Questions about proofs To prove, e.g., the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$, I remembered working, in high school, in the following way. Expanding the LHS gives
\begin{equation}
(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2,\qquad(1)
\end{equation}
while, expanding the RHS gives
\begin{align}
(ac-bd)^2+(ad+bc)^2&=a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2abcd\\
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2,    \qquad(2)
\end{align}
Comparing the RHS of (1) and (2) one deduces that $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$.
First question: Is this considered an acceptable proof at a post-secondary level? Or should one work differently, like, for instance, by completing the square
\begin{align}
(a^2+b^2)(c^2+d^2)&=a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2c^2+a^2d^2-2abcd+b^2c^2+b^2d^2+2abcd\\
&=(ac-bd)^2+(ad+bc)^2
\end{align}
As another example consider the proof that $\sqrt{2}+\sqrt{6}<\sqrt{15}$. Teachers in secondary school (and even some lecturers in engineering schools I know) usually work like this:

*

*Taking the square of each side of $\sqrt{2}+\sqrt{6}<\sqrt{15}$  gives  $2+6+2\sqrt{2}\sqrt{6}<15$.

*Rearranging gives $2\sqrt{2}\sqrt{6}<7$.

*Squaring again gives $48<49$.

*Since $48<49$ then $\sqrt{2}+\sqrt{6}<\sqrt{15}$.

Nevertheless, according to A Concise Introduction to Pure Mathematics of Martin Liebeck, previous argument is not a proof. Indeed, citing the author We have shown that if P is the statement we want to prove, and Q is the statement that 48 < 49, then P⇒Q; but this tells us nothing about the truth or otherwise of P.
The proper proof starts by supposing the veracity of the contrary
\begin{equation}
\sqrt{2}+\sqrt{6}\geq\sqrt{15}
\end{equation}
We have then
\begin{align*}
\sqrt{2}+\sqrt{6}\geq\sqrt{15}&\Rightarrow \left(\sqrt{2}+\sqrt{6}\right)^2\geq\left(\sqrt{15}\right)^2\Rightarrow 2+6+2\sqrt{2}\sqrt{6}\geq 15\\
&\Rightarrow 2\sqrt{12}\geq 7\Rightarrow 4\times 12\geq 49\Rightarrow 47\geq 48
\end{align*}
that is a contradiction.
Second question: Is this lack of rigor on the part of teachers justified?
 A: Your first proof is good, and contains no structural problems whatsoever. If I were presenting it to kids who had just learned how to expand binomials, I might include extra steps. If I were including this in a research paper, I might skip some of these steps (possibly all of them!). But, unlike the second proof, this method of proof (manipulating one side of the proposed equation, then the other) is a perfectly valid way of proving an equality.
I agree with Liebeck about the second proof: it contains a fallacy known as "affirming the consequent". This is a very common problem in beginners' proofs, where it's often much easier to begin with the conclusion (as there's often more to simplify there) and work back to the premises. Here, there isn't really anything in the premises, but the conclusion, $\sqrt{2} + \sqrt{6} < \sqrt{15}$, can be manipulated. You can see the temptation here!
Proving by contradiction, as Liebeck suggests, is one way to handle this. Another is to simply write the steps in reverse order! It is sufficiently obvious that $48 < 49$. If we take the square root of both sides (note: the square root function is strictly increasing, so it preserves even strict inequalities), then we get
$$2\sqrt{2}\sqrt{6} = \sqrt{48} < \sqrt{49} = 7.$$
Adding $8$ to both sides, this becomes
$$(\sqrt{2} + \sqrt{6})^2 = 2 + 2\sqrt{2}\sqrt{6} < 15.$$
Once again, taking the square root of both sides, bearing in mind that $\sqrt{2} + \sqrt{6} \ge 0$, we get
$$\sqrt{2} + \sqrt{6} = \sqrt{(\sqrt{2} + \sqrt{6})^2} < \sqrt{15}.$$
And we are done!
It's the same steps, but written backwards, maybe with a little exposition in between. This is why, even though the teacher has committed a fallacy, it actually kinda works in the end. They've written a perfectly valid proof, just in the wrong order. The steps presented should have been in the teacher's scratch work, but when it came time to present the proof, they should have presented it as I did above.
That said, when teaching students about proofs, it's sometimes (often?) not a good idea to present only the polished, finished proof, and this is a perfect example of why. Imagine you're in the class, the teacher claimed that $\sqrt{2} + \sqrt{6} < \sqrt{15}$, and then said "It is sufficiently obvious that $48 < 49$....". You would spend the whole time trying to figure out how the teacher pulled this inequality out of thin air, and it tells you nothing about how the teacher actually figured out how to prove this in the first place.
So, it's sometimes better that a teacher presents a proof that's less than perfect. And, given time constraints, it's sometimes not possible that they also present the finished product. So, for that reason, I'd say it is justified, for a teacher under time pressure, to present a proof like this. I would just hope that they would also explain to their students that this is the scratchwork for a proof, not the finished product itself (and perhaps, how they'd go about turning it into the finished product).
