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$$\frac{1}{x}-1>0$$

$$\therefore \frac{1}{x} > 1$$

$$\therefore 1 > x$$

However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, but as I was unable to factories the LHS, I did so. How can I get this result algebraically?

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Continuing from the step: $$\frac1x\gt1$$ Now, to multiply the inequality by any non zero number we need to know its sign. So, taking two cases,


Case 1: $x\gt0$

Multiplying by $x$ on both sides will not affect the sign. Thus, $$1\gt x$$ Due to the assumption, $$1\gt x\gt0$$


Case 2: $x\lt0$

Multiplying by $x$ on both sides will reverse the sign. Thus, $$1\lt x$$ But $x\lt 0$. Thus, no solution is there in this case.


Clearly, the case $x=0$ is not defined. The solution then is, $$0\lt x\lt1$$

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  • $\begingroup$ What do you mean after "Due to the assumption"? How did we get there? $\endgroup$ – StopReadingThisUsername Apr 1 '16 at 7:37
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    $\begingroup$ We assumed "Case 1: x>0". The other case was left for case 2. That's the assumption. $\endgroup$ – GoodDeeds Apr 1 '16 at 7:38
  • $\begingroup$ I see. I wasn't able to comprehend the conclusion of case two quite well until now. Thanks a lot! $\endgroup$ – StopReadingThisUsername Apr 1 '16 at 7:41
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You have $$\frac{1}{x} - 1 > 0$$ Forming a common denominator yields $$\frac{1 - x}{x} > 0$$ The inequality is true if the numerator and denominator have the same sign.

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Your answer would be correct were we assuming $x$ is positive. You must keep in mind, though, that the inequality fails for $x<0$ (recall that the inequality sign switches when we multiply by a negative number).

For $x>0$, we have a solution $1>x$. In other words, $0<x<1$ is a solution.

If we have $$\frac 1x > 1$$ and $x<0$, we get solutions $x>1$. But that's an absurdity, and so there are no solutions if $x<0$.

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Consider the function $f(x)=1/x-1=(1-x)/x$ defined for $x\neq0$. As $ f$ is continuous it can only change sign where it is zero or undefined, i.e., at $x=1$ or $ x=0$. Hence the sign of $f$ is constant on $]-\infty, 0[$, $]0,1[$, and $]1,\infty[$.

Finally compute the sign of say $f(-1)$, $f(1/2)$, and $f(2)$.

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Hint: $x(1-x)>0$.Can you continue?

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  • $\begingroup$ Hmm... I don't understand why it is downvoted. Since $x^2\ge 0$,$$x^2\left(1-\frac{1}{x}\right)<0$$ and get $x(x-1)<0$. Thus we can find the correct result: $0<x<1$. $\endgroup$ – choco_addicted Apr 1 '16 at 7:45

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