I am to prove by induction that $$\frac{(2n)!}{n!(n+1)!}$$ is a natural number for all $n\in N$. Checking the base case, assuming that the fraction indeed is a natural number I thought that I could show that the fraction is a natural number for $n+1$ by showing that it is a nonnegative (or positive, if you define naturals without $0$) integer. It is trivial to see that the fraction is never negative, but the part that is bugging me is showing that it is a whole number. Starting for $n+1$ $$\frac{(2n+2)!}{(n+1)!(n+2)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)n!(n+2)(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)(n+2)}\cdot\frac{(2n)!}{n!(n+1)!}$$ Here I use the IH, and I conclude that the first fraction must be whole for the whole product to be whole, since we have a form of $x=ky$ where $y\in N$ and we want $x\in N$ therefore $k\in N$ as well. So i continue to prove that $k$ is whole number and I reduce that to showing that $$\frac{2n+1}{n+2}\in N$$ however this isn't whole? Where is my mistake? How do I continue?
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Please note that I did check out the question here http://math.stackexchange.com/q/1388809/347105 , but the answer seems to be using backwards induction that I do not quite know how is justified, apart form the answer itself being somewhat unclear to me. – bonehead Nov 23 '16 at 15:26
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You forgot the factor $2$. – hamam_Abdallah Nov 23 '16 at 15:30
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I scrapped $2$ since it itself is natural. – bonehead Nov 23 '16 at 15:30
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2$2.\frac{1}{2}$ is natural but. – hamam_Abdallah Nov 23 '16 at 15:31
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Scrapping the $2$ is like dividing by $2$. The result needn't be integral. – MPW Nov 23 '16 at 15:32
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If I don't scrap $2$ I end up with $$\frac{4n+2}{n+2}$$ which isn't whole for $n=2$ – bonehead Nov 23 '16 at 15:34
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2"since we have a form of $x=ky$ where $y∈N$ and we want $x∈N$ therefore $k∈N$ as well". This isn't true. $k$ could be of the form $a/y$ with $a\in N$, in which case $x=a$ – R_D Nov 23 '16 at 15:38
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Let's try with an example. Take $n=5$. Then $$\frac {10!}{5!6!}=42$$ and $$\frac {12!}{6!7!}=132$$
The quotient $132/42$ is not an integer, so you can not prove the statement this way, because the final step is false.
If you still want (or have) to use induction, you should try to change the strategy. (I'm thinking on it...)
ajotatxe
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There's no reason it should be an integer. Just because $a=bc$, and $a$ and $c$ are integers, doesn't mean that $b$ needs to be an integer. In fact, take a simple example; put $n=7$, then $$\frac{(14)!}{7!8!} = 429$$ but $$\frac{(2\cdot7 + 2)(2\cdot7 + 1)}{(7+1)(7+2)}=\frac{15\cdot16}{8\cdot 9}=\frac{240}{72}=\frac{10}{3}\notin \mathbb N$$
MPW
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