By Francis Borceux

ISBN-10: 0521061199

ISBN-13: 9780521061193

Lecture notes in arithmetic No.1348

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**Extra resources for Categorical Algebra and Its Applications**

**Sample text**

Child k gets the last nk presents. It is clear that we can determine who gets what by choosing the order in which the presents are laid out. There are n! ways to order the presents. But of course, the number n! overcounts the number of ways to distribute the presents, since many of these orderings lead to the same results (that is, every child gets the same set of presents). The question is, how many? So let us start with a given distribution of presents, and let’s ask the children to lay out the presents for us, nicely in a row, starting with the ﬁrst child, then continuing with the second, third, etc.

Aren’t we using in the proof the statement that we are trying to prove? Surely this is unfair! One could prove everything if this were allowed! In fact, we are not quite using the assertion we are trying to prove. 2)) that this proves the assertion about the sum of the ﬁrst n odd numbers. In other words, what we have actually shown is that if the assertion is true for a certain value (n − 1), then it is also true for the next value (n). This is enough to conclude that the assertion is true for every n.

If we want to put this in a formula, we get n+k n+2 n+1 n + ··· + + + k 2 1 0 = n+k+1 . 5) To prove this identity, we use induction on k. If k = 0, the identity just says that 1 = 1, so it is trivially true. (We can check it also for k = 1, even though this is not necessary. 5) is true for a given value of k, and we want to prove that it also holds for k + 1 in place of k. In other words, we want to prove that n+k+1 n+k n+2 n+1 n + +· · ·+ + + k+1 k 2 1 0 Here the sum of the ﬁrst k terms on the left-hand side is induction hypothesis, and so the left-hand side is equal to = n+k+2 .

### Categorical Algebra and Its Applications by Francis Borceux

by Charles

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