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?
