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# All Activation Windows 7-8-10 V12.0 (Windows Office Activator) Extra Quality

All Activation Windows 7-8-10 V12.0 (Windows Office Activator)

All Activation Windows 7-8-10 v12 0 (Windows Key Activator). All Activation Windows 7-8-10 v12.0 (Windows +Office Activator). 1 Year+ [ALL ACTIVATION WINDOWS 7-8-10.Q: Proving a recursive definition by induction. Define $\lambda : {\mathbb N} \to {\mathbb N}$ by $\lambda(0) = 0$ and $\lambda(n+1) = 1 + \lambda(n)$, $n \in \mathbb N$. Show by induction that $n \geq 1 \implies \lambda(n) = n$. Definition of $\lambda(n)$ I would think was okay. The following seems to show it too, but I’m not sure. $k=0$ $\Rightarrow$ $\lambda(0) = 0$ = $1 + 0$ $k=1$ $\Rightarrow$ $\lambda(1) = 1 + 0$ $k=2$ $\Rightarrow$ $\lambda(2) = 1 + 0$ and $\lambda(1) = 1 + 1$ $\Rightarrow$ $\lambda(2) = 2$ $k=3$ $\Rightarrow$ $\lambda(3) = 1 + 0 + 1$ and $\lambda(1) = 1 + 1 + 1$ $\vdots$ For the induction step, I’m not entirely sure. I feel like I have an answer but not completely sure. $\lambda(n+1) = 1 + \lambda(n)$. So by induction, $\lambda(n+1) = 1 + 0$ and this makes sense by definition but is this how you show it? A: You’re almost there! To start, for $n=0$, $\lambda(1)=1+0=1$ and $\lambda(0)=0$, both $\lambda(0)=0$ and $\lambda(1)=1$, so $\lambda(0)=0$ and $\lambda(1)=1$. For $n=1$, $\lambda(2)=1+1=2$. Since $\lambda(0)=0$ and $\lambda(1)=1$, $\lambda(1)=1$ and $\lambda(2)=2$. And so,