By Herbert Amann

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**Sample text**

2), that is, 0 0=0, n n∈N. 3) is another commutative operation on N which also satisﬁes 1 = ν(n) and n ν(m) = ν(n m) , n, m ∈ N . 4) For an arbitrary, but ﬁxed, n ∈ N, set M := { m ∈ N ; m n=m n} . 4) that 0 n = n. 3) we get 0 n = n = 0 n, that is, 0 ∈ M . Now suppose that m is in M . 4), ν(m) n=n ν(m) = ν(m n) = ν(m n) = n ν(m) = ν(m) n. Thus ν(m) is also in M . The axiom (N1 ) implies that M = N. Since n ∈ N was arbitrary, we have shown that m n = m n for all m, n ∈ N. 2). 1). 5) . Setting ϕ0 := ν we see that 0 ∈ N .

1 ≤ m ≤ n. n+1 m = n =2 . = n+m+1 n+1 . Remark The formula (ii) makes calculating small binomial coeﬃcients easy when they are written down in the form of a Pascal triangle. In this triangle, the symmetry (i) and the equation (iv) are easy to see. k n=0 n=1 n=2 n=4 n=5 1 q 6 q = 2 k 10 q 3 1 k 4 10 = = 4 k 1 5 = 5 1 q q q Simplify the sum n S(m, n) := k=0 m + n + k n+1−k m + n + k + 1 n−k 2 2 − k k for m, n ∈ N. ) Let p ∈ N with p > 1. Prove that p is a prime number if and only if, for all m, n ∈ N, p | mn = ⇒ (p | m or p | n) .

5 Let and be anticommutative operations on X and Y respectively. Further, let f : X → Y satisfy f (rX ) = rY , f (x y) = f (x) f (y) , x, y ∈ X . Prove the following: (a) x ∼ y :⇐ ⇒ f (x y) = rY deﬁnes an equivalence relation on X. (b) The function f : X/∼ → Y , [x] → f (x) is well deﬁned and injective. If, in addition, f is surjective, then f is bijective. 6 Let (X, ≤) be a partially ordered set with nonempty subsets A, B, C and D. Suppose that A and B are bounded above and C and D are bounded below.