Please help me with the following proof:
Show that if $|x-1|<1$ and $|x-1|<r/4$, where $r \in \Bbb R^+$, then $|x^2+x-2|<r$.
This is the solution that is in my textbook:
First, observe that $$|x^2+x-2|=|(x+2)(x-1)|=|x+2||x-1|.$$
By Theorem 4.17, $$|x+2|=|(x-1)+3|\leq|x-1|+|3|<1+3=4.$$
Therefore, $$|x^2+x-2|=|x+2||x-1|<4(r/4)=r.$$
For reference, Theorem 4.17 is the following: For every real numbers $x$ and $y$, $|x+y|\leq|x|+|y|.$
I understand everything but the second half of the second part of the proof. Where does $1+3$ come from and how is it related to $|x-1|+|3|$?
Thanks.
$2 = -1 + 3$ I hope that is obvious.
We did this because we know something about $x-1$
This one:
$|a+b| \le |a|+|b|$ is called the triangle inequality.
You can't make a triangle if the two shorter sides are not at least as long as the longest side. Or, the shortest distance between two points is a straight line.
You probably should prove this to yourself. The most direct way to prove such a thing would be to create 4 cases.
Both a,b are positive.
a is positive and b is negative
a is negative and b is positive
both are negative.
A more elegant proof would be to square both sides.