# If $x$ is a positive real number , then there exists a natural number $n$ such that $1/n<x<n$.

I am new to proofs and I need help to evaluate my work ...I can easily see that $$x by Archimedean property ,but can I say WLOG ; $$1/x implies that $$1/x and $$x which also implies $$x>1/n$$ and $$x and end the proof as the case where $$x<1/x$$ uses similar argument? (it is clearly cases for $$x<1$$ and $$x>1$$ but I do not know if this is fine to say) Your help will be absolutely appreciated.

• You cannot say that, or at least the way you wrote it doesn't hold. But you can use the same Archimedean property to find an $m$ such that $1/m \le 1/x$. Then $\dots$
– dxiv
May 18 at 0:07
• Even if i firstly claimed that x>1 then said 1/x < x <n? then claimed for the case 0<x<1 and then said x<1/x < n? I do not see how is this wrong. May 18 at 0:09
• There is the case where $1/x=x,$ too. Also, you cannot say “WLOG” here, because you do lose generality. You have to prove the two cases separately. It is essentially the same proof, but WLOG should never be referenced when you need to prove the two cases. Just say, “First, if $1/x<x...$” May 18 at 0:12
• @AbduljabbarOu The "WLOG" in "WLOG $1/x \lt x$" is not justified, and "$x \lt 1/x$ uses similar argument" is not justified, either. Maybe you can salvage the idea itself, but it would need more rigor and clarity.
– dxiv
May 18 at 0:13
• "If $x>=1$ , then $1/x<=x<n$ which implies that $x>1/n$ and $x<n$ hence $1/n<x<n$ " will this do for the first case? Thank you guys for helping me out. May 18 at 0:16

Your instinct is right - you seem headed towards the right idea. It is using the Archimedean Property two times.

Rather than treat the two cases, instead note that the Archimedean property means there is some $$n_1>x$$ and some $$n_2>\frac1x.$$ Then find a suitable $$n,$$ based on $$n_1,n_2.$$

• I honestly am not so sure,but could it be that we choose $n>n_1$? May 18 at 0:27
• That doesn’t ensure $\frac1n<x.$ $n$ will have to depend on both $n_1$ and $n_2.$ May 18 at 0:34
• Would $n>n_1>n_2$ do the job? Sorry for being a hindrance. May 18 at 0:36
• How would we make sure that $n_1>n_2?$ We were given $n_1,n_2$ by that guy Archimedes. He didn’t tell us what order they are in. May 18 at 0:37
• There’s the key word. “Maximum.” That’s it. $n=\max(n_1,n_2)$ You could also use $n=n_1+n_2,$ but that is only because $n_1+n_2\geq \max(n_1,n_2).$ May 18 at 0:58

Those seems like good ideas.

The solution I have seen is to take $$n$$ so that $$n>x$$ and $$n>1/x$$ (a number that is larger than both $$x$$ and $$1/x$$ exists by the Archimedean property. Then you will have $$\frac{1}{n} < x$$ since $$n > \frac{1}{x}$$.

As was noted in the comments the current gap is that you cannot assume that $$1/x< x$$, but we can circumvent this problem as above.