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Well, first of all you mean two maximal $k$-connected subgraphs $G_1$ and $G_2$; as in $G_1$ and $G_2$ are $k$-connected but there is no subgraph $G'_1 \supset G_1$ of $G_1$ nor is there a subgraph $G'_2 \supset G_2$ of $G_2$ that is $k$-connected. The result you want is Claim 1 below: Claim 1: Suppose that $G_1$ and $G_2$ share $k$ or more vertices. Then $... 0 I would try proving this by contradiction. Suppose$V(G_1)\cap V(G_2) \geq k$. Then I would show that$G_1\cup G_2$is$k$-connected. This would contradict the maximality of$G_1$and$G_2$. I would look at the proof that any two distinct blocks in a graph contain at most one common vertex for some more help. 1 This is correct - as far as it goes. But your question Why Rudin used$n(y−x)>1in the first place? is in a way the crux of the matter. The other steps are the routine part - the framework, as it were. This is the place where you need some actual understanding/intuition about "why" the assertion is correct. That you develop over time, by ... 0 This is merely an attempt to inspire, i was not able to conclude this way. So i think you can procede in this way. $$g_{c,d}(x)=\int_{\frac{1}{2}}^xz^{c-1}\sum_{i=0}^{d-1}{{d-1}\choose{i}}(-z)^{d-i-1}dz=$$ $$=\sum_{i=0}^{d-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c-1}(-z)^{d-i-1}dz=\sum_{i=0}^{d-1}(-1)^{d-i-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c+d-i-2}... 1 For the case of sequences, the \varepsilon-\delta type proof is as follows: Given \varepsilon>0, you need to find N\in\Bbb N (depending on \varepsilon) such that for all n\ge N, it holds that |s_n-L|<\varepsilon (in your case, |s_n-1|). For that purpose, observe that$$|s_n-1|=\left|\frac{n}{n+\sqrt n}-1\right|=\left|\frac {n-(n+\sqrt n)... 2 Observe that \begin{align*} \biggl|\frac{n}{n+\sqrt{n}} -1\biggr| = \biggl|\frac{\sqrt{n}}{n+\sqrt{n}}\biggr|\leq \frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}} \end{align*} Hence you have to choose ann$such that$\epsilon >\frac{1}{\sqrt{n}}$happens. 0 Chebyshev bounds are "loose" bounds and useful when you don't know the distribution. Here you know the distribution of$Z$. Why not directly work with the distribution function? For example, let$y = -\frac{\sqrt{\epsilon}}{\Delta t}(1 + \frac{\epsilon - 1}{\epsilon}\Delta t)$. Then, $$N(y) = P(Z < y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y}e^... 1 Your argument is fine, but you are doing more than you need to do in the first part: you are doing the right thing in the second part showing that f = g as sets of pairs if f and g are functions with the same domain that agree on any element of that domain. For the first part, there isn't really anything to prove: if f = g, then any property that ... 2 Your proof is correct. A far shorter proof follows. For 1\le x<kn, b^x\ne e since$$b^x=(b^k)^{\lfloor x/k\rfloor}b^{x\bmod k}=a^{\lfloor x/k\rfloor}b^{x\bmod k}$$Either x\bmod k>0 and b^x\not\in\langle a\rangle because the first factor is in \langle a\rangle and the second is not, or x\bmod k=0 but \lfloor x/k\rfloor\in[1,n) and b^x=a^{... 0 In logic there are generally two kinds of rules: introduction rules and elimination rules. An introduction rules tells us how to prove something, whereas an elimination rule tells us how to use something as a hypothesis. (\forall-introduction) To prove \forall x.\phi(x), you introduce a new variable x, and attempt to prove \phi(x). Note the emphasis ... 0 To prove statements like that you have to take an arbitrary x, and find a y such that p(x,y) holds. If you can't do this, it means that the statement is false. However, the negation of \forall x\,\exists y\,p(x,y) is not usually what you wrote. It is interpreted as \exists x\,\forall y\,\neg p(x,y). Note that quantifiers do not commute in general. 0 Hint: exploit innate symmetry: \ \ \underbrace{\overbrace{-a^4 = 1^{\phantom{|}}_{\phantom{.}}\!\!}^{\!\!\!\!\textstyle \color{#0a0}{(-a^3)}a\!=\! 1}}_{\!\!\!\!\!\!\textstyle(\color{#c00}{-a^2})a^2\! =\! 1}\Rightarrow\ (a+\overbrace{a^{-1}}^{\!\!\textstyle \color{#0a0}{-a^3}})^2 = a^2+\underbrace{a^{-2}}_{\!\textstyle \color{#c00}{-a^2}}+ 2\, =\, 2 1 Assume that the claim is false. Without loss of generality, this implies that there is some m,n\in\mathbb N, m<n, such that there is an injection f from [n]\to[m] ([n] is notation for \{1,...,n\}). Also without loss of generality, we can assume that n is the smallest natural for which this property holds. If n=1, the claim is clearly false ... 1 hint Let \epsilon>0 small enough. Consider the partage$$P=(0,1-\frac{\epsilon}{3},1+\frac{\epsilon}{3},2)$$then$$U(f,P)-L(f,P)=\frac{2\epsilon}{3}<\epsilon$$Its integral is given by$$\int_0^1f+\int_1^2f=\lim_{n\to+\infty}(\int_0^{1-\frac 1n}dx+\int_{1+\frac 1n}^2dx)=\lim_{n\to+\infty}(1-\frac 1n)+(2-1-\frac 1n)=\lim_{n\to+\infty}... 1$n!+n\le n!+n\cdot n!=n!(1+n)=(n+1)!\;$for all$\;n\in\mathbb{N}\cup\left\{0\right\}.$2$\frac{1}{n+1} + \frac{n}{(n+1)!} \leq \frac{1}{n+1} + \frac{n}{(n+1)} = 1$Therefore since$\frac{1}{n+1} + \frac{n}{(n+1)!}= \frac{n!}{(n+1)!} + \frac{n}{(n+1)!} $your inequality follows immediately. 1 Also if$n=0$it is obvious ($1\le 1$), for$n\ge 1$: $$n!+n\le n!+n!=n!2\le (n+1)!$$ 3 If$n \ge 1$, this is equivalent to $$1 \le n(n-1)!$$ which is clear. 0 If every$n_j=1$we have$k\ge k^2$, which is FALSE unless$k=1.$If$n_i\ne n_j$whenever$i\ne j,$we can prove it by induction on$k$. It's obvious when$k=1.$Suppose$A$is a set of$k+1$different members of$\Bbb N.$Let$M=\max A.$Let$C$be the sum of the cubes of the other members of$A$. That is,$C=\sum_{a\in A\setminus \{M\}}a^3.$Let$S$be ... 3 $$a^8-1\equiv 0 \implies (a^4+1)(a^4-1)\equiv 0 \implies a^4 \equiv -1 \pmod p$$ $$c^2=(a-a^3)^2=a^2-2a^4+a^6=a^2(1+a^4)-2a^4\equiv 2 \pmod p$$ $$d^2=(a+a^3)^2=a^2+2a^4+a^6=a^2(1+a^4)+2a^4\equiv -2 \pmod p$$ 1 I don't really get what you mean by simplify it. Anyways, in case the last part of the induction part is required, given your inequality and the induction hipothesys you would have: $$\sum_{j=1}^{n+1}\sqrt[4]{j^5}\ln\biggl(1+\frac{1}{j^2}\biggr)\le 4\sqrt[4]{n}+\sqrt[4]{(n+1)^5}\ln\biggl(1+\frac{1}{(n+1)^2}\biggr)\le$$ $$\le 4\sqrt[4]{n}+4\sqrt[4]{\frac{n+1}{... 1 As regards your final step, just notice that \ln(1+x)\leq x.Therefore it suffices to show that$$\sqrt[4]{(n+1)^5}\cdot \frac{1}{(n+1)^2} \leq 4 \sqrt[4]{\frac{n+1}{n}}$$that is$$\sqrt[4]{n}\leq 4 (n+1)^{\frac{1}{4}+2-\frac{5}{4}}=4(n+1)$$which trivially holds. 0 For n_1=\cdots=n_k=m<k, your inequality is wrong since$$ n_1^3+\cdots+n_k^3=km^3<k^2m^2=(n_1+\cdots+n_k)^2 $$A different approach for correcting the inequality Let's try to solve it through optimization. We solve the following problem$$ {\min n_1^3+\cdots+n_k^3 \\ s.t.\\n_1+\cdots +n_k=u } $$using the Lagrange's method. Skipping the middle steps, ... 0 A subset B implies #A <= #B. Proof. f:A -> B, x -> x is an injection from A into B. 1 You need to show those intervals "continuously overlap". That is for any k \in \mathbb N we have \frac 1{2^{k+1}} < \frac 1{2^k} < \frac 3{2^{k+1}} < \frac 3{2^k}. That follows as \frac 12 < 1 < \frac 32 < 3. This means \cup (\frac 1{2^n},\frac 3{2^n}) = (\lim_{n\to \infty} \frac 1{2^n}, \frac 3{2^1}) = (0, \frac 32) ... 1 Let 0<x<1. Consider the interval (\frac {\ln (\frac 1 x)} {\ln 2},\frac {\ln (\frac 3 x)} {\ln 2}). The length of this interval is easily seen to be \frac {\ln 3} {\ln 2}. Since any interval of length greater than 1 contains an integer there exist a positive integer n in this interval. You can now verify that x \in (\frac 1 {2^{n}},\frac 3 ... 1 I would note that G_1=(1/2,3/2) which covers (1/2,1), and then noting that$$\frac{3}{2^{n+1}} \ge \frac{1}{2^n} \iff \frac{3}{2} \ge 1 \mbox{, for all }n\in \mathbb{N}$$which is true. This gives us that two consecutive intervals have non-empty intersection so that$$\bigcup_{n\le N}G_{n}=\left(\frac{1}{2^N},1\right) \mbox{, for all } N\in \mathbb{N}$$... 0 For the first question, use https://mathworld.wolfram.com/EulersTotientTheorem.html or https://mathworld.wolfram.com/CarmichaelFunction.html and https://en.m.wikipedia.org/wiki/Proof_by_contradiction For the second, if p\nmid a\iff(a,p)=1$$a^{p-1}\equiv 1\pmod p$$If (a,2)=1, using Carmichael or Totient function,$$a^{2^{n-1}}\equiv1\pmod{2^n}\... 0 So I got stuck at that point with my induction: $$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{(n+1)^2}$$ Then, trying to separate elements on the right hand side I got: $$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{n^2(1 + \frac{2}{n} ... 2 Let M := M(\epsilon,\Delta t) = \frac{\sqrt{\epsilon}}{\Delta t}\left(1 + \left(1 - \frac{1}{\epsilon}\right)\Delta t\right). Recall that for any random variable X with a finite second moment and non-negative number x, Chebyshev's inequality states:$$\mathbb{P}(|X| > x) \leq \frac{\mathbb{E}[X^2]}{x^2}.$$I'm guessing your application of the ... 1 You seek to verify the inequality \frac{3}{2} \sqrt[3]{n^2} + \frac{1}{\sqrt[3]{n+1}} \overset{?}{<} \frac{3}{2} \sqrt[3]{(n+1)^2}. Cubing both sides yields$$n^2 + 3 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-1/3} \overset{?}{<} (n+1)^23 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-... 1 Since$k^{-1/3}=\int_{k-1}^kk^{-1/3}dx<\int_{k-1}^kx^{-1/3}dx$,$$\sum_{k=1}^nk^{-1/3}<\sum_{k=1}^n\int_{k-1}^kx^{-1/3}dx=\int_0^nx^{-1/3}dx=\tfrac32n^{2/3}.$$ 2 Hint: the polynomial$\,f(x)\,$is congruent to$\,2x-1\,$both mod$p_1$and$q_1$by little Fermat, so by CRT it boils down to computing its root$\,x\equiv 1/2\equiv (1\!+\!n)/2\pmod{\!n}\,$for odd$\,n = p_1 p_2$. i.e. let$\,p_1,p_2 = p,q.\,\,f(x) = x(x^{q-1}\!-\!1)+\!2x\!-\!1\,$so by Fermat &$\,p\!-\!1\mid q\!-\!1^{\phantom{|^|}}\!\!$&$...

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I assume $\mathbb{Z}_m=\mathbb{Z}/m\mathbb{Z}$. By the chinese remainder theorem, $\mathbb{Z}_m=\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$. So the problem boils down to proving that there is a unique solution to $x^{p_2}+x-1=0$ on each $\mathbb{Z}_{p_i}$. On $\mathbb{Z}_{p_2}$, note that $x^{p_2}=x$ for each $x$ (Fermat's little theorem) and solve the equation....

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Your proof is correct (with one inattentiveness in line -2: $f^{-1}(U) = M(a) \cap m(b)$). For a shorter proof see William Elliot's anwer. It sticks out that in the first part you work with the $\varepsilon$-$\delta$-definition of continuity, whereas in the second part you use a theorem about continuity: A function $f$ is not continuous if for some open ...

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I think the way to think about this is to realize that there's nothing special about any point in $\mathbb P^n$. The same goes for any (n-1)-plane in $\mathbb P^n$. To expand upon @Mummy the turkey's comment: every point in $\mathbb P^n$ is represented by a point $\overline P$ in $\mathbb A^{n+1}$. There is always a linear transformation sending $\overline ... 1 B is a subbase for a space S when {$\cap$F : F finite subset B } is a base for S. Theorem. If f:X -> Y, B is a subbase for Y, and for all U in B, f$^{-1}$(U) is open, then f is continuous. To apply this theorem to your problem, note that { (-$\infty$,x), (x,$\infty) : x in R } is a subbase for R. The proof of the theorem is straight forward and ... 1 Without induction, the problem would be quite simple since you want to prove that $$\frac{2^n }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)<\left( \frac{2n}{e} \right)^{n+1}$$ Taking logarithms and using Stirling approximation $$\log(\text{rhs - lhs})=\left(\log (n)-1+\frac{\log (2)}{2}\right)+\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ $$\text{rhs - ... 1 The first one is what it's saying. In order to satisfy the axiom as presented, one must show: \langle v, v \rangle \in \Bbb{R} for all v \in V, \langle v, v \rangle \ge 0 for all v \in V (note that \langle v, v \rangle can only be considered \ge 0 if it is a real quantity), \langle v, v \rangle = 0 \implies v = \vec{0}, v = \vec{0} \implies \... 1 Just note the notation (even you might know what you are talking about) is not correct$$A \times [(B\land\neg C)\lor (\neg B \land C)]$$Is this a set or a proposition? If it's a set$$A \times (B\oplus C)=A\times[(B\cap C^c)\cup(B^c\cap C)]$$If it's a proposition, for (a,b)\in A \times (B\oplus C) have$$a\in A\land b\in[(B\cap C^c)\cup(B^c\cap C)]... 1 From where you left off, you have a \in A,b\in B,b\notin C \vee a\in A, b \notin B, b \in C. Hence, (a,b) \in (A \times B) \vee (A \times C) and b\in B \Leftrightarrow b\notin C so that (a,b)\in (A \times B) \Leftrightarrow (a,b) \notin (A \times C). Therefore (a,b) \in (A \times B) \oplus (A \times C) and A \times (B\oplus C) \subseteq (A \times ... 1 Suppose that T is the graph of a function, that is, suppose that there exists g : X \to Y such that T = \{(x,y) \in X \times Y : y = g(x)\}. Now, take a \in X and observe that T \cap (\{a\} \times Y) is non-empty since (a,g(a)) \in T \cap (\{a\} \times Y). We claim that T \cap (\{a\} \times Y) = \{(a,g(a))\}. Indeed, if (x,y) \in T \cap (\{a\} ... 1 Assuming that binomial coefficients are integers (they are, by a simple combinatorial argument), you can use them with telescoping to write your expression: \begin{align*} &\binom{j_1+\cdots +j_k}{j_1}\cdot \binom{j_2+\cdots +j_k}{j_2}\cdots \binom{j_{k-1}+j_k}{j_{k-1}}\cdot \binom{j_k}{j_k}\\ &= \frac{(j_1+\cdots +j_k)!}{j_1!\cdot (j_2+\cdots +... 2 For induction on k, the base step k=1 gives a ratio \tfrac{n!}{n!}=1. To go from k=j to k=j+1 in the inductive step, replace \tfrac{1}{j_k^\text{old}!} with \frac{1}{j_k^\text{new}!j_{k+1}!}, wth j_k^\text{new}=j_k^\text{old}-j_{k+1}. This multiplies the ratio by \frac{j_k^\text{old}!}{j_k^\text{new}!j_{k+1}!}=\binom{j_k^\text{old}}{j_{k+1}}... 3 The given expression counts the number of ways to order n objects, with j_1 objects of type 1, j_2 of type 2 and so on until j_k objects of type k, and there are no other objects or types. Thus the expression must be an integer. 2 Write\sum_{i=1}^{n}i\times\frac{^{n-1}P_{i-1}}{n^i}=\sum_{i=1}^n\frac {i(n-1)!}{(n-i)!n^i}=\frac1n\sum_{i=0}^n\frac {i(i!)(n!)}{(n-i)!(i!)n^i}=\frac1n\sum_{i=0}^n\frac {i(i!)}{n^i}\binom n i$$Then refer to Proof Binomial Coefficient Identity: \sum_{k=0}^n\frac{k k!}{n^k}\binom{n}{k}=n 0 Let \pi:\mathbb A^{n+1} \setminus 0 \to \mathbb P^n be the projection map, and let v_0,..., v_n be a basis for the vector space structure V induced on affine space by your choice of coordinates downstairs (the spans \langle v_i \rangle are obtained as$$ \langle v_i \rangle = \pi^{-1}(0:\cdots : 0 : \underbrace{1}_{i\text{-th entry}} : 0 : \cdots : 0)... 0 As shown in Baby Rudin we can apply L'Hopital's Rule as long as the denominator tends to\infty$; it is not necessary that we have$\frac {\infty} {\infty}$form. So apply the rule to$\lim \frac {f(x)} x$to show that$f'(x) \to 0$. Hint for proof using MVT: If$f'(x) \ to l >0$the$f(n+1)-f(n) \geq \frac l 2 $for all$nsufficiently large. Show ... 0 Sketch: 1.$$\lim_{x\to\infty}\frac{e^x f(x)}{e^x}=\lim_{x\to\infty}(f’(x)+f(x))\,,$$ which implies $$L=\lim_{x\to\infty}f’(x)+L\,.$$ 2.$$f(x+h)-f(x)=f’(\xi)h \,,$$ yields $$L-L=\lim_{x\to\infty}f’(x) h\,.$$ 3 Hint: \begin{align} |a_{n+2}-a_{n+1}|&\leq \frac{1}{8} |a_{n+1}^2-a_n^2| \\ &=\frac{1}{8}|a_{n+1}+a_n| |a_{n+1}-a_n| \\ &\leq \frac 12 |a_{n+1}-a_n| \end{align} Hence,|a_{n+k}-a_n|\leq ...\$?

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