3.5 Gaps in Q and why sqrt(2) is not rational

Division by nonzero elements is possible in $Q$, but solving an equation can still require something the system does not contain. Consider

$$q^2=2.$$

Rational arithmetic gives brackets such as

$$1.4^2\lt2\lt1.5^2,\qquad 1.41^2\lt2\lt1.42^2.$$

These calculations compare rational squares. They do not yet produce a number whose square is $2$, nor do they establish that such a number exists in another system. The obstruction can be described entirely within $Q$: a nonempty set of rationals has upper bounds but no least rational upper bound.

Density and completeness ask different questions

Between two rationals $a<b$, the midpoint $(a+b)/2$ is another rational and lies strictly between them. Thus $Q$ is dense in itself as an ordered set. This statement starts with two numbers that already exist in $Q$.

The least-upper-bound form of completeness requires that every nonempty subset bounded above in the ambient ordered set have a least upper bound in that same ordered set. It concerns all the elements of a subset at once; finding another rational between two given rationals does not supply such a bound. The equation $q^2=2$ exposes this difference.

The key order-language

Definition 3.5.1

(Upper bound and supremum)

Let $X$ be a subset of an ordered set $F$.

• An upper bound of $X$ in $F$ is an element $u\in F$ such that $x \le u$ for every $x \in X$.
• A supremum of $X$, written $\sup(X)$, is the least upper bound of $X$.

The word "least" is crucial. A supremum is not just any upper bound. It is the smallest number that still stays above the whole set.

The set that exposes the gap

Theorem 3.5.2

(No rational square equals two)

There is no rational number $s$ with $s^2=2$.

The classic set is

$$S = \{q \in Q \mid q^2 \lt 2\}.$$

This set collects all rational numbers whose square is still less than $2$.

The numbers $1$, $1.4$, and $1.41$ belong to $S$, whereas $2$ and $3/2$ are upper bounds. A least upper bound would have to separate all members of $S$ from all its upper bounds. Naming such a rational separator does not show that it exists; every proposed value must withstand both requirements.

Why $S$ is bounded above in $Q$

The element $1$ belongs to $S$, so $S$ is nonempty. An upper bound can also be verified without knowing a solution of $q^2=2$.

Worked example 3.5.3

(Finding an upper bound for $S$)

Take $u = 2$.

If $q\in S$ and $q>2$, then $q^2>4>2$, a contradiction. Hence every member of $S$ is at most $2$, so $2$ is an upper bound. It is not itself a member, since $2^2=4>2$.

More generally, any rational number $u$ with $u>0$ and $u^2>2$ is an upper bound for $S$. The positivity condition is essential: a negative number can have square greater than $2$ without lying above the positive elements of $S$.

So the issue is not that $S$ has no upper bounds. The issue is that among its rational upper bounds, there is no least one.

Why no rational square equals two

Suppose, for contradiction, that a rational $s$ satisfies $s^2=2$. Write $s=p/q$ with integers $p,q$, $q\ne0$, and the fraction in lowest terms. Substitution gives

$$p^2 = 2q^2.$$

So $p^2$ is even, which forces $p$ itself to be even. Write $p = 2k$.

Substituting back gives

$$4k^2 = 2q^2,$$

so

$$q^2 = 2k^2.$$

Therefore $q^2$ is even, and $q$ is even as well.

Now both $p$ and $q$ are even, which contradicts the assumption that $p/q$ was already in lowest terms.

Therefore no rational $s$ satisfies $s^2=2$. This proves a restriction on $Q$; it does not construct a solution outside $Q$. To show that $S$ has no rational supremum, we must also exclude candidates whose squares are less than or greater than $2$.

The formal gap statement

Theorem 3.5.4

(The set $S = \{q \in Q \mid q^2\lt2\}$ has no supremum in $Q$)

The set $S$ is bounded above in $Q$, but there is no rational number that serves as its least upper bound.

The epsilon argument behind the gap

The phrase “move a little” needs a quantitative proof. The following argument uses only rational arithmetic and the fact that every positive rational has a smaller positive rational.

Proof that no rational candidate can be the supremum

Let $s\in Q$.

If $s<1$, then $1\in S$ and $1>s$, so $s$ is not an upper bound. It remains to consider $s\ge1$, separating the three possibilities for $s^2$.

Case 1: the square is less than two. Suppose $s^2<2$. Put $M=2-s^2>0$ and choose a positive rational $\varepsilon$ with

$$\varepsilon\lt2s \qquad\text{and}\qquad \varepsilon\lt\frac{M}{4s}.$$

Then

$$(s+\varepsilon)^2=s^2+\varepsilon(2s+\varepsilon) \lt s^2+\varepsilon(4s)\lt s^2+M=2.$$

Thus $s+\varepsilon\in S$, so $s$ cannot be an upper bound.

Case 2: the square is greater than two. Suppose $s^2>2$, still with $s\ge1$. Put $M=s^2-2$ and choose a positive rational $\varepsilon$ with

$$\varepsilon\lt s \qquad\text{and}\qquad \varepsilon\lt\frac{M}{4s}.$$

Set $r=s-\varepsilon$. Then $r>0$ and

$$r^2=s^2-\varepsilon(2s-\varepsilon)\gt s^2-2s\varepsilon\gt2.$$

If $x\in S$ and $x\le0$, then $x$ is less than $r$. If $x>0$ and $x>r$, then $x^2>r^2>2$, contradicting $x\in S$. Hence every $x\in S$ satisfies $x\le r$. So $r$ is an upper bound smaller than $s$, and $s$ is not the least upper bound.

Case 3: the square equals two. The remaining possibility $s^2=2$ would make $s$ a rational solution of $x^2=2$, which the parity proof ruled out. Thus every rational candidate fails.

Worked example 3.5.5

(A rational sequence can have a rational supremum)

Consider the sequence

$$f(n)=1+\frac12+\frac1{2^2}+\cdots+\frac1{2^n}.$$

The finite geometric-sum identity gives

$$f(n)=2-\frac1{2^n}.$$

Every $f(n)$ is below $2$, so $2$ is an upper bound in $Q$. Given any rational number $u$ less than $2$, write $2-u=A/B$ with positive integers $A,B$. Choose a positive integer $n>B/A$; for example, $n=B+1$ works since $A\ge1$. Induction gives $2^n>n$ for every positive integer: $2>1$, and $2^{n+1}>2n\ge n+1$ completes the step. Hence $1/2^n<A/B=2-u$, so $f(n)>u$. Thus no smaller rational is an upper bound, and

$$\sup f(N)=2.$$

Rational partial sums therefore do not automatically create a gap: in this example the least upper bound already belongs to $Q$. For $S$, the obstruction was established by excluding every rational candidate, not merely by displaying an endless process of approximation.

Quick check

(In the case where $s^2$ is greater than $2$, why must the smaller candidate $r=s-\varepsilon$ remain positive?)

Use the sign condition needed when comparing squares.

Solution · Answer

The comparison of squares is order-preserving for nonnegative numbers. Choosing $\varepsilon$ less than $s$ ensures $r>0$, so from a positive $x>r$ we may infer $x^2>r^2$. Without that sign check, comparing squares would not control the original order.

Supremum language must name the ambient set

The same subset can have a supremum in one ordered set and fail to have one in another. The geometric sequence above has supremum $2$ in $Q$. If the ambient set is restricted to $\{q\in Q:q<2\}$, the same sequence has no upper bound: the leastness calculation supplies a term above every candidate in that set. For $S$, a later construction will enlarge the ambient system and supply a supremum. Its existence will be a theorem about that construction; it is not assumed in the rational argument above.

Worked example 3.5.6

(An infimum need not be a minimum)

For $Y=Q_{>0}$, zero is a lower bound in $Q$, and every positive rational is larger than zero. If $q>0$, then $q/2$ is another positive rational smaller than $q$, so $Y$ has no minimum. Also, any proposed positive lower bound $u$ is excluded by $u/2\in Y$, which is smaller than $u$. Thus zero is the greatest lower bound: $0=\inf(Y)$ in $Q$, even though $0\notin Y$. This is the same distinction used for the supremum of $S$: least or greatest refers to bounds in the ambient order, not membership in the subset.

A gap inside every rational interval

The same obstruction can be placed inside any nonempty rational open interval. Let $a,b\in Q$ with $a<b$, and define

$$P=\{q\in Q\mid 0<q\text{ and }q^2\lt2\}, \qquad X=\left\{a+\frac{b-a}{2}q\mathrel{\Big|}q\in P\right\}.$$

For $q\in P$, we have $0<q<2$, so $a<a+(b-a)q/2<b$; hence $X\subset(a,b)$.

The set $P$ has no supremum in $Q$: a candidate $t\le0$ is defeated by $1\in P$; if $t>0$ and $t^2\lt2$, the right-perturbation proof above supplies a larger element of $P$ (with $1\in P$ handling $t<1$); if $t^2>2$, the left-perturbation proof supplies a smaller upper bound; and equality is impossible by the parity argument.

If $X$ had a rational supremum $u$, put $v=2(u-a)/(b-a)\in Q$. For every $q\in P$, the upper-bound inequality $a+(b-a)q/2\le u$ implies $q\le v$, because $b-a>0$. Thus $v$ bounds $P$.

Conversely, if $w\in Q$ bounds $P$, then $a+(b-a)w/2$ bounds $X$. The leastness of $u$ gives $u\le a+(b-a)w/2$, hence $v\le w$. Therefore $v$ would be a rational supremum of $P$, a contradiction. This constructs a rational subset of every such interval whose supremum does not exist in $Q$.

Worked applications: parity, gcd, and suprema

Worked example: no rational square equals $2026$

If $x=p/q\in Q$ satisfied $x^2=2026$, choose integers $p,q$ with $q\ne0$ and $\gcd(p,q)=1$. Then $p^2=2026q^2$, so $p$ is even; writing $p=2k$ gives $2k^2=1013q^2$, which forces $q$ to be even. This contradicts $\gcd(p,q)=1$. The parity fact used here is elementary: an odd integer has odd square, so an even square has an even root.

Worked example: the gcd forced by $5x+7y=1$

Let $x,y\in Z$ and $5x+7y=1$. Any positive common divisor $d$ of $x$ and $y$ divides $5x+7y=1$, so $d=1$; therefore $\gcd(x,y)=1$.

Worked example: rule out rational roots of a quadratic

Consider the equation $x^2+2x=1$. Completing the square gives $(x+1)^2=2$. If $x$ were rational, then $x+1$ would be rational, contradicting the proof that no rational square is $2$.

Worked example: a repeating-decimal supremum in $Q$

Let $X=\{1.23,1.233,1.2333,\ldots\}$: each element has finitely many digits $3$ after the initial $1.2$. Writing $x_n=1.2+\sum\limits_{k=2}^{n}3/10^k$ for $n\ge2$, we have

$$x_n=\frac{37}{30}-\frac{1}{3\cdot10^n}, \qquad \frac{37}{30}=1.23333\ldots\in Q.$$

Thus $37/30$ is an upper bound. If $u<37/30$ is rational, write $37/30-u=A/B$ with positive integers $A,B$ and take $n=B+2$. Then $n\ge2$, $n>B/A$, and $10^n\ge2^n>n$, so $1/(3\cdot10^n)<A/B=37/30-u$. Hence $x_n>u$. Therefore $\sup\limits_Q(X)=37/30$.

Common mistake

Common mistake

(Dense does not mean complete)

Between any two distinct rational numbers there is another rational. This only concerns two existing rationals; it does not imply that every nonempty set of rationals bounded above has a rational supremum.

Another common mistake is to think that a supremum must belong to the set itself. That is false. A supremum only has to be the least upper bound in the ambient ordered set.

Quick checks

Quick check

(Why is a rational number whose $s^2$ is less than $2$ not an upper bound for $S$?)

Use the idea that you can move a little to the right and still keep the square below $2$.

Solution · Answer

Because such an $s$ is still too small. There exists a rational $r > s$ with $r^2$ less than $2$, so $r$ also belongs to $S$. That means $s$ cannot be above the whole set.

Quick check

(Does the supremum of a set have to belong to the set?)

Answer in one sentence.

Solution · Answer

No. A supremum only has to be the least upper bound. It may lie outside the set itself.

Exercise

Quick check

(Why does density of $Q$ fail to give the set $S$ a rational supremum?)

Use the difference between density and completeness.

Solution · Guided solution

Density supplies a rational between any two given rationals, for example their midpoint. A supremum must instead bound every member of $S$ and be no larger than any other rational upper bound. The proof excludes every candidate: values below $1$ are not upper bounds; for a candidate at least $1$, a square less than $2$ allows a larger member of $S$, a square greater than $2$ allows a smaller upper bound, and equality is impossible. Density does not remove any of these obstructions.

Optional continuations

The following advanced comparison uses the real and complex number systems; it is not a premise of the rational gap proof.

Read this first

If you want the construction of $Q$ first, read 3.4 Rationals and well-defined operations.