By Bernhard Korte

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**Additional info for Combinatorics, Graphs, Matroids [Lecture notes]**

**Example text**

We consider an example that is motivated by the following question: What immediately to the √ is the √ digit 1980 right of the decimal point in the decimal representation of ( 2 + 3) ? √ √ The approach to solve this problem is to consider more generally the numbers ( 2 + 3)2n for n ∈ N. For small value of n, we get the following numbers: √ √ ( 2 + 3)0 = 1 √ √ √ ( 2 + 3)2 = 5 + 2 6 √ √ √ √ ( 2 + 3)4 = (5 + 2 6)2 = 49 + 20 6 √ Claim: √ sequences (an )n∈N and (bn )n∈N with an , bn ∈ N for n ∈ N such that ( 2 + √ 2n There are 3) = an + bn 6.

We have (5 + 2 6)n = ( 2 + 3)2n = an + bn 6. Therefore, (3) implies: an = an = √ √ 1 an + b n 6 + 5 − 2 6 2 (3) n , which leads to √ n √ an = b n 6 + 5 − 2 6 . √ n √ Since an is integral, this yields {b 6} + {(5 − 2 6) } = 1 (where {x} := x − x for x ∈ R). 11, so for √ nn = 990, the first digits after the decimal point in the decimal representation of 5 − 2 6 are 0. Therefore, the first digits after the decimal point √ in the decimal representation of b 6 must be 9. Since a990 ∈ N, the same is true for 990 √ √ √ ( 2 + 3)1980 = a990 + b990 6, so the answer is “9”.

Gk contains a copy of each of the graphs G1 , . . , Gk−1 as a subgraph. In addition Gk contains a vertex set Ak consisting of |V (G1 )| · |V (G2 )| · · · · · |V (Gk−1 )| vertices. Choose a bijection τk : Ak → {(v1 , . . , vk−1 ) | v1 ∈ V (G1 ), . . , vk−1 ∈ V (Gk−1 )}. Then Gk contains (apart from the edges in the subgraphs G1 , . . , Gk−1 ) for each vertex v ∈ Ak an edge from v to the elements of the (k − 1)-tupel τk (v). By construction, the graph Gk does not contain any cycles of length three (provided that the graphs G1 , .