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, $$\frac{x-1}{x-2} <1$$.

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 $$x-1<-1, x-2<-2$$. So
$$\frac{-1}{2}<\frac{1}{x-2}$$ and $$\frac{x-1}{x-2}<\frac{-1}{x-2}$$.

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$$\frac{1}{x-2} <-1$$ 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, $$\frac{x-1}{x-2} <1$$.

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 $$x-1<-1, x-2<-2$$. So
$$\frac{-1}{2}<\frac{1}{x-2}$$ and $$\frac{x-1}{x-2}<\frac{-1}{x-2}$$.

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$$\frac{1}{x-2} <-1$$ 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.