Interval bisection to find a root of f(x) I'm attempting to understand Interval bisection. I'm given a simple question in my textbook, and I can do the process easily, I just don't know when to stop. The question is "Use Interval bisection to find the positive root of $x^2 - 7 = 0$, correct to one decimal place" (basically find the square root of 7 to 1 dp)
This is the solution I'm given:

How is it known that it is 2.6?
The last line shows that the root is between 2.640625(from (a+b)/2) and 2.65625(from b).
2.640625 rounds to 2.6 but
2.65625 rounds to 2.7
Surely I would have to keep going until both the upper and lower limit of the interval round to 2.6? 
If it's just simple truncation why didn't the solution stop on the second last line?
(this is just a simple question, so it is as if you can't just do root 7 on a calculator)
 A: I would share your concern that this is not complete.  As David Mitra says, one further iteration is needed. 
In fact since at the third line you know $$2.625^2-7 \lt 0 \lt 2.75^2-7,$$ and so you know the rounded answer is $2.6$ or $2.7$ and could save some effort by checking $$2.65^2-7=+0.0225,$$ so $2.6$ is the rounded solution. 
A: Judging from your comment under hardmath's answer (which it would have been a good idea to include in the original question), you're right that the solution is incorrect. If the task is to determine the result of rounding $\sqrt7$ to one decimal place, then, as you say, you'd need to keep going until both bounds of the interval yield the same result upon rounding.
A: It seems likely that the original problem was to repeat the bisection until the length of the enclosing interval was less than half of $0.1$.  That would be one interpretation of finding the root within "one decimal place".
Note that the lengths of the enclosing intervals decrease regularly.  You started with an interval of length $1$, and divided the length by $2$ with each iteration.  Therefore at the sixth line of the table (fifth bisection), the interval $[a,b]$ has been reduced to length $2^{-5} \lt 0.05$, and this is the earliest step at which that criterion is met.
