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It says Proof by Induction, Strong induction or Well-order but I don't know how can i start.

I tried proving for $f(1)= f(0+1)= 1(f(0))=1\times 1$ but $1(1-1)=0$ then I don't know what can I do

scoopfaze
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Kevin Duran
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  • So you've proven your base case. Now try the inductive step - assume $f(n)=n!$ and use that along with your condition for $f$ to show that $f(n+1)=(n+1)!$. – Vedvart1 Jul 04 '19 at 20:02

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$\text{Base Case: }n=0$ $$f(1+0)=(1+0)f(0)=f(0)=1$$

You can also, for your own edification, do it for $n=1$ (recalling the result from above):

$$f(1+1)=f(2)=(1+1)f(1)=2\times 1=2$$

From here, you can do the induction step:

$\text{Induction Step: Given that this holds for $(n-1)$, prove it holds for $n$}$, or:

$$f(n-1+1)=(n-1+1)f(n-2)=n!\rightarrow f(n+1)=(n+1)!$$

Can you take it from here?

scoopfaze
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