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

Question

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.

Answer 1

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