Question: Starting from an initial interval (2, 3) containing 5, we will look for consecutive pairs in tenths, hundredths, thousandths and so on that can

Starting from an initial interval (2, 3) containing 5, we will look for consecutive pairs in tenths, hundredths, thousandths and so on that can bound 5 (See Strategy 1 in Chapter 3, Day 3 lecture). Then we will choose the lower bound of the interval for the next approximation. For instance, let's call a = 2 be the first approximation and a = 2.2 be the second approximation determined by the lower bound among the two consecutive tenths (2.2, 2.3) that bounds 5 (2.2

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