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I have been given the inequality $\frac{1}{x} < x < 1$ and have been told to find the values of $x$ which satisfy this inequality, and I have also been told to find these values using a case-by-case approach. I'd like to know whether my reasoning is valid.

Case 1: $x=0$

In this case we find that the value of $\frac1x$ is undefined, so we know that $x \neq 0$.

Case 2: $0<x<1$

Given the inequality $\frac{1}{x} < x < 1$ and that $x$ lies in the range $0 < x < 1$, we find that $1<x^2$. Therefore, $x>1$ or $x<-1$. However, $x\not>1$ and in this case $x\not<-1$ since we are looking at the case $0<x<1$. Thus we conclude that no values of $x$ in the range $0<x<1$ satisfy the original inequality.

Case 3: $x<0$

From the original equality and the fact that we know that $x<1$ we find:

$$\frac1x < x$$

$$\implies 1 > x^2$$ (since $x<0$) $$\implies -1<x<1$$ but we know that $x$ cannot lie in the range $0<x<1$ so we have that $-1<x$ and that $x<0$ and combining these inequalities we have that $-1<x<0$.

This argument seems to be supported by looking at the graph, but I am unsure whether all the steps I have made are valid and whether this is what is meant by a 'case analysis'.

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    $\begingroup$ This is absolutely fine and there is no part of the solution that is wrong. $\endgroup$ – Prakhar Nagpal Jul 12 '18 at 16:00
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    $\begingroup$ I think the argument is good! ${}{}{}{}{}{}{}$ $\endgroup$ – Andres Mejia Jul 12 '18 at 16:01
  • $\begingroup$ Thank you for the feedback. $\endgroup$ – Benjamin Jul 12 '18 at 16:02
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Overall, I would say that the argument is spot on! Just a couple comments you might want to consider. First of all, to be super rigorous in your answer, you might mention that clearly we know $x \ngeq 1$ so we need not consider that case. Another thing, I felt that your argument via contradiction for Case 2 was a bit lengthy there at the end. Once you reached the point $1 < x^2$ you could have just said that since we have $0 < x < 1$ (via assumption) and no such value squared is greater than 1, we have a contradiction and are done. The whole considering $x > 1$ or $x < -1$ is just overkill.

Also, that is exactly what it means by a case analysis, or case-by-case approach. Use cases to consider every possible value of $x$, and see what happens. Using cases is actually a very common proof technique!

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  • $\begingroup$ Thanks, this was helpful! $\endgroup$ – Benjamin Jul 13 '18 at 8:53

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