In a problem set I found this problem:

Find the $\sup$ and $\inf$ of $\frac mn$ such that $m,n$ are natural numbers and $m<2n$.

All I could do is this:

Since $m$ and $n$ are natural numbers, we have $m/n > 0$. And since $m<2n$ we have $\frac mn < \frac{2n}n = 2$, thus $\frac mn<2$.

I got stuck on showing that $2$ is the supremum rigorously.

The answer provided in the solutions sheet is not helping at all, it goes like this (in French, translated to English):

Il est clair que $0 < \frac mn < 2$, et on peut facilement en déduire que $\sup = 2$ et que $\inf=0$.

In english:

It is obvious that $0<\frac mn<2$, from this we can easily deduce that $\sup = 2$ and $\inf=0$

  • $\begingroup$ Do you know the Archimedean property of real numbers? $\endgroup$
    – Koro
    Commented Dec 22, 2021 at 10:36
  • $\begingroup$ @Koro Yes, for every real number 'r' we can find a natural number 'n' such that 'n>r'. $\endgroup$
    – Denis
    Commented Dec 22, 2021 at 10:41
  • $\begingroup$ The bounds alone do not give the values of inf and sup. They also have to be the best possible bounds. What we have to show is that we can get arbitary close to $0$ and also arbitary close to $2$. $\endgroup$
    – Peter
    Commented Dec 22, 2021 at 10:44
  • $\begingroup$ @Koro I am curious. What is wrong about the above comment ? And what do you consider to be the Archimeadian property ? $\endgroup$
    – Peter
    Commented Dec 22, 2021 at 10:50
  • $\begingroup$ @Peter: By Archimedean property, I had this in mind: For every $\epsilon\gt 0$, there is an $n\in \mathbb N$ such that $\frac 1n<\epsilon$. $\endgroup$
    – Koro
    Commented Dec 22, 2021 at 10:51

2 Answers 2


I think what your solution sheet is saying is straight up wrong. We cannot conclude, just from the inequality $0<\frac mn < 2$, that $\sup =2$. You need some more steps and more assumptions.

To prove that $s$ is the supremum of a set $A$, you need to prove two things.

  1. That $s$ is the upper bound of $A$.
  2. That any number smaller than $s$ is not the upper bound of $A$.

Let $$A=\left\{\frac mn| m,n\in\mathbb N, m<2n\right\}$$

You already proved that $2$ is the upper bound of $A$. Now, you need to prove that any number smaller than $2$ is not the upper bound of $A$. You can start doing this by saying one of the most common phrases in calculus: **Let $\epsilon > 0$.

Now, you need to prove that $2-\epsilon$ is not an upper bound of $A$. You can do this by proving that there exists some element $a\in A$ such that $a>2-\epsilon$.

To do this, think about what property $a$ needs to fulfill. You know that $a=\frac mn$ for some pair $m,n$. You also know that $a>2-\epsilon$.

You can write this out a bit and get

$$\frac mn > 2-\epsilon\\ m > 2n - \epsilon n$$

and note that the deduction above goes both ways. In other words, if you can find some $m,n\in\mathbb N$ such that $m>2n-\epsilon n$, then you will also have $a=\frac mn > 2-\epsilon$.

However, you have one more limitation, and that is that if you want $\frac mn$ to be an element of $A$, then $m<2n$ must also be true. This means that $m$ must be some value that is greater than $2n-\epsilon n$ and smaller than $2n$.

So, the question remains, can you find such a pair of values?

Hint: Think about what happens to the interval $(2n-\epsilon n, 2n)$ when $n$ is very large.

  • $\begingroup$ When ϵ > 2, 2n-ϵn becomes negative, and since 2n is greater than 1, we can always find a value for m in the interval (2n−ϵn,2n). When ϵ<2 is causing me problems, if n=1 and ϵ = 1 we get this interval (1,2), so m must be a natural between 1 and 2, which is not possible! $\endgroup$
    – Denis
    Commented Dec 22, 2021 at 11:03
  • $\begingroup$ @ShinobiSan But you don't need to limit yourself to $n=1$! You are completely free to choose any $n$ you want! The only thing you cannot choose is the value of $\epsilon$. Once that is selected, you can take any $n$ you want, and any $m$ you want. $\endgroup$
    – 5xum
    Commented Dec 22, 2021 at 11:04
  • $\begingroup$ OOOOOOOOOOOOOOOOOOOH yes of course, I am chosing ϵ = 1, then I have to find two naturals m and n that satisfy 2n- ϵn < m < 2n, which is always possible (in this case we can take n=3 and m=4. Thank you! $\endgroup$
    – Denis
    Commented Dec 22, 2021 at 11:11
  • $\begingroup$ @ShinobiSan whoah whoah! YOU CANNOT CHOOSE $\epsilon$!!! $\endgroup$
    – 5xum
    Commented Dec 22, 2021 at 11:12
  • 1
    $\begingroup$ @Koro No problem, happens to everyone :). $\endgroup$
    – 5xum
    Commented Dec 22, 2021 at 11:31

You have already noted that $2$ and $0$ are upper and lower bounds respectively for $S$.

Let's show that $2$ is indeed the supremum of $S$. . Note that $2n-1<2n$ for any $\mathbb N$ and therefore $\frac{2n-1}{2n}\in S$ for all $n\in \mathbb N$. $\tag 1$

By Archimedean property of the reals, given any $\epsilon\gt 0, $ there exists $N\in \mathbb N$ such that $1/N\lt \epsilon\implies 2-\epsilon< 2-1/N=\color{blue}{\frac{2N-1}{2N}}<2$. By $(1)$, the quantity in blue is in $S$ and hence by uniqueness of $\sup S$, it follows that $2$ is the supremum of $S$ (For if,suppose on the contrary $x:=\sup S<2$ then by the inequality established earlier, there must be some $m\in \mathbb N$ such that $2-(2-x)=x<\color{red}{\frac{2m-1}{2m}}<2$. But the quantity in red is in $S$ and this means that $S$ has an element larger than $\sup S$, which is a contradiction.).

$\inf S$ is the "greatest" lower bound for $S$. Now, if $\inf S=:y>0$, then by Archimedean property there is an $n\in \mathbb N$ such that $\frac 1n<y$ but $1/n\in S$ and hence $y$ is not a lower bound for $S$. This is a contradiction. Hence $y$ must be $0$.


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