MaWi 1 Skript
... 1. We only allow different elements. We pick, e.g. 2 or 9 of the 10 given elements 0, 1, 2, ...9; or generally k different elements. Obviously k ≤ 10 applies. For k = 3, we may thus pick {1,2,3}, or {0, 5,7}, but not {1,1,2} or {3,3,5}. However, it just means that you can pick a given element only ...
... 1. We only allow different elements. We pick, e.g. 2 or 9 of the 10 given elements 0, 1, 2, ...9; or generally k different elements. Obviously k ≤ 10 applies. For k = 3, we may thus pick {1,2,3}, or {0, 5,7}, but not {1,1,2} or {3,3,5}. However, it just means that you can pick a given element only ...
week5
... Congruent segments have the same length, whereas equal segments share the same set of points. If AB = CD, then A and C must be the same points, and B and D must be the same points. If AB = CD, then the length of AB is the same as the length of CD. ...
... Congruent segments have the same length, whereas equal segments share the same set of points. If AB = CD, then A and C must be the same points, and B and D must be the same points. If AB = CD, then the length of AB is the same as the length of CD. ...
1 The Inclusion-Exclusion Principle
... partition a set. Here, we are not looking at distinguishable elements, so while we are still performing partitions, instead of partitioning a set into smaller sets, we are partitioning an integer into smaller integers. Let us denote by pn (k) the number of ways of expressing k as an unordered sum of ...
... partition a set. Here, we are not looking at distinguishable elements, so while we are still performing partitions, instead of partitioning a set into smaller sets, we are partitioning an integer into smaller integers. Let us denote by pn (k) the number of ways of expressing k as an unordered sum of ...
compact - Joshua
... For the example statement about odd numbers and squares, the intuition behind the principle is first that the base step directly verifies the statement for the initial number 0. Next, because we have shown that the implication (∗) holds in all cases, applied to the k = 0 case it gives that the state ...
... For the example statement about odd numbers and squares, the intuition behind the principle is first that the base step directly verifies the statement for the initial number 0. Next, because we have shown that the implication (∗) holds in all cases, applied to the k = 0 case it gives that the state ...
On the determination of sets by the sets of sums of a certain order
... if and only if P8(X) = P8(Y). Let Fs(n) be the greatest number of sets X which can fall into one equivalence class. Our purpose in this paper is to study conditions under which Fs(n) > 1. Clearly Fs(n) — oo if n ^ s so that we may restrict our attention to n > s. In [5] Selfridge and Straus studied ...
... if and only if P8(X) = P8(Y). Let Fs(n) be the greatest number of sets X which can fall into one equivalence class. Our purpose in this paper is to study conditions under which Fs(n) > 1. Clearly Fs(n) — oo if n ^ s so that we may restrict our attention to n > s. In [5] Selfridge and Straus studied ...
PROBLEM SET 7
... have the same number of hairs on their head. This is because humans have < 1, 000, 000 hairs and there are > 1, 000, 000 people in NYC. The pigeonhole principle is particularly powerful in existence proofs which are not constructive. For example in the previous example we proved the existence of two ...
... have the same number of hairs on their head. This is because humans have < 1, 000, 000 hairs and there are > 1, 000, 000 people in NYC. The pigeonhole principle is particularly powerful in existence proofs which are not constructive. For example in the previous example we proved the existence of two ...
