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

This note is where the construction story of the number systems starts to change direction. Up to now, you have been building $N$, $Z$, and $Q$ and checking that their operations are well-defined. Here the question is different:

Does $Q$ already contain every number you need for order and limit arguments?

The answer is no. The standard example comes from $\sqrt{2}$.

Intuition first: dense is not the same as complete

A common first reaction is:

"But there are so many rational numbers. Surely there is no gap."

That reaction mixes up two different ideas.

• Dense means that between two different rational numbers, you can always find another rational.
• Complete means that certain bounded sets really do have least upper bounds inside the number system you are working in.

The set $Q$ shows that density does not guarantee completeness. Rationals can be packed closely together and still miss an important boundary point.

A warning from geometric series

Infinite processes can have rational partial information while pointing toward a boundary that is not captured by any finite stage. A simple geometric series already shows the pattern:

$$1+\frac12+\frac14+\cdots+\frac{1}{2^n}$$

has rational partial sums, and those partial sums approach $2$. Each finite sum is still below $2$, but the boundary is read from the whole infinite process, not from one finite partial sum.

This example is not a proof that $\sqrt{2}$ is missing from $Q$. It prepares the right habit: distinguish a sequence of better rational approximations from the limit or least upper bound that those approximations are trying to reach.

The key order-language

Definition

Upper bound and supremum

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

• An upper bound of $X$ is a number $u$ 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.

Definition

Irrational number

An irrational number is a real number that does not belong to $Q$.

The set that exposes the gap

Theorem

The square root of two is not rational

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 $2$.

At first glance, it feels as if this set ought to have a rational boundary. After all, numbers such as $1$, $1.4$, and $1.41$ belong to it, while numbers such as $2$ or $3/2$ sit above it.

The problem is that the "correct" boundary is $\sqrt{2}$, and that number is not in $Q$.

Why $S$ is bounded above in $Q$

Before proving that $S$ has no supremum in $Q$, you should first check that $S$ really is a bounded-above set.

Worked example

Finding an upper bound for $S$

Take $u = 2$.

Since $2^2 = 4 > 2$, the number $2$ does not belong to $S$. More importantly, every rational $q$ with $q > 2$ also satisfies $q^2 > 2$, so such a $q$ lies above every element of $S$.

That means $2$ is an upper bound for $S$.

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 $\sqrt{2}$ is not rational

First recall the standard contradiction proof.

Proof that $\sqrt{2}$ does not belong to $Q$

Assume, for contradiction, that $\sqrt{2}$ is rational.

Then there exist integers $p$ and $q$ with $q \ne 0$ such that

$$\sqrt{2} = \frac{p}{q}$$

and the fraction is already in lowest terms.

Squaring both sides 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 $\sqrt{2}$ is not rational.

This matters because if $\sqrt{2}$ were rational, it would be the obvious candidate for $\sup(S)$ in $Q$. The contradiction tells you that the obvious candidate is missing from the rational number system.

The formal gap statement

Theorem

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 candidate lies below the boundary. 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 candidate lies above the boundary. 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 candidate lies on the boundary. 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

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

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

This is a useful contrast: rational partial sums do not automatically create a gap. The gap appears when the boundary required by the set is the irrational number $\sqrt{2}$.

Checkpoint

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. For example, the geometric sequence above has supremum $2$ in $Q$, and the set $S$ has supremum $\sqrt{2}$ when regarded as a subset of $R$. But when the ambient set is restricted to $Q$, $\sqrt{2}$ is not an admissible candidate.

Worked example

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. Nevertheless, $0=\inf(Y)$ in the ambient ordered set $Q$, 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 irrationality proof.

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

The following worked applications use three different methods: parity detects impossible rational squares, common divisors constrain integer equations, and leastness identifies a supremum.

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: reduce a quadratic to the irrationality proof

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 $x_n=1.2+\sum_{k=2}^{n}3/10^k$ 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 $\sup_Q(X)=37/30$.

Common mistake

Common mistake

Dense does not mean complete

It is true that between any two rational numbers there is another rational. But that fact only talks about what happens between two existing rationals. It does not say that every bounded set of rationals 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

Checkpoint

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.

Checkpoint

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

Checkpoint

Why do infinitely many rationals near $\sqrt{2}$ still fail to fix the gap in $Q$?

Use the difference between density and completeness.

Solution · Guided solution

Having many rationals near $\sqrt{2}$ only shows that $Q$ is dense. It tells you that you can approximate the missing boundary very well. But approximation is not the same as possession. The boundary point that should play the role of the least upper bound is $\sqrt{2}$, and that point is not rational. So the set can have arbitrarily close rational approximations and still have no supremum in $Q$.

Optional continuations

Read this first

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