# True or False: If $x\notin \mathbb{Q}$ then $\sum_{m\geq 0} mx^{m-1}\notin \mathbb{Q}$

True or False: If $$x\notin \mathbb{Q}$$ then $$\sum_{m\geq 0} mx^{m-1}\notin \mathbb{Q},$$ where $$|x|<1.$$

So I considered the contra-positive of the above statement: If $$\sum_{m\geq 0} mx^{m-1}\in \mathbb{Q}$$ then $$x\in \mathbb{Q}.$$

Now if the contra-positive is true/false then the statement is true/false.

Thus, if $$\sum_{m\geq 0} mx^{m-1}\in \mathbb{Q}$$ $$\implies 1+2x+3x^2+... \in \mathbb{Q}$$ $$\implies 1'+x'+(x^2)'+(x^3)'+... \in \mathbb{Q}$$ $$\implies (1+x+x^2+x^3+...)'\in\mathbb{Q}$$ $$\implies (\frac{1}{1-x})' \in \mathbb{Q}$$ $$\implies \frac{-1}{(1-x)^2}\in \mathbb{Q}$$

Now from the last step we can conclude that $$x\in \mathbb{Q}$$, since $$|x|<1.$$ Therefore the contrapositive is true, so the given statement is also true.

Is my analysis correct?

Thanks for any kind of help.

• You seem to use $x$ both for a fixed number and a variable (you cannot differentiate with respect to a fixed number). – darij grinberg Feb 22 at 3:38
• That said, there is another, deeper problem with your argument: $\dfrac{-1}{\left(1-x\right)^2} \in \mathbb{Q}$ does not imply that $x \in \mathbb{Q}$. – darij grinberg Feb 22 at 3:39
• @darijgrinberg can you please point me out about the reasoning of the last argument? – Kushal Bhuyan Feb 22 at 3:41
• Why don't you try to write out that last argument in detail? – darij grinberg Feb 22 at 3:41
• Yeah got it now. @darijgrinberg – Kushal Bhuyan Feb 22 at 3:44

Take $$x=\sqrt2-1$$, you have $${1\over{(1-x)^2}}\in\mathbb{Q}$$.