# Question on the logic of a proof involving complex numbers

The title is a bit vague, but really boils down to not the steps of a proof--I've been able to reason through what I know, via reviewing the solution and the suggestion in the text, that the steps are correct--but 'why' they'e correct.

I want to show that, for any complex number, $z$, $\sqrt{2} \lvert z \rvert \geq \text{Re$z$} + \text{Im$z$}$. The suggestion in the text is to manipulate this expression until we end up with $\left(\lvert x \rvert + \lvert y \rvert\right)^2 \geq 0$, which is necessarily true.

I was able to find this, which concludes the proof. My question is, really, why this concludes the proof. I believe the idea is that, via manipulating the inequality, we conclude that this original inequality is true for all $x$ and $y$ that satisfy the inequality that the square of the sum of their absolute values is non-negative. Because that is necessarily true for any real numbers, this clearly holds for all reals and thus for all complex numbers, which are created from reals.

The other idea I had was that we could retrace our steps to produce the original expression, but that would require reversing one step that required us to square both sides: we maintained the inequality sign because both sides of the inequality were positive, but it need not be true, unless I'm mistaken, that we end up with a 'positive' result via reversing the inequality, so we need not end up with our original result.

I'd appreciate any insights on this problem. Thanks in advance. If necessary, I could write out the steps of the proof, especially if there is any merit to the second explanation from above.

Revision: I have revised this question to include the steps of this proof. \begin{align*} & \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Re $z$} \rvert + \lvert \text{Im $z$} \rvert \\ & \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \\ & 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \\ & 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2 \\ & \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert y^2 \geq 0 \\ & \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0. \end{align*}

• This sounds dubious: "manipulating an expression to get an obvious truth" is generally an unsound mode of inference. Jul 5 '18 at 2:07
• As long as the steps are individually reversible (check that as you proceed), you don't need to reverse it at the end. In other words, if all steps are can be obtained from the previous one using "if and only if", then it's legitimate to start with the conclusion and end with something which is obviously true. Jul 5 '18 at 2:10
• Perhaps edit your question to show the actual proof about which you have doubts. Jul 5 '18 at 2:13
• that would require reversing one step that required us to square both sides You can safely square (or take square roots) on both sides of an inequality if they are both non-negative, such as for example $\,2(x^2+y^2) \ge (|x|+|y|)^2\,$.
– dxiv
Jul 5 '18 at 2:16
• @Matt.P $a^2 \ge b^2 \iff |a| \ge |b|\,$, and if $\,a,b\,$ are known to be non-negative then you can drop the $\,|\,\cdot\,|\,$.
– dxiv
Jul 5 '18 at 2:33

As long as the steps are reversable then "Conclusion $\iff .... \iff .... \iff$ something obviously true" is valid as we could just as easily do it the other way "something obviously true $\iff ... \iff ... \iff$ Conclusion".
\begin{align*} & \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0\text{(which is obviously true)}\implies\\ & \lvert x \rvert^2 - 2 \lvert x \rvert \lvert y\rvert +|y^2| \geq 0 \implies\\ & 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2\implies \\ & 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \implies\\ & \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \implies\\ & \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Rez$} \rvert + \lvert \text{Im$z} \rvert \\ \end{align*}
\begin{align*} & \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Rez$} \rvert + \lvert \text{Im$z} \rvert \Leftarrow\\ & \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \Leftarrow\\ & 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \Leftarrow\\ & 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2 \Leftarrow\\ & \lvert x \rvert^2 - 2 \lvert x \rvert \lvert y \rvert +|y^2| \geq 0 \Leftarrow\\ & \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0\text{(which is obviously true)}. \end{align*}