By Lovasz L., Vesztergombi K. L., Pelikan J.

This ebook is geared toward undergraduate arithmetic and laptop technological know-how scholars drawn to constructing a sense for what arithmetic is all approximately, the place arithmetic may be necessary, and what different types of questions mathematicians paintings on. The authors speak about a couple of chosen effects and techniques of discrete arithmetic, quite often from the components of combinatorics and graph concept, with a bit quantity concept, chance, and combinatorial geometry. anyplace attainable, the authors use proofs and challenge fixing to aid scholars comprehend the ideas to difficulties. moreover, there are lots of examples, figures, and workouts unfold in the course of the publication.

**Read Online or Download Discrete Mathematics: Elementary and Beyond PDF**

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**Extra resources for Discrete Mathematics: Elementary and Beyond**

**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 .

### Discrete Mathematics: Elementary and Beyond by Lovasz L., Vesztergombi K. L., Pelikan J.

by John

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