Trouble understanding absolute value inequalities

Question

I am working on epsilon delta limit proofs and am having trouble understanding inequalities with absolute value.

here are a few examples:

if $7 < x + 4 < 9$ than $|x + 4| < 9$

I do not fully understand how they infer this. What if x is $0$? Then $|0 + 4| < 9$ but $0 < 7$, and therefore not $7 < 0 + 4 < 9$.

2 other examples I have come up with but and am having trouble thinking about are:

1. say we have $ x + 4 > 9$. What can we say about $|x + 4|$ ?

2. say we have $-4 < x + 2 < -2$, what can we say about $|x + 2|$?

Answer 1

The first question is simply a propositional logic issue. You have a statement: $[7<x+4<9] \to [|x+4|<9]$, where I've put each of the mathematical expressions in brackets for clarity. Now remember that a proposition $A\to B$ is true whenever $A$ is false, regardless of whether $B$ is true or not. In your test case, $x=0$, we get $[7<0+4<9] \to [|0+4|<9]$, which we can simplify to $False \to True$. Since the expression on the left side of the arrow is false, it doesn't matter what the right side is, the whole expression will be true. I could have written $[7<0+4<9] \to \text{I am an alien from Mars}$ and the statement would still be true.

As for all of your other examples, in all of the cases you have chosen, the range where the expression is true is either always positive or always negative.

• If we look at $x+4>9$, all $x$ which satisfy this inequality are positive numbers. Thus, for all $x$ which satisfy that inequality, $|x+4|=x+4$. This makes it easy to see that $|x+4|>9$
• If we look at $-4 < x + 2 < -2$, all $x$ which satisfy this inequality are negative numbers. Thus, for all $x$ which satisfy that inequality, $|x+2|=-(x+2)$. If we take the original equation and multiply each side by -1, we get $2 < -(x+2) < 4$ (note: multiplying all sides by a negative number reversed the direction of each of the signs). Thus, we can say $2 < |x+2| < 4$

Answer 2

1. In this case, $|x+4|=x+4$, so it's larger than $9$ too.
2. In this case, $|x+2|=-(x+2)$, so $$2<|x+2|<4$$ But we can say more: as $\;-4<x+2<-2\iff -6<x<-4$, we can say $$|x-5|<1.$$