I am asking this not so much to get the "right" answer, but more to understand the approach and thinking that the seasoned mathematicians of this board follow.
So, the problem
If $a,b \geq 0$ and ${a}^{2} < {b}^{2}$ then $a < b$. (Use (ix) backwards)
As a reminder, (ix) asks
If $0 \leq a < b$ then ${a}^{2} <{b}^{2}$ (Use (viii))
As a commenter once noted with Spivak, he leans quite heavily on previously proven problems ( which are probably axioms important to future use).
So the proof to this problem ( from Spivak's answer book) says
If $a < b$ were false, then either $a=b$ or $a>b$. But if $a=b$ then $a^{2} = b^{2}$ and if $a > b \geq 0$ then $a^{2} > b^{2} $ by part (ix)
So, here are my two questions.
(1) What part of that proof is the "backwards" part?
(2) Is it common to use a false assumption as a proof? (Or perhaps this is part of the "backwards" part).
I hope this all makes sense.
Best Answer
Spivak's proof simply takes cases and derives a contradiction by using (ix)—forwards, not backwards.
Here's a clearer proof, also invoking (ix), that actually does involve working backwards (note that $P\Rightarrow Q\equiv$ (not $P$) or $Q):$
We want to show that \begin{align}a \geq b\implies \text{not}\,\big(a,b \geq 0\;\text{and}\;a^2 < b^2\big),\end{align} which is equivalent to each of the following \begin{align}a < b\quad\text{or}\quad \big(a<0\quad\text{or}\quad b<0\quad\text{or}\quad a^2 \geq b^2\big)\\\text{not}\,\big(a > b\quad\text{and}\quad a>0\quad\text{and}\quad b>0 \big)\quad\text{or}\quad a^2 \geq b^2 \\0<b<a\implies a^2 \geq b^2 \\0<a<b\implies a^2 \leq b^2.\tag{*}\end{align}
$(*)$ is a consequence of $$0<a<b\implies a^2< b^2,$$ which is an immediate consequence of (ix).