1. ## Absolute value proofs

I need to prove the implication,

$|a| < |b| \Rightarrow a^2 < b^2$ I know that $x^2 \mbox{is equal to} |x|^2$ but how can I use that in a formal proof?

Also I need to show that $|a|<|b| \Leftrightarrow a^2 < b^2$

Once again all I can think of is the thing I wrote above. Thank you!

2. Originally Posted by meg0529
I need to prove the implication,

$|a| < |b| \Rightarrow a^2 < b^2$ I know that $x^2 \mbox{is equal to} |x|^2$ but how can I use that in a formal proof?
note that an alternate definition for $|x|$ is $\sqrt{x^2}$. also note that the square root function is a strictly increasing function, that is, for positive $x$ and $y$, $x < y \implies \sqrt{x} < \sqrt{y}$.

Now, to prove our implication, we can use the contrapositive: assume $a^2 \ge b^2$, then we have $\sqrt{a^2} \ge \sqrt{b^2}$. But that means $|a| \ge |b|$.

Also I need to show that $|a|<|b| \Leftrightarrow a^2 < b^2$

Once again all I can think of is the thing I wrote above. Thank you!
in light of what i did before, try this one

3. Originally Posted by Jhevon
note that an alternate definition for $|x|$ is $\sqrt{x^2}$. also note that the square root function is a strictly increasing function, that is, for positive $x$ and $y$, $x < y \implies \sqrt{x} < \sqrt{y}$.

Now, to prove our implication, we can use the contrapositive: assume $a^2 \ge b^2$, then we have $\sqrt{a^2} \ge \sqrt{b^2}$. But that means $|a| \ge |b|$.

in light of what i did before, try this one

This might be really stupid, but by proving using contrapositive A -> B are we not proving B->A using inverse?

4. Why not just apply simple rules?
$\begin{gathered}
\left| a \right| < \left| b \right|\, \Rightarrow \,\sqrt {a^2 } < \sqrt {b^2 } \, \Rightarrow \,a^2 < b^2 \hfill \\
\hfill \\
a^2 < b^2 \, \Rightarrow \,\left| a \right|^2 < \left| b \right|^2 \, \Rightarrow \,\left| a \right| < \left| b \right| \hfill \\
\end{gathered}$