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sav08:proof_of_first_lecture01_example [2008/02/17 19:56] vkuncak |
sav08:proof_of_first_lecture01_example [2008/02/20 14:53] vkuncak |
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| f(x,y) = \left\{\begin{array}{rl} | f(x,y) = \left\{\begin{array}{rl} | ||
| 0, & \mbox{ if } y = 0 \\ | 0, & \mbox{ if } y = 0 \\ | ||
| - | 2 f(x,\lfloor\frac{y}{2}\rfloor), & \mbox{ if } y > 0, \mbox{ and } x=2k \mbox{ for some } k \\ | + | 2 f(x,\lfloor\frac{y}{2}\rfloor), & \mbox{ if } y > 0, \mbox{ and } y=2k \mbox{ for some } k \\ |
| - | x + 2 f(x,\lfloor\frac{y}{2}\rfloor), & \mbox{ if } y > 0, \mbox{ and } x=2k+1 \mbox{ for some } k \\ | + | x + f(x,y-1), & \mbox{ if } y > 0, \mbox{ and } y=2k+1 \mbox{ for some } k \\ |
| \end{array}\right. | \end{array}\right. | ||
| \] | \] | ||
| Line 33: | Line 33: | ||
| * **Inductive hypothesis.** Assume that the claim holds for all values less than $y$. | * **Inductive hypothesis.** Assume that the claim holds for all values less than $y$. | ||
| * Goal: show that it holds for $y$ where $y > 0$. | * Goal: show that it holds for $y$ where $y > 0$. | ||
| - | * **Case 1**: $y = 2k$. Note $k < y$. By definition | + | * **Case 1**: $y = 2k$. Note $k < y$. By definition and I.H. |
| \[ | \[ | ||
| f(x,y) = f(x,2k) = 2 f(x,k) = 2 (x k) = x (2 k) = x y | f(x,y) = f(x,2k) = 2 f(x,k) = 2 (x k) = x (2 k) = x y | ||
| \] | \] | ||
| * | * | ||
| - | * **Case 2**: $y = 2k+1$. Note $k < y$. By definition | + | * **Case 2**: $y = 2k+1$. Note $y-1 < y$. By definition and I.H. |
| \[ | \[ | ||
| - | f(x,y) = x + f(x,2k+1) = x + 2 f(x,k) = x + 2 (x k) = x (2 k + 1) = x y | + | f(x,y) = f(x,2k+1) = x + f(x,2k) = x + x \cdot (2k) = x (2 k + 1) = x y |
| \] | \] | ||
| This completes the proof. | This completes the proof. | ||