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4

The statement wishing to be proven is as mentioned equivalent to whether or not for $n>2$ and $k\in\Bbb N$ we have the following relation: $$k^n < (k+2)^n - (k+1)^n$$ This is false. Consider the validity of the statement for $k=6, n=3$. One has $6^3=216$ and $(6+2)^3 - (6+1)^3 = 512 - 343 = 169$ $$6^3 = 216> 169 = (6+2)^3 - (6+1)^3$$ Why ...


4

In the simple case of $A>0$, it suffices to just look at the taylor series definition: $$1+A\leq 1+A+A^2/2!+A^3/3!+\cdots=e^A$$


3

Since $(e^x)' = e^x$ for all $x$, and $e^x > 1$ for $x > 0$, we have, for $x > 0$, $\begin{array}\\ e^x-1 &=\int_0^x e^t dt\\ &> \int_0^x 1 dt \qquad\text{since } e^x > 1 \text{ for } x > 0\\ &=x\\ \end{array} $


3

$\lim_{x\to\infty} \frac{x^a}{x^b}=\lim_{x\to\infty}x^{a-b}=\lim_{x\to\infty}\frac{1}{x^{b-a}}$ Since $a<b$, we have $b-a>0$. Since the function $x^{b-a}$ is monotonically increasing and x goes to infinity, $x^{b-a}$ also goes to infinity. Therefore $\lim_{x\to\infty}\frac{1}{x^{b-a}}=0$


2

Let us clarify it. If $(X,\Sigma_1)$ and $(Y,\Sigma_2)$ are measurable spaces, then a function $f:X \to Y$ is $\Sigma_1$-$\Sigma_2$-measurable if for all $A\in \Sigma_2$, $f^{-1}(A) \in \Sigma_1$. Now, take $X=Y=[0,1]$ and let us answer your questions 1. Composition of Lebesgue measurable functions may not be Lebesgue measurable What is called a ...


2

Consider a sequence of natural numbers $n - 1$, $n$, and $n + 1$. Since the question wants you to add them up, do so. Check if this gives you a multiple of 3. The smart choice of $n - 1$, $n$, $n + 1$ is "easier" than the choice of $n$, $n + 1$, $n + 2$ in many cases, and is a useful trick to know


2

Easier: let $(a,b)$ be an arbitary open interval, and $x \in A$. Then both $(a,x)$ and $(x,b)$ are non-empty open intervals, so they both contain a rational point, say $p$ in the former, $q$ in the latter. Then $x \in (p,q) \subset (a,b)$ as required.


2

For $n\in\mathbb{N}$ and $x\gt-1$, we have $$(1+x)^n\ge1+nx$$ which is easily proved by induction on $n$: $$\begin{align} (1+x)^n\ge1+nx\implies(1+x)^{n+1}&=(1+x)(1+x)^n\\ &\ge(1+x)(1+nx)\\ &=1+(n+1)x+nx^2\\ &\ge1+(n+1)x \end{align}$$ It follows that, if $A\gt0$, then $$\left(1+{A\over n}\right)^n\ge1+n\left(A\over n\right)=1+A$$ and ...


2

Given the prime $p$ the integer number you are looking for is $n(p) = 3^2p^2$. It has $3^2 = 9$ factors ($1$, $3$, $p$, $3p$, $3^2p$, $3p^2$, $3^2$, $p^2$, $3^2p^2$) and $$ \mbox{Factof}(n(p)) = \frac{3^2p^2}{3^2} = p^2 $$


2

'A prime no. always has two factors.' So the product of two primes $p$ and $q$ will have $4$ factors: $1$, $p$, $q$, $pq$ Now if the primes are distinct then the product $pq$ is not a multiple of $4$ since $2$ is the only even prime (and hence the other prime will be odd). So, $\text{Factof}(pq)$ is not an integer. For (b) part, Number of factors of ...


2

If $\mathcal{B}'$ is a basis, then in particular every element of $\mathcal{B}$ is a union of elements of $\mathcal{B}'$. But a singleton cannot be a union of proper subsets, so $\mathcal{B} \subset \mathcal{B}'$ and $\mathcal{B}'$ has at least $n$ elements. As an alternative proof, we could observe that the number of possible unions that we can for from a ...


1

Hint on how to compute this integral: Write $x$ (I assume $x>0$) as $[x]+\{x\}$ and now $$\int_0^x=\int_0^1+\dots+\int_{[x]-1}^{[x]}+\int_{[x]}^{x}$$ Observe that on each interval $[k,k+1)$, $k\in\mathbb{Z}$, you have $[t]^2=k^2$.


1

The notation is already defined in your "theorem". What you start with would thus be unmangling the definitions used in the theorem. It says The function $\ln : \mathbb R_{>0} \to \mathbb R$ is surjective (=onto). (note $\mathbb R_{\ge 0}$ is not the domain of $\ln$ since $\ln 0$ is undefined) Thus you need to find two definitions: $\ln$ is ...


1

So, you need to show $f(x)=\ln(x)$ for $(x>0)$ is onto. That means, you need to show that for every $y\in \mathbb{R}$, you can find an $x\in \mathbb{R}_{>0}$ such that $f(x)=\ln x=y$. This tells me $x=e^{\ln x}=e^{y}$. Here's how I'd write the proof. Fix $y\in\mathbb{R}$. Let $x=e^{y}$. Since $e^{y}>0$ for all $y\in\mathbb{R}$, it follows that ...


1

The first sentence will be a definition of "surjective" in this particular context. For example: We need to show that for every $y \in \mathbb{R}$, there is $x \in \mathbb{R}^{>0}$ such that $\ln(x) = y$.


1

The ($\Rightarrow$:) part is almost complete, except where you are not sure. We pick $x \in X$ and $y \in Y$, and we know that there is a path between them. Let the vertices in this path be $x, a_1, a_2, \dotsc, a_n, y$. The first vertex is in $X$, and the last vertex is in $Y$; all vertices are either in $X$ or $Y$ and never in both. Consequently, there ...


1

Perhaps you would like to consider the following: the inequality $$\frac{2n^2}{n^3 + 3}< \frac{2n^2}{n^3}=\frac{2}{n}.$$ In this case, if $N$ is an integer larger than or equal to $\frac{2}{\epsilon}$, the sequence would converge to $0$.


1

Your argument is correct, but it could be presented more clearly. Suppose that $h(x)=h(y)$. Then $x^2=y^2$ and $g(x)=g(y)$, so in particular $x=y$ or $x=-y$. If $x=y$, we’re done, so suppose that $x=-y$ and $x\ne y$. Suppose without loss of generality that $x\le y$. Then $x\le 0\le y$, so $g(x)=0\le g(y)$, and we must have $g(x)=0=g(y)$ and hence ...


1

You proof is basically correct. For the sake of proving that $h$ is injective, you don't need to say anything about why $f$ and $g$ are not. I would write "let $x,y\in \mathbb{R}$ such that $h(x)=h(y)$". Writing $(x,y)\in\mathbb{R}^2$ causes confusion. if $x=-y$ then $0=g(x)=g(y)=y$ or $x=g(x)=g(y)=1$ thus if $x=0$ then $x=y=0$ if $y=0$ then ...


1

I disagree with Hagen that you have the right idea. It is a serious logical error you made concerning induction, and I encourage you to read and fully grasp http://matheducators.stackexchange.com/a/10034/1550 before attempting induction problems. After you read that then read my answer. I give you any tree $T$ and graph $G$ such that every vertex in $G$ has ...


1

You have the right idea, but you should not modify $G$, esp. you cannot expect to obtain a given degree-$e$ graph from an unspecified degree-$(e-1)$ graph. Instead try the following: Let $T=(V_T,E_T)$ be a tree with $|E_T|=e$ edges (and hence $|V_T|=e+1$ vertices) and let $G=(V_G,E_G)$ be a non-empty, simple graph be given such that all vertices of $G$ ...


1

Here's one way to phrase this: Let $x \in \overline{A}$. Then every open set containing $x$ must intersect $A$ nontrivially. Since $X$ is under the discrete topology, we know that $\{x\} \subseteq X$ is open.


1

Assume that $n>6$. We first prove that $n$ is not odd. Assume, for contradiction, that $n$ is odd. Then $\left\lfloor\frac{n}{2}\right\rfloor\geq 3$ and $\left\lfloor\frac{n}{2}\right\rfloor-1\geq 2$. Therefore, both $\left\lfloor\frac{n}{2}\right\rfloor$ and $\left\lfloor\frac{n}{2}\right\rfloor-1$ divide $n$, but one of them is even, so $n$ is ...


1

On the bright side $$\sum_{i=0}^{n}\binom{n}{i}k^{n-i}\left ( 2^i-1 \right )=\sum_{i=0}^{n}\binom{n}{i}k^{n-i}2^i - \sum_{i=0}^{n}\binom{n}{i}k^{n-i}=(k+2)^n-(k+1)^n=$$$$(k+2 - (k+1))\cdot((k+2)^{n-1}+(k+2)^{n-2}(k+1)+...+(k+1)^{n-1})>$$$$(k+1)^{n-1}+(k+1)^{n-1}+..+(k+1)^{n-1}=n(k+1)^{n-1}$$ This is definitely true for $n>k$.


1

Yes, $\Bbb Z\setminus Z(m,b)$ is a union of arithmetic progressions. HINT: $$\Bbb Z=\bigcup_{r=0}^{|m|-1}Z(m,r)\;,$$ and the $|m|$ sets in that union are pairwise disjoint.


1

Archaick has already given you a simple exposition of your correct approach. I would like to point out an interesting fact, which is that it is actually crucial to use the Eulerian property of $G$, which proceeds via the theorem that every finite graph with even vertex degrees has an Eulerian loop, and hence any two vertices are in some cycle. The reason is ...


1

Perhaps this is just me being pedantic but the argument I think you're trying to make wasn't very clear to me upon the first reading. Is this what you mean? Note that since every such graph has an Eularian cycle, there necessarily exist two paths for each vertex to every other vertex which have no edges in common. Thus removing any single edge cannot ...


1

You're right, that's not very convincing. :-) There are essentially two different trees with $3$ edges. You can have all three edges incident at the same vertex; this is the star graph $S_3$. Or you can have at most $2$ edges incident at any vertex – then the tree is the path graph $P_4$ (why?). $S_3$ is easy. The claim isn't quite right since it ...


1

STF, Your notation of using superscript to differentiate between variables is confusing - I originally thought they are exponents. Below I will add parentheses to all superscripts. Your problem has an infinite number of solutions as loup blanc pointed out. There is actually a nice geometric interpretation of your system and you would be able to write ...



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