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5

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


5

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 $$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$ $$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$ $$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$ $(1)+(2)+(3)$ ...


5

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 ...


4

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 write $$B = P^{-1}AP$$ 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 $$ a_{2k}=1\rightarrow1, \quad a_{2k+1}=0\rightarrow 0. $$


1

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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