In Tom Apostol's Calculus I book, theorem $1.31$ says
If three real numbers $a, x$ and $y$ satisfy the inequalities ($1.14$) $$a \leq x \leq a + \frac yn$$ for every interger $n \geq 1$, then $x=a$.
The proof in the book goes like such
If $x>a$, Theorem $1.30$ tells us that there is a positive integer $n$ satisfying $n(x-a) >y$, contradicting ($1.14$). Hence we cannot have $x>a$, so we must have $x=a$
Theorem $1.30$ states that
If $x>0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx>y$
I first thought that he proved theorem $1.31$ by contradiction. Theorem $1.31$ can be written logically as $\forall a, x, y \in \mathbb{R}:(\forall n \in \mathbb{Z}^+ : a \leq x \leq a + \frac yn) \rightarrow x=a$ (I got this logic statement from this answer) and the negation of it is $\exists a,x,y \in \mathbb{R} : (\forall n \in \mathbb{Z}^+ : a \leq x \leq a + \frac yn \wedge x \neq a$. So I don't really understand how he proves theorem $1.31$ or what proof method he uses.
I'd appreciate if it can be explained what proof method he uses (and how he uses it) and if there's another way to prove theorem $1.31$ using algebra.
I am assuming you are comfortable with theorem 1.30. The statement you have to prove is: If $a ≤ x ≤ a + \frac{y}{n}$ for every integer $n ≥1$ , then x = a. That is $$ ∀ a,x,y ∈ \mathbb R ((∀n ∈ \mathbb N ( a≤x≤a+ \frac{y}{n}) ) \implies x=a)$$ Now we shall prove this statement using contradiction. I am proving it will more details.
Proof: Assume to the contrary that the conclusion of the statement is false, i.e. for some integer$n≥1$, $a ≤ x ≤ a + \frac{y}{n}$ and $x ≠ a$. Since $x≠a$, thus we have two cases namely when $x>a$ and $x<a$.
Case (1) : If $x<a$, then we wil get a contradiction as $x≥a$.
Case (2) : If $x>a$ that is same as $x -a > 0$, then by theorem 1.30 there exist a positive integer $n$ s.t $n(x-a) > y$, for every real real number $y$. This implies $x > a + \frac{y}{n}$, this is a contradiction since $x ≤ a+\frac{y}{n}$. As in each case we got a contradiction, hence our assumption is false and the statement is true.