I just explained in a DM how math proofs must be read. I think it is a nice, though very simple example of why it is a big deal to proofread scientific texts. It is not as if you could read them like an essay. Here is my example:
Proposition: The square root of a prime number is an irrational number.
Proof: If there is a prime number ##p## such that ##\sqrt{p}\in\mathbb{Q}##, then we have an equation
$$
\sqrt{p} = \dfrac{r}{s}\quad (*),
$$
where we may assume that ##r,s## are coprime. Hence, ##p \cdot s^2=r^2.## Thus ##p## divides ##r,## say, ##r=q\cdot p.## Then ##r^2=q^2\cdot p^2## and ##s^2=q^2\cdot p.## This, however, implies that ##p## divides ##s## and ##r/s## wasn't coprime, contradicting our assumption. This means that the equation ##(*)## does not exist and ##\sqrt{p}\not\in \mathbb{Q}.\quad\blacksquare##
This is one of the shortest proofs I know. Yet, it is full of questions whose answers are not included: Why can we make those assumptions, the existence of such a ##p##, the fraction, coprimality? Where did we use that ##p## is prime? It must have been used, since ##\sqrt{36}## is definitely rational.
Now for the extended version of how it has to be read:
Proof: If there is a prime number ##p## such that ##\sqrt{p}\in\mathbb{Q},## ...
[Indirect proof by contradiction. We want to show non-existence, so we assume existence and require that our conclusions will inevitably lead to a contradiction. Since FALSE cannot follow from TRUE, the assumption had to be FALSE.]
... then we have an equation
$$
\sqrt{p} = \dfrac{r}{s}\quad (*),
$$
[By the definition of rational numbers, and our assumption.]
where we may assume that ##r,s## are coprime.
[Why may we assume that? Imagine ##\sqrt{p}=\dfrac{r'}{s'}## and ##r',s'## weren't coprime. This means that there is an integer ##t'## such that ##r'=r''\cdot t'## and ##s'=s''\cdot t'.## Next, we may cancel ##t'## and receive an expression ##\sqrt{p}=\dfrac{r'}{s'}=\dfrac{r''\cdot t'}{s''\cdot t'}=\dfrac{r''}{s''}.## If ##r'',s''## are still not coprime, then we continue with that procedure. But why does it have to come to an end? ##r',s'## are finite numbers and as such can only have finitely many, proper (not ##\pm 1##) divisors. Every cancellation leads to a smaller value of ##|r''|,|s''|## by the Euclidean division since we excluded ##\pm 1.## Note that we used that the integers are a Euclidean domain. (Maybe there are other ways to show that the procedure has to halt. I just took what I had anyway.) But the positive integers ##|r''|,|s''|## are bounded from below by ##0,## so we cannot go on forever. At the end, we have a coprime representation in ##
(*).##]
Hence, ##p \cdot s^2=r^2.##
[Simple algebra.]
Thus ##p## divides ##r,## ...
[Why? This is where we use that ##p## is prime. We have ##p \,|\,p\cdot s^2=r\cdot r## and a prime number is one, that whenever it divides a product, in our case ##r\cdot r,## then it has to divide one of its factors. We have only the factor ##r,## so ##p## has to divide ##r.## This argument fails for non-prime numbers. E.g., ##36\,|\,12\cdot 12## but ##36\,\nmid\,12.## Pretty hidden, isn't it?]
... say, ##r=q\cdot p.## Then ##r^2=q^2\cdot p^2##
[Algebra.]
and ##s^2=q^2\cdot p.##
[Substitution plus cancellation of the common factor ##p.## Note that the cancellation can only be done since ##\mathbb{Z}## is an integral domain. We have ##a\cdot x= b\cdot x.## How can we conclude ##a=b##? We first get ##a\cdot x- b\cdot x=(a-b)\cdot x=0.## If there are no zero-divisors, we may conclude ##x=0## OR ##a-b=0## which means, if ##x\neq 0## that ##a=b.## That's how cancellation works. It does not automatically work if there are zero-divisors. Consider ##\mathbb{Z}/6\mathbb{Z}.## Then ##2\cdot 3=0## although ##2\neq 0## and ##3\neq 0.## ]
This, however, implies that ##p## divides ##s## ...
[By the same argument as before. We use that ##p## is prime and divides a product, this time ##s\cdot s.##]
... and ##r/s## wasn't coprime, contradicting our assumption.
[This is the contradiction we needed: (##r,s## are coprime) AND (##r,s## have a common factor ##p##) is FALSE. Note that we used ##p>1## here. We use that units cannot be primes!]
This means that the equation ##(*)## does not exist ...
[as we falsely assumed]
... and ## \sqrt{p}\not\in \mathbb{Q}.##
[Again, by definition of ##\mathbb{Q}.##]
##\blacksquare##
Sure, this example is more or less trivial. Nevertheless, it shows all the hidden properties of primes and integers that I have used without mentioning. One sees that the written-out thoughts are significantly longer than the proof itself. Now, that happens with every scientific text, only that the parts in brackets are way longer, often requiring looking up many references, scribblings to convince the reader, and so on. This is why it is not a matter of kindness to read a paper. It is hard work!