Given $n$ a positive integer, how would you prove from scratch that there exists a rational number $q$ such that $n<q^2<n+1$?

By "from scratch" I mean by not using any "advanced" tools like the density of the rational numbers in the real numbers. Just using the definition of rational numbers, how to prove that?

I faced this problem while trying to verify that the Dedekind cut $(A,B)$ cannot be determined by a rational number, where:

  • $B=\{x \in Q^+: x^2>2\}$

  • $A=Q\setminus B$

where $Q^+$ denotes the positive rationals.

So, for the purposes of the problem, I still don't even know what the real numbers are.

  • $\begingroup$ Are we allowed to use the archimedian property? $\endgroup$ – gambler101 Sep 28 '16 at 18:21
  • $\begingroup$ @JonathanRichardLombardy yes $\endgroup$ – Cauchy Sep 28 '16 at 18:22

As $1<(\frac54 )^2<2<(\frac32 )^2<3$, we may assume wlog. that $n\ge 3$.

With $q=\frac ab$, our task is to find $a,b$ such that $nb^2<a^2<(n+1)b^2$. Pick $b=2n^2$; so we want $4n^5<a^2<4n^5+4n^4$. The set $\{\,k\in\Bbb N\mid k^2>4n^5\,\}$ is a non-empty (contains $3n^3$) subset of $\Bbb N$, hence has a minimal element $a$. Clearly, $a>2n^2>1$. Then $$(a-1)^2=a^2-2a+1>a^2\left(1-\frac 2a\right)>a^2\left(1-\frac 1{n^2}\right) $$ If we assume $a^2\ge 4n^5+4n^4$, this leads to $$ (a-1)^2>4n^5+4n^4-4n^3-4n^2=4n^5+4n^2((n-1)^2-2)>4n^5$$ contradicting minimality of $a$. Hence $a^2<4n^5+4n^4$, as desired.


One idea: It's easy enough to find a $q_0 \in \mathbb{Q}$ that satisfies $q_0^2 > n$. Now use Newton's method to approximate a solution to $q^2 - n = 0$. This gives a recurrence $$ q_{i+1} = q_i - \frac{q_i^2 - n}{2q_i}. $$ It can be shown that $$ q_{i+1}^2 - n = \left(\frac{q_i^2 - n}{2q_i}\right)^2 \le \frac{q_i^2 - n}{4} $$ so that eventually $q_i^2 - n < 1$.

  • $\begingroup$ This method is very good in OP's context since it avoids refering to irrational numbers. $\endgroup$ – Olivier Moschetta Sep 28 '16 at 18:50
  • $\begingroup$ @OlivierMoschetta agreed. I picked Hagne's answer because it's very easy to follow. Though I up-voted this (thank you by the way) $\endgroup$ – Cauchy Sep 28 '16 at 18:56

Notice that for $n > 4$

$$ n\sqrt{n + 1} - n \sqrt{n} = \frac{n}{\sqrt{n + 1} + \sqrt{n}}\geq \frac{n}{2\sqrt{n + 1}} > 1$$

so there is an integer $k$ such that

$$n\sqrt{n} \leq k <n \sqrt{n + 1}$$ so

$$n \leq \frac{k^2}{n^2} <n + 1$$ hence $q := \frac{k}{n}$ is a rational number with the property you are looking for.


The problem is reduced to showing that there exists a rational number between $\sqrt{n}$ and $\sqrt{n+1}$.

Let $q=a/b$ where $a,b\in\mathbb{N}$ (as $\sqrt n >0)$.

Archimedian property (modified) : Given any real number $y>0$ there exists an $n\in\mathbb{N}$ such that $1/n<y$.

Thus there exists an integer $b$ with $\frac 1b <\sqrt{n+1}-\sqrt n$. We can pick another integer $a$ such that $a-1\le b\sqrt n<a$. Now $\sqrt n<\sqrt{n+1}-1/b\Rightarrow b\sqrt n<b\sqrt{n+1}-1\Rightarrow b\sqrt{n+1}>b\sqrt{n}+1\ge a$. Thus $\sqrt{n+1} >a/b$.

Now from choice of $a$ we conclude that $a/b>\sqrt{n}$.

Please notify about any errors committed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.