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Why is it defined that $(-1)!!$ equal to $1$, where $!!$ is the double factorial?

I've only seen it defined that $(-1)!!=1$, but I don't see why it should be so.

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    $\begingroup$ A better question might be "Why is $(-1)!!$ defined to be $1$?" $\endgroup$ – robjohn Feb 27 '15 at 9:03
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    $\begingroup$ @robjohn I agree. That would be a much better question. I remember I used to get annoyed that people would say $0!=1$ because it's defined that way--why? Because it's convenient, they would say. The reason behind the definition here is much more interesting I think. :) $\endgroup$ – Daniel W. Farlow Feb 27 '15 at 9:10
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Note that $(n+2)!!=(n+2)\,n!!$. Since $1!!=1$, the logical extension would be $(-1)!!=1$.

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    $\begingroup$ This can be found, as $n!!=n\times(n-2)!!$, in this Wikipedia article. $\endgroup$ – robjohn Feb 27 '15 at 9:34
  • $\begingroup$ Since the OP is gone and the question has been improved, can we stop what seems to be the vengeful downvoting? $\endgroup$ – robjohn Mar 14 '15 at 0:18
  • $\begingroup$ Would the recent downvoter care to comment? The question is old enough that it doesn't seem this is vengeful. Certainly, one can say this is the definition, but it also follows from the recurrence for $n!!$ $\endgroup$ – robjohn Oct 24 '15 at 20:30
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First, the post is low quality. Please write it in a better style.

Also, by definition, $(-1)!!=0!!=1$

This is by definition, as $0!=1$ by definition.

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  • $\begingroup$ There are reasons that these definitions are made. In this case, the rule defining $n!!$ can be logically extended to $(-1)!!$. There is a similar extension for $0!$, but that one is easier to understand since it can be seen as the product of the first $0$ positive integers, and empty products are usually taken to be $1$. $\endgroup$ – robjohn Mar 13 '15 at 12:08
  • $\begingroup$ @robjohn Everyone's downvoting because he actually changed his question. I answered it when it was first posted, and one might want to look at his edits. $\endgroup$ – Hasan Saad Mar 13 '15 at 12:10
  • $\begingroup$ Note that the OP is no longer around. Someone else was trying to improve the question (taking care of the low quality aspect). However, I was commenting on the claim that the reason was "by definition" when there are logical reasons behind that definition. $\endgroup$ – robjohn Mar 13 '15 at 12:15
  • $\begingroup$ Indeed. My answer was downvoted, too, even though it applies equally to the improved question. $\endgroup$ – robjohn Mar 13 '15 at 12:15
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    $\begingroup$ No worries. Although I made a comment about why I did not find your answer complete, I did not downvote your answer. $\endgroup$ – robjohn Mar 13 '15 at 22:46

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