shohvanilu wrote: Given a set such that every positive integer n $I=($n^2$,$n^2$+$n$+3*$3^0,5) Prove that every positive integer n have ab/c such that a,b,c in I These statements are meaningless Is $I$ depending on $n$ ? and so $I_n=\{n^2,n^2+n+3\sqrt 3\}$ ? Is $I=\bigcup_{n\in\mathbb N}\{n^2,n^2+n+3\sqrt 3\}$ ? What is the meaning of "have" in the sentence " n have ab/c " ???

n equal to ab/c

shohvanilu wrote: n equal to ab/c Ok, thanks. And what is I ? Is $I=\{n^2,n^2+n+3\sqrt 3\}$ (set depending on $n$) ? Is $I=\bigcup_{n\in\mathbb N}\{n^2,n^2+n+3\sqrt 3\}$ ? Is $I=[n^2,n^2+n+3\sqrt 3]$ (interval depending on $n$) ? Is $I=\bigcup_{n\in\mathbb N}[n^2,n^2+n+3\sqrt 3]$ ? Is $I=(n^2,n^2+n+3\sqrt 3)$ (interval depending on $n$) ? Is $I=\bigcup_{n\in\mathbb N}(n^2,n^2+n+3\sqrt 3)$ ?

$ I=\bigcup_{n\in\mathbb{N}}(n^{2},n^{2}+n+3\sqrt 3) $

Ok, thanks. So the problem is : Prove that $\forall n\in\mathbb N$, $\exists a,b,c\in\bigcup_{k\in\mathbb N}(k^2,k^2+k+3\sqrt 3)$ such that $n=\frac {ab}c$

pco wrote: Prove that $\forall n\in\mathbb N$, $\exists a,b,c\in\bigcup_{k\in\mathbb N}(k^2,k^2+k+3\sqrt 3)$ such that $n=\frac {ab}c$ $1=\frac {2\times 2}4$ where both $2,4\in(1,1+1+3\sqrt 3)$ Let then $n\ge 2$ and let $a_k=\left\lfloor 10^k\sqrt{n-1}\right\rfloor$ The rational $p_k=\frac {a_k^2+1}{10^{2k}+1}$ is such that $\lim_{k\to+\infty}p_k=n-1$ Then, for $k$ great enough, we get $\frac {a_k^2+1}{10^{2k}+1} <n<(2+3\sqrt 3)\frac {a_k^2+1}{10^{2k}+1}$ Setting $a=a_k^2+1$ and $c=10^{2k}+1$ and $b=\frac{cn}a$, this inequality is $1<b<2+3\sqrt 3$ And so we got $n=\frac {ab}c$ with : $a=a_k^2+1\in(a_k^2,a_k^2+a_k+3\sqrt 3)$ $b=\frac{cn}a\in(1^2,1^2+1+3\sqrt 3)$ $c=10^{2k}+1\in((10^k)^2,(10^k)^2+10^k+3\sqrt 3)$ Q.E.D.