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sav08:proof_of_first_lecture01_example [2008/02/19 14:15] vkuncak |
sav08:proof_of_first_lecture01_example [2015/04/21 17:30] (current) |
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By translating Java code into math, we obtain the following mathematical definition of $f$: | By translating Java code into math, we obtain the following mathematical definition of $f$: | ||
- | \[ | + | \begin{equation*} |
f(x,y) = \left\{\begin{array}{rl} | f(x,y) = \left\{\begin{array}{rl} | ||
0, & \mbox{ if } y = 0 \\ | 0, & \mbox{ if } y = 0 \\ | ||
Line 27: | Line 27: | ||
x + f(x,y-1), & \mbox{ if } y > 0, \mbox{ and } y=2k+1 \mbox{ for some } k \\ | x + f(x,y-1), & \mbox{ if } y > 0, \mbox{ and } y=2k+1 \mbox{ for some } k \\ | ||
\end{array}\right. | \end{array}\right. | ||
- | \] | + | \end{equation*} |
By induction on $y$ we then prove $f(x,y) = x \cdot y$. | By induction on $y$ we then prove $f(x,y) = x \cdot y$. | ||
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* **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. |
- | \[ | + | \begin{equation*} |
f(x,y) = f(x,2k) = 2 f(x,k) = 2 (x k) = x (2 k) = x y | f(x,y) = f(x,2k) = 2 f(x,k) = 2 (x k) = x (2 k) = x y | ||
- | \] | + | \end{equation*} |
* | * | ||
- | * **Case 2**: $y = 2k+1$. Note $k < y$. By definition | + | * **Case 2**: $y = 2k+1$. Note $y-1 < y$. By definition and I.H. |
- | \[ | + | \begin{equation*} |
- | f(x,y) = f(x,2k+1) = x + f(x,2k) = x + 2 f(x,k) = x + 2 (x k) = x (2 k + 1) = x y | + | f(x,y) = f(x,2k+1) = x + f(x,2k) = x + x \cdot (2k) = x (2 k + 1) = x y |
- | \] | + | \end{equation*} |
This completes the proof. | This completes the proof. | ||