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