I'm working on introductory chapters of Lang's "A First Course in Calculus", and in the part on inequalities I didn't quite understand why, when the solution of an inequality consists of more than one interval, the answer is sometimes written with the conjunction "and", and sometimes with the conjunction "or". For example:
For the inequality $|x-5|>2$, at the end of the book Lang gives the answer "$x<3$ or $x>7$", for the inequality $(x-1)(x+1)>0$ the answer is "$x<-1$ or $x>1$", but for $|x-3|>7$ it is "$x<-4$ and $x>10$". Is there a substantial difference between these?
Thank you in advance.
Let's just use the inequality $|x - 5|>2$, for which the solution set consists of any $x$ such that $x \lt 3$ OR $x > 7$, that is $$x \in (-\infty, 3) \cup (7, +\infty).$$
Technically speaking, the "or" is correct here, since we are talking about a solution set that is the union of sets. So there is no difference in any of the inequalities you post: they should all be "or's.
I suspect the author got a wee bit sloppy when he used the word "and" in the last example. To be clear about this: if the solutions set is given by any $x$ such that $x\lt -4$ AND $x>10$, then no solution would exist, because there is no $x$ satisfying both constraints. We'd have the solution set $$(-\infty, -4)\cap (10, +\infty) = \emptyset$$
Note that in the case that a solution to an inequality is any $x$ such that $-1\lt x \lt 1$, for example, then the term "and" is needed: $x$ must be greater than $-1$ AND it must be less than $1$: $$x \in (-1, +\infty) \cap (-\infty, 1) = (-1, 1)$$