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How do I solve $\frac{1}{x-1}>0$ for $x$?

If I multiply both sides with $x-1$ then becomes $1\gt 0$. I know it's wrong. How do I solve it?

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    $\begingroup$ $1$ and $x-1$ have the same sign. $\endgroup$ – Yves Daoust Apr 22 '15 at 12:50
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    $\begingroup$ "I know it's wrong" You know the result is wrong, but do you know why the procedure is wrong? You can multiply both sides of an inequality by an expression (and keep the inequality) only if the expression is guaranteed to be positive. $\endgroup$ – leonbloy Apr 22 '15 at 20:54
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    $\begingroup$ Taking things strictly, $1>0$ is not wrong... :-) $\endgroup$ – CiaPan Apr 23 '15 at 8:11
  • $\begingroup$ Note that the title of the question uses a different expression than the one in the text. x < 1 is the right answer if the question is about 1/(1-x) > 0. x > 1 is the right answer if the question is about 1/(x-1)>0. $\endgroup$ – Pete Becker Apr 23 '15 at 13:40
  • $\begingroup$ @CiaPan that's what I was gonna write! :-) $\endgroup$ – Mr Reality Oct 18 '17 at 13:25
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Divide this into cases. The expression is not defined if $x = 1$. If $x > 1$, you can multiply both sides by $x - 1$ to get $1 >0$ So, if $x > 1$ the inequality is satisfied.

If $x < 1$, multiplying both sides by $x - 1$ reverses the inequality and you hae $1 < 0$. This is never true, so if $x < 1$, the inequality does not hold.

Hence the solution is $x > 1$.

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  • $\begingroup$ The answer given us x<1 $\endgroup$ – Mathxx Apr 22 '15 at 12:45
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    $\begingroup$ put in $x = 0$ and you will see that that answer is in error. $\endgroup$ – ncmathsadist Apr 22 '15 at 12:46
  • $\begingroup$ lets see 1/(x-1)<0 we dont know if x-1 is more or less than zero but we do know that (x-1)^2 is more than zero and wont affect the inequality hence, multiply both sides by (x-1)^2 giving x-1 < 0 rearranging gives x<1. My friend sent this to me . $\endgroup$ – Mathxx Apr 22 '15 at 13:08
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    $\begingroup$ @Mathxx How come > in your friend solution became <? Btw your friend solution is correct (with the < changed) by chance. Actually with that procedure you study when the denominator is > 0, which, along with studying when the numerator is > 0 (in this case always), is the correct way of studying when the whole inequality is > 0 $\endgroup$ – Narmer Apr 22 '15 at 13:43
  • $\begingroup$ @Mathxx Your question reads $\frac{1}{x-1} > 0$ but your comment here starts with $\frac{1}{x-1} < 0$. Which one is it? $\endgroup$ – Kyle Apr 22 '15 at 16:31
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Instead of multiplying both sides by $x-1$, multiply instead by $(x-1)^2$

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    $\begingroup$ This is a lot simpler than splitting into cases $\endgroup$ – Ben Aaronson Apr 22 '15 at 14:47
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    $\begingroup$ That discards information when $(x-1)^2 = 0$, and the only way to deal with that is to split into cases anyway . . . $\endgroup$ – ruakh Apr 22 '15 at 15:11
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    $\begingroup$ @ruakh: It's pretty trivial to see that the LHS of the original inequality is undefined iff $x-1=0 \iff (x-1)^2 = 0$. $\endgroup$ – Ilmari Karonen Apr 22 '15 at 16:31
  • $\begingroup$ @IlmariKaronen: OK, fair enough. $\endgroup$ – ruakh Apr 22 '15 at 19:20
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    $\begingroup$ while this approach doesn't eliminate the need for splitting into cases, it reduces its number (in this way we need only consider the cases where $x-1$ is zero or nonzero). $\endgroup$ – John Joy Apr 22 '15 at 23:03
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Hint: ${x\over{y}}>0$ where $x>0$ if and only if $y>0$ (if it confuses you, try to see what happens if $y\le0$!)

In other words, ${1\over{x-1}}>0$ if and only if $x-1>0$

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A basic fact is that a number $a\neq 0$ and its inverse have the same sign, hence $$\frac1{x-1}>0\stackrel{x\neq 1}\iff x-1>0.$$

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The sign of the fraction is given by the multiplication of the sign's of the numerator and the denominator. In your case you want the fraction be positive, and since the numerator is $1 > 0$, then the denominator must also have the same sign. Thus, you want $ x-1 > 0$, which turns out to give you $ \boxed{x >1}$.

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Remember that all fractions $\dfrac{a}{b}$ is greater than zero, if $a>0$ and $b>0$ or $a<0$ and $b<0$. In the case of your problem, we can take the same approach. In $\frac{1}{x-1}$, $a=1$ and $b=x-1$. We know that $1>0$ therefore the answer is $x-1>0$

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    $\begingroup$ Beware! $\frac{a}{b}$ is greater than zero also when $a<0$ and $b<0$! $\endgroup$ – Narmer Apr 22 '15 at 13:51
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The quotient of real numbers is positive $\iff$ numerator and denominator are of the same sign. Numerator is $1$, that's positive, so denominator must be positive, too: $$x-1>0$$ so the solution is $$x>1$$ Done.

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