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In solving a larger problem, I ran into the following inequality which I must solve:

$$ 1+\frac{1}{x} \gt 0.$$

Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.

Gorg
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4 Answers4

15

There are two cases. If $x>0$ we may multiply both sides by $x$ and the inequality is unchanged, i.e. $x+(1/x)x>0(x)$, which simplifies to $x+1>0$ or $x>-1$. This case is the combination of $x>0$ and $x>-1$, which is $x>0$.

If $x<0$ we multiply both sides by $x$ and the inequality reverses, i.e. $x+(1/x)x<0(x)$, which simplifies to $x+1<0$ or $x<-1$. This case is the combination of $x<0$ and $x<-1$, which is $x<-1$.

vadim123
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  • (+1) as this shows how $x$ must be considered positive and negative to obtain all solutions to the inequality. – Daryl May 15 '13 at 12:34
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    Can I multiply both sides by $x^2$? – Gorg May 15 '13 at 12:36
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    @Gorg, that's another way to go. It has the advantage of always being positive, so there aren't two cases. It has the disadvantage of leaving you with a slightly trickier polynomial inequality to solve. – vadim123 May 15 '13 at 12:43
11

You asked how to show this formally, so a picture is possibly not very useful, but I include one because I need practice with tikz. Vadim's answer above is excellent.

enter image description here

in_mathematica_we_trust
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  • I'm jealous - trying to get good with tikz myself. Great graph! – DanZimm May 15 '13 at 12:51
  • I'm jealous, too. Can you post the code for creating the graph? – Elmar Zander May 15 '13 at 12:52
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    Thanks! It really does just take time and practice and some help from tex.stackexchange. – in_mathematica_we_trust May 15 '13 at 12:53
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    \begin{document}

    \begin{tikzpicture}[]

    \draw[blue, thick, domain=-5:-1, samples = 250] plot (\x,{1/\x + 1}); \draw[red, thick, domain=-1:-0.3, samples = 250] plot (\x,{1/\x + 1}); \draw[blue, thick, domain=0.3:5, samples = 250] plot (\x,{1/\x + 1}) node [right] {$f(x) = 1 + \frac{1}{x}$};

    \draw[black, thick, ->] (-6,0) -- (6,0) node[right] {$x$}; \draw[black, thick, ->] (0,-3) -- (0,5) node[above] {$y$};

    \draw[black, thick, dashed] (-1,-3) -- (-1,0) node [below=0.25cm, left] {$-1$} -- (-1,5);

    \end{tikzpicture}

    \end{document}

     – in_mathematica_we_trust May 15 '13 at 12:53
  • If I could upvote more, I would.

    Why do you suggest that a picture is not formal proof though?

     – joneshf May 15 '13 at 17:11
  • @joneshf It needs a little bit of work to show that this really is the right way to draw $1 + 1/x$, and that it doesn't dip under the axis again at a later stage. – in_mathematica_we_trust May 16 '13 at 05:13
  • Interesting, I thought it'd be fairly straight forward that it's the same as the graph of $1/x$ just shifted. So you could infer the same shifted asymptotes and general behavior. – joneshf May 16 '13 at 05:16
  • Yes, but how do we know what $1/x$ looks like? We should really take is derivative, show that it's decreasing everywhere, evaluate the limit at infinities and zero, look for intercepts and all that. – in_mathematica_we_trust May 16 '13 at 05:28
5

After reading your comment here is how to go about it when you multiply by $x^2$. So we need to solve

$$x^2 + x > 0.$$

That can be written as

$$x(x + 1) > 0.$$

Now the product of two numbers is positive when both numbers are positive or when both are negative. Therefore we have two cases to solve:

Case 1:

$x >0$ and $x + 1> 0$

Both of these inequalities are satisfied when $x > 0$.

Case 2:

$x < 0$ and $x + 1 < 0$

Both of these inequalities are satisfied when $x < -1$.

Hope that helps!

3

note that $\frac{x+1}{x}>0$ and $(x+1)x>0$ have the same solution(or, as you mentioned, multiply them both with $x^2$)

generally speaking, bringing the denominator over the bar doesn't change the answer, but be careful that the denominator can't be zero.

arax
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