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.

  • 2
    $\begingroup$ You are given that $x\geq a$ and you want to exclude the possibility that $x>a$. To this end, if $x>a$, then Theorem 1.30 implies the existence of a positive integer $n$ such that $n(x-a)>y$, contrary to the assumption that $k(x-a)\leq y$ for all positive integers $k$. $\endgroup$
    – Andrew
    Jul 7, 2023 at 7:19
  • $\begingroup$ I don't find it particularly helpful to think about proofs in a rigid logical structure. However, to be formal, what's going on here is really a proof by contrapositive. It becomes clear if you state the claim as: Suppose $x\geq a$. If $x\leq a+1/n$ for all $n$, then $x=a$. $\endgroup$
    – Andrew
    Jul 7, 2023 at 7:22
  • $\begingroup$ I'm not sure if there is an actual definition of proof by contradiction, and if there is, most in mathematics are not concerned with it. For me, proof by contradiction means you assume AND USE the hypothesis, while assuming the conclusion is false and deriving a logical inconsistency. However, many authors will conclude a proof by contrapositive with "contrary to our hypothesis" or "a contradiction." I suppose that they are not necessarily saying that it was a proof by contradiction, it is just an elegant (though highly misleading) way of concluding. $\endgroup$
    – Andrew
    Jul 7, 2023 at 7:25

1 Answer 1


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.

  • 1
    $\begingroup$ Shouldn't the statement be written as $\forall a, x, y \in \mathbb{R}:(\forall n \in \mathbb{Z}^+ : a \leq x \leq a + \frac yn) \rightarrow x=a$ instead of $∃a,x,y ∈ \mathbb R ∀n ∈ \mathbb N ( a≤x≤a+ \frac{y}{n} \implies x=a)$? I got this from this question $\endgroup$ Jul 7, 2023 at 12:33
  • $\begingroup$ Yes you are correct. I have made a mistake, will edit it. $\endgroup$
    – Afzal
    Jul 7, 2023 at 15:36
  • $\begingroup$ We got those 2 cases because $x \neq a$ correct? $\endgroup$ Jul 10, 2023 at 20:53
  • $\begingroup$ @Mohammadmuazzamali Yes. One case ( when $x<a$) is trivial so author left it and worked on other. $\endgroup$
    – Afzal
    Jul 11, 2023 at 0:34

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