The map $x \mapsto (x + 2/x)/2$ is a contraction of $[1,2]$ with fixed point $\sqrt{2}$, and the a priori bound gives the error after $n$ steps

Let $X := [1,2] \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y) = |x-y|$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and define

$f(x) \;:=\; \frac{x + 2/x}{2} \;=\; \frac{x}{2} + \frac{1}{x} .$

Then:

1. $(X,d)$ is a nonempty complete metric space (Complete metric space: every Cauchy sequence converges in the space).
2. $f$ maps $X$ into $X$.
3. $f$ is a contraction with constant $q = 1/2$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction).
4. $f$ has exactly one fixed point in $X$, namely $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$), and the iterates $x_{n+1} = f(x_n)$ converge to it from every starting point $x_0 \in X$ (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point).
5. Starting from $x_0 = 1$ one has $x_1 = 3/2$ and $d(x_1,x_0) = 1/2$, so the a priori bound of The a priori bound $d(x^{*}, x_n) \le q^n d(x_1,x_0)/(1-q)$ and the a posteriori bound $d(x^{*}, x_{n+1}) \le q\,d(x_{n+1},x_n)/(1-q)$ reads $\big|\sqrt{2} - x_n\big| \;\le\; \frac{(1/2)^n \cdot (1/2)}{1 - 1/2} \;=\; \Big(\tfrac{1}{2}\Big)^{n},$ and the a posteriori bound reads $\big|\sqrt{2} - x_{n+1}\big| \le |x_{n+1} - x_n|$.

• No derivative is used anywhere. The contraction constant comes from the algebraic identity of step 1.4, $f(x) - f(y) = (x-y)\big(1/2 - 1/(xy)\big)$, and from the bound $1 \le xy \le 4$ that the interval supplies. This matters here: the mean value theorem, which is how this estimate is usually obtained, belongs to a later page of this library and is not available at this point.
• The lower endpoint is what makes the estimate close. The bound $|1/2 - 1/(xy)| \le 1/2$ of step 2.2 needs $xy \ge 1$, and that is exactly what $x, y \ge 1$ supplies; on a set of small positive reals the factor $1/(xy)$ is large and the same computation gives nothing. The self-mapping property uses both endpoints: step 1.3 gives the lower bound $\sqrt 2$ and step 2.1 the upper bound $2$.
• What the bounds say numerically. From $x_0 = 1$ the a priori bound guarantees $|\sqrt 2 - x_n| \le 2^{-n}$, so ten steps give an error at most $2^{-10}$. This is a guarantee, not the truth: the iteration is Newton's method for $x^2 - 2$, and it converges far faster than the bound admits. The a posteriori bound of claim 5, which costs one subtraction, is what a computation would actually report.
• The restriction to $[1,2]$ is convenience, not necessity. On $[1,\infty)$ one still has $x,y \ge 1$, hence $0 < 1/(xy) \le 1$ and $|1/2 - 1/(xy)| \le 1/2$, so the same computation makes $f$ a contraction with the same constant there, and step 1.3 already shows $f$ maps $[1,\infty)$ into itself. The shorter interval is used only to keep the arithmetic of step 2.2 explicit.