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?

• Regarding squaring both sides of an equation, my comments to this question are relevant. (Also, somewhat related are my comments to this answer). Incidentally, a "school-level proof" that the squaring function for positive inputs is injective (i.e. is a one-to-one function) is the fact that the right half of the graph of $y=x^2$ (i.e. the graph of $y=x^2$ for $x \geq 0)$ satisfies the horizontal line test. Nov 27, 2021 at 14:59
• To continue, what the teachers and authors should do in the case of squaring is say something like "because both sides are positive" (if the teacher/author feels that only "a reminder" is needed) or say something like "which is reversible because both sides are positive" (if the teacher/author feels a bit more is needed to be said). Of course, this assumes that the intended audience will know the relevance of saying this, but unless one is specifically trying to explain the underlying ideas (rather than simply solving/proving something), I think saying more is too much of a tangent. Nov 27, 2021 at 15:17
• Closely related: matheducators.stackexchange.com/questions/12586/… Nov 27, 2021 at 15:20
• Actually you should say what a,b,c,d are. So say 'let a,b,c,d be element in R...'. Or maybe some arbitrary commutative field (I think commutative Ring is enough though). If this sounds nitpicking, are you sure this proof goes through in non -commutative fields like the Quarternions? Maybe yes, maybe no, but one of the steps would not hold like you wrote it. You do not need to prove it for the most general case, but you shd say for which case you prove it. Nov 27, 2021 at 16:55

In general, it is very difficult to answer the question about whether or not the lack of rigor is "justified", as the ultimate goal is competence and literacy in math and logical reasoning, so what is justified for one student may not be justified for another, as each student's situation and current understanding will be different. What constitutes something being justified on the part of the teacher is whether the teacher can correctly identify the gaps and poor habits on the side of the student, and then choose the best method of explaining and fixing them.

That's the ideal case anyways. In the case of teaching a class containing many students, often times the teacher has a goal that they need to meet and cannot cater to every single student at every single step to make sure they don't fall behind. The teachers make their best effort, but their time and resources are limited. There is always some push and pull there. Putting these generalities aside, I'll respond to the specific questions now.

The first question of proving $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$$ is pretty straightforward. There's not that much more you can do to make it more rigorous without getting into some very technical machinery that is not expected of students until they reach formal proof-writing courses in university. This is mostly because everything involves equality and basic arithmetic properties like distributive and commutative.

The second question is significantly trickier. The proof by contradiction method certainly is rigorous and one way of approaching the problem, but the other method shown can also be made rigorous if one carefully constructs the reasoning used to justify each statement.

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

One can prove that if $$a$$ and $$b$$ are both positive, and $$a < b$$, then $$\sqrt{a} < \sqrt{b}$$. If this statement is proved, then it follows that as long as $$2+6+2\sqrt{2}\sqrt{6}<15$$ is true, then $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$ must also be true.

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

One can also show that if $$a, then $$a+c, where all the variables are real numbers. This would allow for the "rearranging" of values being done here.

Once these intermediate concepts are proven, they can be used as the building blocks of proving this statement in this way, but as you can see, this involves a lot more steps if one chooses to take this approach in order to be rigorous. Most of the time, when teachers ignore these steps to make this rigorous, they are assuming that the students already understand the truth of the concepts being applied and are able to intuitively understand the logic behind it.

The reason why rigor is important for problems like these is that students need to be careful not to build bad habits and apply reasoning in fallacious ways. For instance, just because $$a > b$$ does not imply that $$a^2 > b^2$$ (assuming real values). Just because $$a^2 > b^2$$ does not imply that $$a > b$$. There are important restrictions to $$a$$ and $$b$$ in order for these statements to hold true, specifically that they both must be positive.

But even if we understood that given positive real values, $$a > b$$ implies that $$a^2 > b^2$$, that doesn't mean that this is the reason why we can square both sides of an inequality. Instead of proving $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$ by showing that $$2+6+2\sqrt{2}\sqrt{6}<15$$ is true, we would be assuming $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$ to prove that $$2+6+2\sqrt{2}\sqrt{6}<15$$ is true, the opposite of what we are trying to do.

As you can see, it is very easy to slip and make wrong arguments when writing proofs. The point of rigor is making sure that we aren't just loosely applying things that feel right but aren't actually valid. Whether it is justified or not depends on if the teacher is conveying these concepts properly, and if it is within the purview of the teacher's goals.

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?

It's perfectly valid to meet in the middle like this.

It's even valid to start with the desired result then work towards $$1=1$$ or some established identity, as long as the logical connective between each step is justifiably and explicitly displayed as $$'{\iff}'.$$ (Typically, the implicit punctuation between lines is “therefore”, in which case—since anything is a consequence of pigs flying—it is invalid to start with the desired result.)

Second question: Is this lack of rigor on the part of teachers justified? Teachers in secondary school (and even some lecturers in engineering schools I know) usually work like this:

• Taking the square of each side of $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$
• Rearranging gives $$2\sqrt{2}\sqrt{6}<7$$.
• Squaring again gives $$48<49$$.
• Since $$48<49$$ then $$\sqrt{2}+\sqrt{6}<\sqrt{15}$$.

The proper proof starts by supposing the veracity of the contrary ..... that is a contradiction.

The proof-by-contradiction comparison is immaterial, since the above proof attempt is related to it in neither intent nor substance.

The above proof is actually fine and valid if

• it is explicitly noted that both values of each inequality being squared are nonnegative, and
• $$'{\iff}'$$ is explicitly displayed between steps.

$$(|p|<|q|\iff p^2

While these may not be stressed in writing, are they perhaps at least being communicated orally? It is also unclear whether their omission is deliberate or due to ignorance.

Echoing CosmoVibe, it's hard to categorically say aye or nay to your question, since the desirable level of rigour muchly depends on context and circumstances.

It's anyway easy and always valid to present just the required direction of argument after having used the 'cheat' direction to devise the proof's meat. This—just like filtering out extraneous solutions at the end of equation-solving—also sidesteps having to continually ensure that the $$'{\iff}'$$s are genuine.
• @Daniel R. Collins: In high school, for harder trig. identity proofs, I would often do them backwards and then reverse the order of the steps. A couple of my mathy friends also did this when they took the class (after I did, as I took it earlier), having discovered it independently (i.e. I didn't tell them what I had done), so I would imagine some others here did the same thing. Of course, this can lead to some highly non-intuitively discoverable proofs, where (for example) you might wind up replacing $\sin x$ with $1\cdot \sin x,$ followed by $(\sin^2 x + \cos^2 x)\cdot \sin x \; \ldots$ Nov 27, 2021 at 17:30
• @DaveLRenfro: For whatever reason, I have never needed to write a proof that goes via a chain of equivalences to something trivial like "$1=1$". And regarding doing proofs backwards, I think it is pedagogically more instructive to learn the concepts of canonicalization and reductions, which actually suffice all high-school trigonometry. But of course, I completely agree with the point that both of you make, namely that correct proofs must be recognized as correct even if it looks silly, and likewise wrong proofs shouldn't be accepted. =) Feb 11, 2022 at 7:41