# Prove $\frac{x-1}{x}$ has exactly two real solutions [closed]

How do I prove $\frac{x-1}{x}=0$ has two real solutions?

I know it has one solution at $x=1$, but the question said "find the 2 real solutions", so I was confused if I was overlooking something. Maybe it's an error in the question.

Does anybody have any ideas to this problem?

## closed as unclear what you're asking by user137731, Jack D'Aurizio, ajotatxe, qbert, Jean MarieOct 2 '16 at 21:43

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• Usually one talk about solutions when there is an equation (or inequation) involve, but in your question there was only one function. Are you sure you should not include some $=$, or some inequality? – Darío G Oct 2 '16 at 21:18
• Shouldn't this be equal to something if it is to have "solutions?" I don't see an equation. Unless if you're asking if $f(x) = \frac{x-1}{x}$ has exactly two real roots. – Sean Roberson Oct 2 '16 at 21:19
• This question is not well posed just yet. To what is it supposed to be equal? – B. Pasternak Oct 2 '16 at 21:19
• If roots are meant, there is only one. At the root of the denominator , the function has a pole. Maybe, the OP means "asymptotics". In this case, we actually have two , namely $x=0$ and $y=1$ – Peter Oct 2 '16 at 21:22
• Ah yes I meant (x-1)/x = 0 , I know it has one solution but the question said find the 2 real solutions so I was confused if I was overlooking something. Maybe it's an error in the question. – Curtis Cleary Oct 2 '16 at 21:39

Well, assuming you want to find the roots of $f(x)=\frac{x-1}{x}$: $$f(x)=1-\frac{1}{x}$$ The roots are given when $f(x)=0$, so $$1=\frac{1}{x}\Rightarrow x=1$$
You could even plot $f(x)$ to see it: