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.

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

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

  • $\begingroup$ Oh my bad. So my approach is no good. How can I arrive at a conclusion? $\endgroup$ – Kushal Bhuyan Feb 22 at 3:43
  • $\begingroup$ ok got it.never mind. $\endgroup$ – Kushal Bhuyan Feb 22 at 3:44

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