# Error analysis - Bisection algorithm

I have a brief question related to an example in my textbook. In my book, the following theorem on Bisection Method is presented:

If $[a_0,b_0], [a_1,b_1],. . .,[a_n,b_n]. . .$ denote the intervals in the bisection method, then the limits $\lim_{n \to \infty} a_n$, and $\lim_{n \to \infty} b_n$ exist, are equal, and represent a zero of $f$. If $r=\lim_{n \to \infty} c_n$ and $c_n = \frac{1}{2}(a_n + b_n)$, then

$$|r-c_n| \leq 2^{-(n+1)}(b_0 - a_0)$$

Next, there is the following example:

Suppose that the bisection method is started with the interval $[50,63]$. How many steps should be taken to compute a root with relative accuracy of one part in $10^{-12}$?

OK, so if I were going to solve this, I would have used the theorem above and thought that we must have:

$$2^{-(n+1)}(63-50) \leq 10^{-12}$$

and then solve this for $n$. However, the book example says:

The stated requirement on relative accuracy means that

$$|r-c_n|/|r| \leq 10^{-12}$$

We know that $r \geq 50$, and thus it suffices to secure the inequality

$$|r-c_n|/50 \leq 10^{-12}$$

By means of the theorem above, we infer that the following condition is sufficent:

$$2^{-(n+1)}\cdot (13/50) \leq 10^{-12}$$

Solving this for $n$, we conclude that $n \geq 37$.

OK, so what I don't understand here is why the example begins by writing $|r-c_n|/|r| \leq 10^{-12}$ instead of just $|r-c_n| \leq 10^{-12}$. What is the motivation for including the $|r|$ in the denominator on the left side of the inequality?

If someone could explain this to me, I would be very grateful!

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