# Ideas in Geometry/Instructive examples/Section 3.2

3.2 #26 5/3 We have 2= 6/3 > 5/3 > 3/3 = 1 So 5/3 = 1+ (2/3) Here, 2 is the whole number part of 5/3 and 2/3 is the mixed number part. We cannot stop here because we do not have 1 in our numerator. To get 1 in the numerator we must work with its reciprocal so we get: 2+[1/(3/2)] Repeat the same steps We have 2= 4/2 > 3/2 > 2/2 = 1 So 3/2 = 1 + (1/2) 5/3 = 2 + [ 1/[1/(1/2)]] We can stop here because we have a 1 in the numerator after configuring 3/2