# $x-\left\lfloor x \right\rfloor +\frac { 1 }{ x } -\left\lfloor \frac { 1 }{ x } \right\rfloor =1\implies x$ is irrational. [duplicate]

If $x-\left\lfloor x \right\rfloor +\frac { 1 }{ x } -\left\lfloor \frac { 1 }{ x } \right\rfloor =1$, then $x$ is irrational.

I am thinking of using the contrapositive: If $x$ is rational, then $x-\left\lfloor x \right\rfloor +\frac { 1 }{ x } -\left\lfloor \frac { 1 }{ x } \right\rfloor \neq 1$

However, I don't know how I would approach this after that point. My first guess it to create a rational number, $\frac { a }{ b }$ such that $a,b\in\mathbb{Z}$, but I don't know how to manipulate it from there to get to the right hand side. I guess the floor function is throwing me off.

I would appreciate a hint/nudge in the right direction so that I can arrive at the solution by myself.

• This question looks awfully familiar... – copper.hat Nov 4 '15 at 6:31
• – lab bhattacharjee Nov 4 '15 at 6:33
• @labbhattacharjee I think you are talking about this link- math.stackexchange.com/q/1504381/177357 By the way, good find – Rajat Nov 4 '15 at 6:36
• @labbhattacharjee: Good find. I spent a few minutes looking but couldn't find it... – copper.hat Nov 4 '15 at 6:36
• Searching for latex works surprisingly good. Try for example searching for \lfloor \frac 1x \rfloor and all the similar questions comes up! – Winther Nov 4 '15 at 6:39