So the lemma I had trouble understanding was this:
For all real numbers r, $−|r| ≤ r ≤ |r|$
The solution was something like this(used the definition of absolute values + division into cases)
Suppose r is any real number. We divide into cases according to whether r ≥ 0 or r < 0.
Case 1 (r ≥ 0):
$|r| = r$. (By definition of absolute value)
$−|r| < r$.(since $r$ is positive and $−|r|$ is negative)
Thus it is true that $−|r| ≤ r ≤ |r|$.
Qn: But based on what it said above, shouldn't it be more of $−|r| < r$ and $r=|r|$ not $−|r| ≤ r ≤ |r|$?
Case 2 (r < 0):
$|r|=−r$. (By definition of absolute value)
Multiplying both sides by −1 gives that $−|r| = r$.
$r < |r|($since $r$ is negative and $|r|$ is positive)
Thus it is also true in this case that $−|r| ≤ r ≤ |r|$.
Qn: But based on what it said above, shouldn't it be more of $r < |r| $ and $-r=|r|$ not $−|r| ≤ r ≤ |r|$ ?
Hence, in either case, $−|r| ≤ r ≤ |r|$
Is the answer to my question in anyway linked to abusing inequalities?
Eg: Something like $1≤2$, although correct, it is imprecise. (sorry if this is totally off)
Case $(1)$:
We have $$-|r|< r =|r|$$
but $-|r| < r$ would imply that $-|r|\le r$ and $r=|r|$ would imply that $r \le |r|$.
Hence we can write $-|r| \le r \le |r|$.
Note that $\le $ means less than or equal to.
Similarly for case $(2)$.
Explicit numerical example:
If $r=2$, then we have $-2 < 2 = 2$, we also have the inequalities $-2 \le 2 \le 2$.
If $r=-2$, then we have $-2 = -2 < 2$, which implies that $-2 \le -2 \le 2$.
Notice that for $-|r| \le r \le |r|$, equality is attained on at least one side. If both equalities are attained, then we have $r=0$.