qquad u_n=t_n+10^{-n},\qquad n\ge0.$$

The sum is empty at $n=0$. Let $S=\{t_n:n\ge0\}$ and $T=\{u_n:n\ge0\}$. Finite place-value arithmetic shows the lower fences are nondecreasing, the upper fences are nonincreasing, and $t_m\lt u_n$ for all finite $m,n$. Define

$$A=\{q\in Q:\exists s\in S,\ q\lt s\},\qquad B=Q\setminus A=\{q\in Q:\forall s\in S,\ s\le q\}.$$

Thus $B$ is the set of rational upper bounds of all lower fences. It is not the set of rationals strictly above some upper fence: that rule would lose a rational boundary such as $1.25000\ldots$ from both sides.

The set $A$ is nonempty since $k-1\lt t_0$; it is proper since $u_0=k+1$ is above every truncation. If $q\lt t_n$ and $y\lt q$, then $y\lt t_n$, giving downward closure. Finally $(q+t_n)/2$ is a larger rational still below $t_n$, so $A$ has no greatest element. Consequently $(A,B)$ is a cut, including for terminating decimals and tails of nines. The mesh $10^{-n}$ can be made arbitrarily small: choose $n\gt 1/\varepsilon$ and use $10^n\ge n$. This justifies the shrinking-fence interpretation without assuming an infinite string already denotes a real number.

Why decimal strings are useful but not the primary definition

The previous note already hinted at the main problem with taking decimal strings as the primary definition of real numbers:

• the same real number can have more than one decimal expansion,
• for example $0.999...=1$.

So decimal notation is excellent for intuition and approximation, but it needs a deeper structural interpretation. Dedekind cuts supply that interpretation.

Common mistake

A finite truncation is not the real number itself

The decimal $1.414$ is not $\sqrt{2}$. It is only a rational approximation from below. Likewise $1.415$ is not the exact number either; it is an upper fence. The real number is the boundary captured by the whole infinite approximation process.

Refining the rational interval

Use the builder below to reveal one more decimal digit at a time and watch the lower bound, upper bound, and interval width update together.

Irrational numbers inside $R$

Once the real numbers are constructed, define:

Definition

Irrational number

An irrational number is an element of $R\setminus Q$.

This definition is short, but its meaning is deep. An irrational number is not “mysterious” or “unfinished”. It is a perfectly legitimate real number that is simply not represented by any rational cut.

The real number $\sqrt{2}$

Set

$$C=\{r\in Q:r\le0\text{ or }r^2\lt 2\},\qquad Q\setminus C=\{r\in Q:r\gt 0\text{ and }r^2\gt 2\}.$$

The equality in the complement uses the earlier proof that no rational square is $2$. The set $C$ is nonempty ($1\in C$) and proper ($2\notin C$). It is downward closed: below a nonpositive member everything is included; below a positive member, a positive rational has a smaller square. A nonpositive member is exceeded by $1$, and a positive member is exceeded by the rational perturbation from 4.3. Thus $C$ has no greatest element and is a cut. It is strictly positive since it contains $0_R$ and also $1$.

The cut really satisfies C times C equals 2_R

Use the nonnegative product definition from 4.5. A negative rational is in $2_R$. For nonnegative $a,b\in C$, if $a\le b$, then $ab\le b^2\lt 2$; the other order is symmetric. Hence $C\cdot C\subseteq2_R$.

For the reverse inclusion, negative rationals are already included. Given $z\in Q$ with $0\le z\lt 2$, choose a rational $0\lt \delta\lt \min\{1,(2-z)/5\}$. Apply the rational stepping argument from 4.5 starting at $1\in C$: there are $a\in C$ and $b=a+\delta\notin C$, with $a\ge1$. Then $a\lt 2$, $b\lt 3$, and $b^2\gt 2$, so

$$2-a^2\lt b^2-a^2=\delta(a+b)\lt 5\delta\lt 2-z.$$

Thus $z\lt a^2$. Put $c=z/a$; then $0\le c\lt a$, so $c\in C$ and $z=ac$ belongs to $C\cdot C$. Therefore $C\cdot C=2_R$.

Worked example

Why this is an irrational cut

If $C=q_R$, its boundary must have $q\gt 0$. If $q^2\lt 2$, then $q\in C$, contradicting $q\notin q_R$. If $q^2\gt 2$, the downward perturbation from 4.3 gives a positive rational $r\lt q$ with $r^2\gt 2$, so $r\in q_R$ but $r\notin C$. Equality $q^2=2$ is impossible in $Q$. Hence $C$ is not a rational cut. We call the positive cut with the displayed square $\sqrt{2}$.

What decimals and irrationals are teaching you together

There are two compatible claims here.

1. A real number can be approximated arbitrarily well by rationals.
2. Some real numbers are nevertheless not rational.

Those statements do not conflict. In fact, they are exactly what makes the real number system powerful. Irrational numbers can be reached by rational approximations without ever turning into rational numbers themselves.

Decimal nonuniqueness and periodicity

Decimal notation is a representation, so it must be tested for uniqueness before it is used as a definition. The fundamental example is

$$0.999\ldots=1.$$

For the truncations $0.9$, $0.99$, $0.999$, and so on, the gap to $1$ is $10^{-n}$ after $n$ digits. Given any $\varepsilon\gt 0$, choose $n$ with $10^{-n}\lt \varepsilon$. More precisely, $1$ bounds all truncations; for every $u\lt 1$, choose $n$ with $10^{-n}\lt 1-u$, so $u\lt 1-10^{-n}$. Hence $1$ is their supremum, and the cut is exactly $1_R$. A decimal convention must therefore exclude tails of all $9$s if a unique string representation is desired.

Theorem

Eventually periodic decimals are rational

If a decimal has a repeating block of length $k$ after $m$ initial digits, then it represents a rational number. Multiplying by $10^{m+k}$ and by $10^m$ shifts the repeating tail into the same position; subtracting cancels the infinite tail and leaves an integer equation with a nonzero integer coefficient. Solving that equation gives a quotient of integers.

Theorem

A terminating decimal has a periodic alternative

Every positive terminating decimal is rational and can also be written with a tail of $9$s. For instance, $1.25000\ldots=1.24999\ldots$. This follows by applying the same finite-place-value calculation behind $0.999\ldots=1$; it is another reason to treat decimal strings as names whose equality must be justified.

Worked example

Converting $0.\overline{27}$ to a fraction

Let $x=0.272727\ldots$. Then $100x=27.272727\ldots$, so subtraction gives $99x=27$, hence $x=27/99=3/11$. The argument uses only finite subtraction after aligning the repeated block; it does not claim that every infinite decimal is periodic.

Worked example

Why a nonrepeating decimal can still be real

The decimal expansion of $\sqrt{2}$ is not eventually periodic, but its finite truncations and one-step upper fences form nested rational intervals. Their widths tend to zero by the Archimedean estimate, and the associated cut gives one real boundary. Irrationality means that this boundary is not a rational element, not that it fails to be a real number.

The converse: rationals produce periodic decimals

The cancellation argument proves “eventually periodic implies rational.” The converse is also elementary. Write a rational number in lowest terms as $a/b$, with $b\gt 0$. Long division repeatedly records a remainder in $\{0,1,...,b-1\}$. If a remainder becomes zero, the decimal terminates. If not, some remainder repeats because there are only finitely many possibilities. The digits from the first occurrence of that remainder then repeat forever, so $a/b$ has an eventually periodic decimal expansion.

Theorem

Rationality and eventual periodicity

A real number represented in base ten is rational exactly when its decimal expansion is terminating or eventually periodic, after identifying the two representations such as $0.999\ldots$ and $1$.

Reducing a general real number to its fractional part

For an arbitrary $x\in R$, the Archimedean property gives an integer $k$ with $k\le x<k+1$. Total order makes this integer part unique: if both $k$ and $\ell$ work, then neither can be strictly smaller than the other. Put $y=x-k$, so $0\le y<1$. Construct the decimal digits of $y$ and then add $k$ back to every truncation. This reduces the general case to the unit interval while preserving the same error width.

Worked example

Constructing fractional digits by floors

For $y\in[0,1)$ and $n\ge1$, define $d_n=\lfloor10^n y\rfloor-10\lfloor10^{n-1}y\rfloor$. Then $d_n\in\{0,\ldots,9\}$ and the truncation $t_n=\lfloor10^n y\rfloor/10^n$ satisfies $t_n\le y\lt t_n+10^{-n}$. Therefore $k+t_n\le x<k+t_n+10^{-n}$; the digits construct nested rational fences for every real $x$, with an explicit error bound independent of any picture of an infinite string.

The digit formula needs two checks. Put $m=\lfloor10^{n-1}y\rfloor$. Then $m\le10^{n-1}y<m+1$, so $10m\le10^ny<10m+10$. Taking floors gives $10m\le\lfloor10^ny\rfloor\le10m+9$; subtracting $10m$ proves that $d_n$ is one of the ten permitted digits. Next the weighted digit sum telescopes:

$$\sum_{j=1}^n d_j10^{-j} =10^{-n}\lfloor10^ny\rfloor-\lfloor y\rfloor =10^{-n}\lfloor10^ny\rfloor=t_n.$$

The last equality uses $0\le y<1$. Thus the digit construction really recovers the lower fences, rather than merely producing a list of allowed digits. For negative $x$, the formula $x=k+y$ is an integer-plus-fraction decomposition: for example $-1.25=-2+0.75$. It must not be read as concatenating the minus sign of $k$ with the digits of $y$.

Worked example

A rational but nonterminating expansion

The fraction $1/3$ has no terminating decimal because a terminating decimal has denominator, after reduction, containing only factors $2$ and $5$. Long division gives $0.333\ldots$; the repeated remainder $1$ explains the period.

Common mistake

Nonterminating does not mean irrational

$1/3=0.333\ldots$ is rational because the digits are eventually periodic. Irrationality requires failure of eventual periodicity (or an independent proof that the real is not in $Q$), not merely infinitely many digits.

Quick checks

Checkpoint

For the terminating decimal $1.25000\ldots$, which side of the cut contains $1.25$: $A$ or $B$? Explain from the definitions.

Use all lower fences, including the ones that equal the boundary.

Solution · Answer

Every lower fence is at most $1.25$, so $1.25$ belongs to $B$, the set of rational upper bounds of all lower fences. It does not belong to $A$: no lower fence is strictly greater than $1.25$. This is why defining $B$ as the complement of $A$ handles a rational boundary correctly.

Exercises

Checkpoint

Write the first four lower and upper rational fences suggested by the decimal $3.14159...$.

Begin with the integer interval and then reveal one more digit at each step.

Solution · Guided solution

One possible chain is

$$3\lt x\lt 4,$$$$3.1\lt x\lt 3.2,$$$$3.14\lt x\lt 3.15,$$$$3.141\lt x\lt 3.142.$$

Each step traps the target in a narrower rational interval.

Checkpoint

Starting from the informal decimal $2.718...$, write four lower fences, four upper fences, and the corresponding cut $(A,B)$.

Follow the same construction used above.

Solution · Guided solution

One valid choice is

$$S=\{2,\ 2.7,\ 2.71,\ 2.718,\ldots\},\qquad T=\{3,\ 2.8,\ 2.72,\ 2.719,\ldots\}.$$

Then

$$A=\{q\in Q:\exists s\in S,\ q<s\},\qquad B=Q\setminus A=\{q\in Q:\forall s\in S,\ s\le q\}.$$

The exact later digits do not affect the argument: lower fences increase, upper fences decrease, and the two families enclose one boundary.

Checkpoint

Why does the cut $C=\{r\in Q:r\le0\text{ or }r^2\lt2\}$ represent an irrational number?

Use what was already proved about rational squares.

Solution · Guided solution

If the cut were rational, then its boundary would be some rational number $q$ with $q^2=2$. But earlier results show that no rational has square $2$. Therefore the cut cannot be rational, so the corresponding real number is irrational.

Checkpoint

If the first $k$ decimal digits of a cut are already known, how should the next digit be selected?

Describe the rule in terms of the embedded rational cuts.

Solution · Guided solution

Test the ten candidates $0,1,\ldots,9$ in the next place. Append each candidate to the current truncation and choose the largest candidate $q$ whose embedded cut satisfies $q_R\le A$. That candidate is the next lower fence; the matching upper fence is one place-unit larger.

Related notes

Read this after 4.5 Dedekind cuts and the embedding of Q and 3.5 Gaps in Q and why sqrt(2) is not rational. Then continue to 5.1 Sequences and epsilon-N limits.