Solutions to selected exercises about induction
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چکیده
Then, for all x ∈ R, g(x) = x−1 < x = i(x) and i(x) = x < x+1 = h(x). Hence g < i < h. (b) To prove that 〈Φ, <〉 is an irreflexive ordering, it suffices to prove that < on Φ is irreflexive and transitive. (We do not need to prove that < on Φ is strictly antisymmetric because, according to Exercise 20.2, this already follows from irreflexivity and transitivity.) To prove that < on Φ is irreflexive, let f : R → R be an arbitrary mapping in Φ and suppose that f < f ; we derive a contradiction. Note that from the assumption that f < f it follows by the definition of < on Φ that f(x) < f(x) for all x ∈ R. But this contradicts the irreflexivity of < on R. We conclude that ¬(f < f) for all f ∈ Φ, so < on Φ is irreflexive. To prove that < on Φ is transitive, let f : R → R, g : R → R and h : R → R be arbitrary mappings in Φ, and suppose that f < g and g < h; we need to prove that f < h. To this end, let x ∈ R. Then, since f < g, f(x) < g(x), and, since g < h, g(x) < h(x). Moreover, since < on R is transitive, f(x) < h(x). We have established that f(x) < h(x) for all x ∈ R, and hence f < h. We conclude that, for all f, g, h ∈ Φ, if f < g and g < h, then f < h, so < on Φ is transitive.
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متن کاملEXERCISES ( without solutions ) for Chapter 7 ( symmetric functions ) of
1. [2] Find the number f (n) of pairs (λ, µ) such that λ ⊢ n and µ covers λ in Young's lattice Y. Express your answer in terms of p(k), the number of partitions of k, for certain values of k. Try to give a direct bijection, avoiding generating functions, recurrence relations, induction, etc.
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تاریخ انتشار 2012