Suppose $a$ is a positive real number. Prove that there exists a natural number $n$ such that $0 < 1/n < a$.

Could you guys help with this? I'm awful at writing proofs, I know why this is true but I don't know how to write the proof in a way that others could follow. Any help would be greatly appreciated. Thanks.

  • $\begingroup$ I guess it it obvious that the proposition holds. If I understand you correctly you are interested in the way to express this logically. $\endgroup$ – Andreas Oct 3 '16 at 9:09
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    $\begingroup$ It follows immediately from the Archimedean property of the field $\mathbb{R}$: en.wikipedia.org/wiki/Archimedean_property. For more information you may want to know about the construction of the real numbers. $\endgroup$ – Colescu Oct 3 '16 at 9:12
  • $\begingroup$ The desired result $0 < 1/n < a$ is equivalent to $n > 1/a > 0$. So effectively you need to show that given any real number there is a positive integer greater than it. Use the fact $(1)$ that for every rational number there is a positive integer greater than it and $(2)$ that for every real number there is a rational number greater than it. $\endgroup$ – Paramanand Singh Oct 3 '16 at 9:24

To derive this we need to use the completeness of $\mathbb{R}$:

Every nonempty set of real numbers that is bounded above has a supremum in $\mathbb{R}$.

There are two approaches to this, the axiomatic approach (where it is an axiom) and the constructive approach (where it is a theorem). More information can be found in analysis texts.

Now we prove the Archimedean property of $\mathbb{R}$:

$\mathbb{N}$ is not bounded above in $x\in\mathbb{R}$. That is, for each $x\in\mathbb{R}$ there exists $n\in\mathbb{N}$ such that $n>x$.

For $x<0$ it is trivial. Suppose $x\geq0$. Let $A=\{n\in\mathbb{N}\ ;n\leq x\}$. By completeness we have $\sup A\in\mathbb{R}$. Obviously, there is $a\in A$ such that $\sup A-1/2<a$ (Why?), then let $n=a+1$ and $n>x$ and the theorem is proved.

Finally, we turn to the question:

For any real number $a>0$, there exists a natural number $n$ such that $0<1/n<a$.

We proceed using proof by contradiction. Let $0<1/n<a$ for all $n\in\mathbb{N}^{\times}$, then $n\leq 1/a$ for all $n\in\mathbb{N}^{\times}$. Thus $\mathbb{N}$ is bounded in $x\in\mathbb{R}$, contradicting the Archimedean property.

  • $\begingroup$ I think the axiomatic approach is so so roundabout a way to prove a trivial property (archimedean property) of real numbers. In the constructive approach the archiemdean property of reals follows from the same property of rationals trivially. $\endgroup$ – Paramanand Singh Oct 3 '16 at 10:19

You can argue that the proposition requires $n > 1/a$. Since $1/a$ is fixed, and since natural numbers have no upper bound, there will alwas be an $n$ which satisfies $n > 1/a$.

You could also argue by contradiction: Suppose there is no such natural number $n$ which satisfies $n > 1/a$. Then for all natural number we must have $n \le 1/a$. This is a contradiction, since there is no upper bound for natural nunbers.


If $a$ is a real number, there must exist some positive rational number $q$ such that $0<q<a$ (why?).

Now, $q$ is rational, so $q=\frac{a}{b}$ where $a,b\in\mathbb N$.

Can you find a natural number $n$ such that $\frac1n \leq \frac ab$?

Another way to solve this is to take the decimal expansion of $a$, so $$a=0.a_1a_2a_3\dots$$

(we can assume that $a<1$, since if $a\geq 1$, we can take $n=2$ and be done with it).

Now, take $m=\min\{k|a_k\neq 0\}$, and it should be easy to show that $$b=\frac{1}{10^{k+1}}$$

is strictly smaller than $a$.

And another way: $1/n < a$ is true if and only if $n>\frac1a$, so you can take any natural number greater than $\frac1a$ (why does that exist?) and be done.

  • $\begingroup$ why the detour to rational numbers? This was not asked in the question. $\endgroup$ – Andreas Oct 3 '16 at 9:02
  • $\begingroup$ @Andreas Because the question is easier in rationals, i.e. he can take $\frac1b$ as the answer. $\endgroup$ – 5xum Oct 3 '16 at 9:04
  • $\begingroup$ It's far easier to prove it, see my answer below. Rationals add a complexity not required here. $\endgroup$ – Andreas Oct 3 '16 at 9:06
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    $\begingroup$ Looks like someone has chosen to downvote each and every one of the answers here. $\endgroup$ – barak manos Oct 3 '16 at 9:13
  • $\begingroup$ Yeah, the arguments go more "Trumpian" here... $\endgroup$ – Andreas Oct 3 '16 at 9:17

If $a\geq1$, then choose $n=2$

If $a<1$, then choose $n=\lceil1/a\rceil+1$

Note that this is true because $0<a<1\implies0<\frac{1}{\lceil1/a\rceil+1}<\frac{1}{\lceil1/a\rceil}\leq\frac{1}{1/a}=a$

  • $\begingroup$ This is circular. This problem is equivalent to showing that for all real $a>0$ there exists a natural $n$ satisfying $0<a<n$. If you haven't proven this, you can't prove that $\lceil x \rceil$ is well defined for all $x$. $\endgroup$ – MathematicsStudent1122 Oct 3 '16 at 9:10
  • $\begingroup$ @MathematicsStudent1122: I have added an explanation. $\endgroup$ – barak manos Oct 3 '16 at 9:12
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    $\begingroup$ @MathematicsStudent1122: It is only circular if we use the axiomatic approach to real numbers in which case we prove the Archimedean property using completeness property. Otherwise it is a perfectly valid and perhaps more useful answer. Unfortunately most modern analysis texts have chosen to use a canon ball (completeness property) to kill an ant (Archimedean property). $\endgroup$ – Paramanand Singh Oct 3 '16 at 9:16
  • $\begingroup$ @ParamanandSingh This doesn't "kill the ant" though. It takes it for granted. $\endgroup$ – MathematicsStudent1122 Oct 3 '16 at 9:43

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