How can we prove that \frac{x-1}{x-2} < 1 for x < 0?

[doubleaxel195]

Homework Statement

I just want to show that given x<0, \frac{x-1}{x-2} &lt;1.

The Attempt at a Solution

I don't know why I am having trouble with this! I feel like this is so easy!

So if x<0, then we know x-1&lt;-1, x-2&lt;-2. So
\frac{-1}{2}&lt;\frac{1}{x-2} and \frac{x-1}{x-2}&lt;\frac{-1}{x-2}.

I can't seem to get a good upper bound on \frac{1}{x-2} that makes the entire thing less than one. Am I doing something illegal? Because now it looks like I should want to get\frac{1}{x-2} &lt;-1 to make it all less than one, but clearly that is not true.

 

[SammyS]

Homework Statement

I just want to show that given x<0, \frac{x-1}{x-2} &lt;1.

The Attempt at a Solution

I don't know why I am having trouble with this! I feel like this is so easy!

So if x<0, then we know x-1&lt;-1, x-2&lt;-2. So
\frac{-1}{2}&lt;\frac{1}{x-2} and \frac{x-1}{x-2}&lt;\frac{-1}{x-2}.

I can't seem to get a good upper bound on \frac{1}{x-2} that makes the entire thing less than one. Am I doing something illegal? Because now it looks like I should want to get\frac{1}{x-2} &lt;-1 to make it all less than one, but clearly that is not true.

\displaystyle \frac{x-1}{x-2}=\frac{x-2+1}{x-2}=1+\frac{1}{x-2}

Can you show that 1/(x-2) < 0 ?

 

[doubleaxel195]

Yes I can. Thank you so much! Was I doing anything illegal or just picking bad bounds? I can't seem to find a mistake in what I was doing before.