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Take your claim But since $\forall x \in A: x \in B$ and $\forall x \in B: x \in C$ by the statement above, every $x \in A$ must also be in C. That's already a proof of your result! So packaging that proof inside a longer argument which starts off by assuming the opposite of what you want to establish, and then aims for a contradiction, is just ...

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Use the induction method: First, take $n=3$, $3! = 6$ and $3^3 =27$, $3! < 3^3$. Second, assume the inequality holds for $n = K$, $K \in \mathbb{N}$, $K>3$, i.e. $K! < K^K$. Then consider $n= K+1$, $(K+1)! = (K+1) K! < (K+1) K^K < (K+1) (K+1)^K = (K+1)^{K+1}$, which is $(K+1)! < (K+1)^{K+1}$. Proved.

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Observe that: $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$. Thus: $\displaystyle \sum_{k=1}^{n+1} k\binom{n+1}{k} = \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k} + \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k-1} = \displaystyle \sum_{k=1}^n k\binom{n}{k} + \displaystyle \sum_{k=0}^n (k+1)\binom{n}{k} = 2\displaystyle \sum_{k=1}^n k\binom{n}{k} + ... 3 We know that$n$and$m$divide both$k$and$q$. Since$k$is the lowest common multiple,$k \leq q$. If$k = q$then the result is obvious. Let's assume that$k < q$Then we can do the integer division between$k$and$q$:$q = t·k + r$so that$0 \leq r < k$. Since$n$and$m$divide$q$and$k$then they divide$q - t·k = r$. Therefore$n$... 3 Hint: $$T=\{1,8\}\cup\{2,7\}\cup\{3,6\}\cup\{4,5\}\ .$$ 3 Via induction it's a bit tiresome but Base case$n=2$,$2!=2<4=2^2$is pretty straightforward. Then multiplying both sides by$n+1$gives $$(n+1)!< (n+1)n^n$$ Considering $$(n+1)n^n < (n+1)(n+1)^n=(n+1)^{n+1}$$ so by induction we are done. Again, a direct proof is infinitely easier, for$n\ge 2$, so I include it for comparison's sake. ... 3 If$x > 1 \quad \Rightarrow \quad x^2 > x \quad \Rightarrow x^3 > x^2 \ \ldots$. By induction,$x^{n+1} > x^n$. Thus,$x^m \geq x^{n+1} > x^n$. 2 It is okay. You could also do it directly: Take$x \in A$. Since$A \subseteq B$, we get$x \in B$. But$B \subseteq C$implies that$x \in C$. So we conclude that$A \subseteq C$. 2 Inductive hypothesis: For$n = k$, $$\color{blue}{2^k \geq k^2},\quad k \geq 4.$$ $$2^{k+1} = 2\cdot \color{blue}{2^k} \geq 2(\color{blue}{k^2}) \geq k^2 + k^2 \geq k^2 + 2k + 1 \overset{\large k>2} = (k+1)^2$$ 2 Edited: I see why this is false but in general, why every closed subset of a compact set is compact? Another proof: Let$S \subset T$be a closed set, where$T$is compact. Let$\{\mathcal{U}_\alpha\}$be an open cover of$S$. Then$\{\mathcal{U}_\alpha\} \cup \{S^c\}$, where$S^c$is the complement of$S$w.r.t. to$X$, covers$T$. Since$T$is ... 2 Lemma: If$0<a<1$I claim for any$b>0$that$0<ab<b$Proof:$b(1-a)$is a product of positive numbers hence is positive. Corollary:$a^n<1\implies a^{n+1}<1$when$0<a<1$Proof: By induction, the base case being given, then let$b=a^n$in the lemma. Alternatively, a direct proof: Write $$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\ldots ... 2 Yes, you want to show that given a^n\le 1, then a^{n+1}\le 1. Well since a>0, we know that a^{n+1}\le a. But a<1. So we have a^{n+1}\le a\le1 as desired. This proof would clearly not work if a were not less than 1. 2 You are very close. For brevity, denote R composed with R by R\circ R. By the definition of composition, the pair (a,c) is in R\circ R if and only if there exists a b such that (a,b) and (b,c) are both in R. But if there is such a b, then by transitivity (a,c) is in R. Thus if (a,c) is in R\circ R, then (a,c) is in R. ... 2 You can simplify your argument. You should look at g(i) compared to g(i_0), which reduced the number of cases to 2: if g(i_1)=g(i_2)=g(i)<i_0, then i<i_0 as g is strictly increasing. Then you know that i_1=g(i_1)=g(i_2)=i_2. same goes for g(i_1)=g(i_2)=g(i)>i_0 i_1+1=g(i_1)=g(i_2)=i_2+1. there is no case g(i_0)=i_0 by definition ... 1 Here is a way to derive this result. By the binomial theorem,$$(1+x)^n =\sum^{n}_{k=0}\binom{n}{k}x^k$$Differentiate both sides.$$n(1+x)^{n-1} =\sum^{n}_{k=0} k\binom{n}{k}x^{k-1}$$Substitute x=1$$n2^{n-1} =\sum^{n}_{k=0} k\binom{n}{k} =\sum^{n}_{k=1} k\binom{n}{k} $$1 The first thing you need to do is to get the definitions absolutely clear. A (binary) relation R on a set A is not a subset of A, it is a subset of A\times A. In other words, R does not consist of elements of A, it consists of pairs of elements of A. A relation R on a set A is transitive means: for all a,b,c\in A, if (a,b)\in R and ... 1 0 < a < 1 \ \Rightarrow \ 1 = a + x with x > 0. Thus,$$ 1 = 1^n = (a + x)^n = a^n + \sum_{k=0}^{n-1}{n \choose k}a^kx^{n-k} \geq a^n \quad \Rightarrow \quad a^n \leq 1 $$1 Let's take a look at what we're trying to prove:$$ \forall \epsilon > 0 \; \exists N \in \mathbb{N} : \lvert a_n + c - (a_\infty + c) \rvert < \epsilon \; \forall n \ge N \tag{1} $$Now let's take a look at what we know: Since a_n \to a_\infty we have$$ \forall \epsilon > 0 \; \exists N \in \mathbb{N} : \lvert a_n - a_\infty \rvert < ... 1 It is a matter of doing things in an orderly manner. Yes, this covers all cases: either the image is contained in$[k]$, or it maps to$k+1$. Induction is made over$k$; not$m$. Careful. This is wrong. Induction is being made over$k$, and your induction hypothesis is that if$[m]\to [k]$is an injection,$m\leqslant k$. You don't know anything about ... 1 Basically you want to show that$f(x) = c^x > 1$when$c>1$for all$x>0$. Consider$\frac{d}{dx}f(x)$to show that$f$is a strictly increasing function, and evaluate$f(0)$. 1 To show$g \circ f$is continuous, you need to show it satisfies the property of continuity that you stated. So, you need to prove if$V \in \tau_{3}$, then$(g \circ f)^{-1}(V) \in \tau_{1}$. (Note that$(g \circ f)^{-1}(V) = f^{-1}(g^{-1}(V)).)$Here is a hint: Use the fact that both$f$and$g$are continuous. Start with$g$. If we have$V \in ...

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We know that $n!<n^n$ for some $n$, via our inductive hypothesis. We want to show that $(n+1)!<(n+1)^{n+1}$. Your first step is good, multiplying both sides of our inductive hypothesis by $n+1$ to get $(n+1)!<(n+1)n^n$. But $n^n<(n+1)^n$ (we can assume this, if not, it is very easy to prove), so $(n+1)(n^n)<(n+1)(n+1)^n$ and we have that ...

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Think of it this way. You know the inequality $|x-2| \cdot |x+2| < |x-2| \cdot 5$ is true. You are trying to show that the inequality $|x-2| \cdot |x+2| < \epsilon$ is true. By the transitive property of inequality, you will have successfully shown that inequality to be true if you are able to show that $|x-2| \cdot 5 < \epsilon$. This is ...

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The sum $w$ covers the first $m$ terms, where the two sums agree. What is left over is the sum from the $m$th term on. We have $\alpha_m = 0$ so we can write $$x = w + \sum_{i=m+1}^\infty \frac{\alpha(i)}{3^i}$$ But of course since $\alpha(i)$ is either $0$ or $2$, $\alpha(i) \leq 2$ $$x \leq w + \sum_{i=m+1}^\infty \frac{2}{3^i}$$ The infinite part of ...

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I like your idea that if $U(n)$ has an element of even order, then the order of $U(n)$ is even by Lagrange's Theorem. On the other hand, for $n>2$, the order of $n-1$ in $U(n)$ is 2. Another approach to this problem is to work with properties of the Euler phi function since $o(U(n))=\varphi(n)$.

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