Absolute value inequality

[syncmaster913n]

I have this as the solution to a problem that I was doing:

-1 < x+1 < 1 leads to -2 < x < 0 so that 1 < |x-1| < 3

And I'm unable to understand last representation results from the second one. Can someone please explain where 1 < |x-1| < 3 came from?

 

[pka]

I have this as the solution to a problem that I was doing:
-1 < x+1 < 1 leads to -2 < x < 0 so that 1 < |x-1| < 3

Why not do this?
\(\displaystyle \begin{align*}-1 &<x+1<1 \\-3 &<x-1<-1\\&|x-1|<3 \end{align*}\)

 

[syncmaster913n]

This seems to be the part that I'm not understanding. How do we get from:

-3 < x-1 < -1

to:

|x-1| < 3

I mean, if I were to 'expand, the latter, it would become:

-3 < x-1 < 3

Which is not the same as the original (-3 < x-1 <-1).

What am I missing?

 

[pka]

This seems to be the part that I'm not understanding. How do we get from:-3 < x-1 < -1 to:
|x-1| < 3
What am I missing?

If \(\displaystyle x< 1 \) then \(\displaystyle x< 10 \). By logic that is a true statement.

So if \(\displaystyle -1<x<1 \) then \(\displaystyle -4<x<2 \). By logic that is also a true statement.

Thus if \(\displaystyle -3<x-1<-1 \) then \(\displaystyle -3<x-1<3 \), by the same logic that as above. Moreover, it is equivalent to \(\displaystyle |x-1|<3 \)

 

[HallsofIvy]

You may be, without realizing it, trying to conflate an "if then" statement to an "if and only if" statement.

Notice that if -3< x- 1< 1 then |x- 1|< 3.

It is not true that "if |x- 3|< 3 then -3< x- 1< 1" but that is not claimed.

 

[syncmaster913n]

Thank you for explaining, pka.

 

[syncmaster913n]

You may be, without realizing it, trying to conflate an "if then" statement to an "if and only if" statement.

Notice that if -3< x- 1< 1 then |x- 1|< 3.

It is not true that "if |x- 3|< 3 then -3< x- 1< 1" but that is not claimed.

Yes, I think that was it. Thanks to you too.