| ▼ | 1 | Chapter 1 |
| 1.1 | Exercise 1.1 |
| 1.2 | Exercise 1.2 |
| 1.3 | Exercise 1.3 |
| 1.4 | Exercise 1.4 |
| 1.5 | Exercise 1.5 |
| 1.6 | Exercise 1.6 |
| 1.7 | Exercise 1.7 |
| 1.8 | Exercise 1.8 |
| 1.9 | Exercise 1.9 |
| 1.10 | Exercise 1.10 |
| 1.11 | Exercise 1.11 |
| 1.12 | Exercise 1.12 |
| 1.13 | Exercise 1.13 |
| 1.14 | Exercise 1.14 |
| 1.15 | Exercise 1.15 |
| 1.16 | Exercise 1.16 |
| 1.17 | Exercise 1.17 |
| 1.18 | Exercise 1.18 |
| 1.19 | Exercise 1.19 |
| 1.20 | Exercise 1.20 |
| 1.21 | Exercise 1.21 |
| 1.22 | Exercise 1.22 |
| 1.23 | Exercise 1.23 |
| 1.24 | Exercise 1.24 |
| 1.25 | Exercise 1.25 |
| 1.26 | Exercise 1.26 |
| 1.27 | Exercise 1.27 |
| 1.28 | Exercise 1.28 |
| 1.29 | Exercise 1.29 |
| 1.30 | Exercise 1.30 |
| 1.31 | Exercise 1.31 |
| 1.32 | Exercise 1.32 |
| 1.33 | Exercise 1.33 |
| 1.34 | Exercise 1.34 |
| 1.35 | Exercise 1.35 |
| 1.36 | Exercise 1.36 |
| 1.37 | Exercise 1.37 |
| 1.38 | Exercise 1.38 |
| 1.39 | Exercise 1.39 |
| 1.40 | Exercise 1.40 |
| 1.41 | Exercise 1.41 |
| 1.42 | Exercise 1.42 |
| 1.43 | Exercise 1.43 |
| 1.44 | Exercise 1.44 |
| 1.45 | Exercise 1.45 |
| 1.46 | Exercise 1.46 |
The sqrt-iter procedure using the new-if procedure never terminates.
When evaluating a function in applicative order, the operands are evaluated first, and then the function is applied to the evaluated operands. This means that, when evaluating a function using new-if, both branches of the conditional will always be evaluated, meaning that even when the recursion should end, it won’t. Thus the procedure will never terminate.