Htykuut

Here's what my textbook asks me to prove:

From algebra, recall that, if a prime $p$ divides a product $ab$, then
$p$ must divide either $a$ or $b$. Use this to prove that $sqrt{p}$
is irrational for any prime $p$.

I know the proof follows exactly like the proof of irrationality of $sqrt{2}$. But my question is, what if $b=a$? Then the statement would follow like "if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" which is false.

I wonder if I am doing some mistake here because the following video https://youtu.be/uQ6KSt94jVY seems to prove the same statement. I think it should be an "inclusive or" there.

Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.

Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.

"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.

The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.

for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.