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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: enter image description here

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)

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    Please post the exact wording of the question. Currently you've posted the part of the question that you already understand verbatim but only paraphrased the part of the question that you're asking about. That's a bad idea, since one likely cause of the problem is that you misunderstood the question, and we can't help you with that unless you quote the question. The phrase "to 1 decimal place" in the answer could refer either to $x^2-7=0$ or to $x=2.6$; in the former case, the solution would be correct, since the error in the equation rounds to $0.0$ for the first time in that iteration.2012-05-07
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    By the way, if this is an exact copy of the answer, there's another error in it; $\frac{f(a+b)}2$ should be $f\left(\frac{a+b}2\right)$; that's what the column actually contains.2012-05-07
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    @Joriki, sorry I managed to leave off the "correct to one dp", but now the question is exaclty what I've quoted.2012-05-07
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    Hmm, actually, you know the root is in $[2.640625, 2.65625]$; so indeed the root is $2.6$, correct to the first decimal place (although the root may be closer to 2.7 than to 2.6, for all we know at this point).2012-05-07
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    @David, I see how you got that, but it seems more common sense than mathematical. As if that was the way to get the answer then it should have stopped on the third line from the bottom when it is know to in [2.625, 2.6875]2012-05-07
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    It seems the solution is off no matter how you interpret it. If they mean "find the digit in the tenth's place of the decimal expansion of $\sqrt 7$", then, as you said, they've done too much work. If they mean "find $\sqrt 7$ with error at most $\pm0.05$", the answer"x=2.6" is not justified by the process, and another iteration is needed to justify it. Perhaps Joriki is right...2012-05-07

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