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$$\frac{1}{n + 1} + \frac{1}{n(n + 1)} = \frac{n}{n(n + 1)} + \frac{1}{n(n + 1)} = \frac{n + 1}{n(n + 1)} = \frac{1}n$$

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Hint $$\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{x},x>0$$ it is easy to prove by derivative. so $$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$ $$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$ $$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$ $(1)+(2)+(3)$ ...

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This result holds in the general case with $2m$ objects of total weight $4m$, with none of the weights exceeding $2m$. If a list P of positive integers sums to $n$ then P is called a partition of $n$. Let |P| denote the number of elements in P. Let S' be the set of partitions P' of $2n$ with |P'| = $n$ (i.e., each partition P' in S' has $n$ elements and ...

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That's odd; you shouldn't need that. One can proceed as follows: Assuming that we know $$F^2_n-(-1)^n = F_{n+1}F_{n-1}$$ we can write \begin{align} F^2_{n+1}-(-1)^{n+1} & = F_{n+1} (F_n+F_{n-1}) + (-1)^n \\ & = F_{n+1}F_n + F_{n+1}F_{n-1} + (-1)^n \\ & = F_{n+1}F_n + F^2_n \\ & = (F_{n+1}+F_n)F_n ... 4 Since B = P^{-1} A P, \tag{1} B^T = P^T A^T (P^{-1})^T; \tag{2} Now PP^{-1} = I, \tag{3} whence (P^{-1})^T P^T = I \tag{4} as well; but (4) implies (P^{-1})^T = (P^T)^{-1}; \tag{5} thus (2) becomes B^T = P^T A^T (P^T)^{-1}, \tag{6} so A^T is similar to B^T, putting the whole proof together! QED. 4 If B is similar to A, then we writeB = P^{-1}AP$$for some invertible P. Then transposing everything, we get:$$B^T = (P^{-1}AP)^T = P^TA^T(P^{-1})^T = P^TA^T(P^T)^{-1}.$$Since P is invertible, P^T is also invertible. So B^T is similar to A^T, and the matrix who plays the role of P in the definition is P^T now. 2 One thing that come to mind is that you have to think carefully about using \in and \subset. Unless your function is bijective it is a very real possibility that f^{-1}(y) is not an element, but a set. In this event the statement f^{-1}(y) \in Y is not true. 2 Two sets A and B are equal, if and only if for all x we have x\in A \Leftrightarrow x\in B. Alternatively you can show A\subseteq B and B\subseteq A. We have:$$x\in A \cup \emptyset \Leftrightarrow x\in A \vee x\in \emptyset \Leftrightarrow x\in A$$so A\cup \emptyset = A. (Note, that x\in A\Rightarrow x\in A\vee \phi is true for any ... 2 Your base case holds. Suppose we have a cycle-free, connected graph G with n vertices. and n-1 edges. Because G is cycle-free, there is a vertex with degree 1. Delete this vertex. We are left with a connected graph G' with n-1 vertices and n-2 edges, which is a tree by induction. Reattaching the former pendant vertex does not introduce a ... 2 I would suggest a small correction on the domain of your functions, namely to take x\leqslant y <x^2- 2 and to put "p<x" on the sum of f_p(x). This way t(x) seems correct, since it counts exactly for how many y\in[x,x^2-2) holds \text{lpf}(y(y+2)) > x, and it clearly implies y,y+2 both primes, otherwise there would be a divisor of ... 2 Given a set D=\{d_1,\ldots,d_m\}\subset\mathbb N of coin denominations, for n\in\mathbb N_0 let f(n) denote the minimum number of coins (with repetition) in D needed to obtain sum n (or f(n)=\infty if it is ompossible). Then clearly f(n)\ge 0 for all n and f(n)=0\iff n=0. If n>0 and a way to obtain n with f(n) coins uses at least ... 1 You want to show that A\cup \phi = A. Then you have to show A\subseteq A\cup \phi  A\cup \phi \subseteq A If x\in A then x\in A\cup \phi  obviously. If x\in A\cup \phi  then either x\in A or x\in \phi but the latter case is impossible since \phi  is the set with no members. So x miust be in A. The proof for intersections is ... 1 Hint: Consider, as an example, the function f(x) defined by:$$\begin{array}{lll} f(1)=3\\ f(2)=-\pi\\ f(x)=0,\mbox{ for all other } x\\ \end{array}$$Now define functions f_i(x) so that f_i(x)=1 if x=i and f_i(x)=0 if x\neq i. Can you write f as a linear combination of f_1 and f_2? 1 Let$$ f_y(x)=\begin{cases}0\text{ if }x\neq y\\1\text{ if }x=y\end{cases}. $$Further, let$$ B=\{f_y(x):y\in\mathbb{R}\}. $$Try to see if you can show B is a basis for this space. 1 You can just apply your previous question to the inequality \kappa(G)\leq \lambda(G)\leq \delta(G) which holds for any graph G. 1 There is not really a need to quote that inequality in this case: you can directly show that both values are n-1. Certainly \delta(K_n) is n-1 as you know, and now just apply the definition of \kappa to see that \kappa(K_n)=n-1: Since K_n contains at least (well, exactly) (n-1)+1 vertices and there is no set of (n-1)-1 vertices whose removal ... 1$$\begin{align} x\in f^{-1}(X) & \implies f(x)\in X\\ & \implies f(x)\in Y\\ & \implies x\in f^{-1}(Y) \end{align}$$1 These are the standard definitions: Let X, Y be any sets whatsover.$$ f^{-1}(X) = \{ e \in E \ \mid \ f(e) \in X \} f^{-1}(Y) = \{ e \in E \ \mid \ f(e) \in Y \}$$Your requirement is to prove that the former is a subset of the latter, given X \subseteq Y. So we prove the implication$$ a \in f^{-1}(X) \implies a \in f^{-1}(Y) $$So to ... 1 The proof looks pretty good! There are some things that I can say though. For starters, we want to show the implication x\in f^{-1}(X)\implies x\in f^{-1}(Y). Because of this, we don't really need to look for y\in Y. You made this claim, and then the corresponding claims about the preimages with a subtle implication to if y\in f^{-1}(X)\cap f^{-1}(Y) ... 1 You basically got it. To make it slightly more rigorous you may want to introduce an \varepsilon and an N such that$$\left|\sin(an!\pi)\right|<\varepsilon$$for all n\geq N and \varepsilon>0. As you say, once n! gets large enough it'll cause n!a to be an integer. We can guarantee this by choosing N = q. Then we can say without question ... 1 The idea is correct. You can write it as you said it, more or less. Suppose a is rational. There exists an integer p and a positive integer q such that a= p/q. Clearly, for all integers 1 \le m \le n one has m \mid n!. Thus, for all n \ge q we have q \mid n! and thus a \ n! is integral. Consequently for all n \ge q we have \sin ( a ... 1 You're pretty much done. Here's how to finish up. Given a=p/q where q \in \mathbb{N}, let N=q. Then if n \geq N, then n! a \in \mathbb{Z} (why?), so if n \geq N then \sin(n! a \pi)=0. Hence the limit is zero. 1 \{a_k\} converges if and only if \{a_{2k}\} and \{a_{2k+1}\} converge to the same limit. Here$$ a_{2k}=1\rightarrow1, \quad a_{2k+1}=0\rightarrow 0. 

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You have to note that in Enderton's system $\exists$ is not primitive; thus, there are no rules for "managing" it. We have use contraposition : 1) $∀x(α→β)$ --- assumed 2) $α→β$ --- from 1) and Ax.2 by mp 3) $\lnot \beta \to \lnot \alpha$ --- from 2) and Rule T 4) $∀x(\lnot \beta \to \lnot \alpha)$ --- from 3) and Gen Th 5) $∀x\lnot \beta \to ∀x\lnot ... 1 For all$n \in \mathbb{N}$,$\frac{n}{n+1} < 1$. The slick algebraic proof of this would be$\frac{n}{n+1} = 1 - \frac{1}{n+1} < 1$since$\frac{1}{n+1}>0$for all$n \in \mathbb{N}\$. Induction would be much messier...

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