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6

You don't need induction, since it is enough to show $$n(n+2)<(n+1)^2$$


6

If you already know that $G$ is a group, then to prove that $a$ and $b$ are inverses, it is enough to check $ab = e$. If $G$ is not a group or you don't yet know that $G$ is a group (for example, if you are trying to prove that $G$ is a group), then it is not enough to show $ab =e$. You also need to show that $ba = e$. For matrices over a field, $AB = I$ ...


5

Yes. For instance $$ A = \{1\}, B = \{2\}\\ A = \text{All the negative integers},B = \text{All the positive integers}\\ A = \{2n \mid n \in \Bbb Z\}, B = \{2n+1\mid n \in \Bbb Z\}\\ A = \Bbb Z \setminus \{1\}, B = \Bbb Z \setminus \{2\}\\ A = \{5n \mid n \in \Bbb Z\}, B = \{3n \mid n \in \Bbb Z\} $$


5

Remember that the difference between a partial order and a total order is that in a partial order, you can have incomparable elements. Can you think of sets $A$ and $B$ such that $A\not\subseteq B$ and $B\not\subseteq A$? To clarify: such an example will merely show that $(\mathcal{P}(\mathbb{Z}), \subseteq)$ is not a total order. You still need to show ...


4

$y^2$ is positive, so $y^2 \leq 3y^2$. Thus, if $x,y$ are such that $x^2 + 3y^2 = 1$ then $x^2+y^2 \leq 1$.


4

In any field (well, at least in math and physics, the two fields I am interested in) one of the skills acquired over time by a serious researcher is a feel for what problems are open, what problems are interesting enough to be published, who is currently working on what, and so forth. In general, one doesn't have this fell until at least the second year of ...


4

Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that $$ \lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta| =\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2, $$ that is, $$ ...


4

Alternative idea: $$\bigcap_{n=1}^\infty E_n = \emptyset\implies \bigcup_{n=1}^\infty(X\setminus E_n) = X$$


4

Actually, you need both to hold; in classical logic, $P \vee Q \to R$ is equivalent to $(P \to R) \wedge (Q \to R)$. The intuition is as follows: assuming $P \vee Q$, you know at least one of $P$ or $Q$ is true, but you don't know which. Thus, to deduce $R$ from the assumption $P \vee Q$, you need to be able to deduce $R$ from whichever one is true—since ...


3

In any group (including groups of invertible matrices) it is sufficient to check that something is either a left inverse or a right inverse. This is because in any group, the inverse necessarily exists (by definition of something being a group) and is unique (since $ag=e$ implies $g=a^{-1}e=a^{-1}$ by multiplying on the left by $a^{-1}$, and this implies ...


3

For square matrices over many rings, it is indeed sufficient to check that $AB=I$ to conclude $A=B^{-1}$. It is certainly true for matrices over fields like the real or complex numbers. You can read about it in the link If $AB = I$ then $BA = I$ The class of rings for which this works is called the class of stably finite rings. It is a very broad class of ...


2

To show an inverse for both a (square!) matrix and a group element, you must show that it is both a left and a right inverse. Matrix multiplication and general group operations are not commutative. For homomorphisms sending inverses to inverses, you have more information, which you employed.


2

If you want to prove by contradiction that ($A$ AND $B$)$\implies C$, you assume that $A$ AND $B$ AND (NOT $C$) and show that this leads to a contradiction. Here's why: $a\implies b$ means $b$ OR (NOT $a$). So [($A$ AND $B$)$\implies C$] means [$C$ OR (NOT ($A$ AND $B$))] ie [$C$ OR (NOT $A$) OR (NOT $B$)]. If you negate this, you get $A$ AND $B$ AND (NOT ...


2

It's fine.   The predicate is a material implication.   This is evaluated as true if either its antecedant is false or the consequent is true. So for any two arbitary integers, we claim either they are not both odd or their product is odd. Thus, in the case where at least one of our arbitrary integers is not odd, then our claim is clearly true. ...


2

HINT: If you assume that $F_k\ge 2^{(k-2)/2}$ for all $k\le n$, then you have $$\begin{align*} F_{n+1}&=F_n+F_{n-1}\\ &\ge 2^{(n-2)/2}+2^{(n-3)/2}\\ &=2^{(n-3)/2}\left(2^{1/2}+1\right)\;, \end{align*}$$ so you’d be done if you could show that $$2^{(n-3)/2}\left(2^{1/2}+1\right)\ge 2^{(n-1)/2}\;.$$ Now use the fact that $2^{(n-1)/2}=2\cdot ...


2

Let $a_1=a_2=a_3=a_4=\frac{1}{16}$. Then $A=\frac{1}{16}$. Note that $B=(1/4)^{1/4}\gt A$. Perhaps $B$ is supposed to be $(a_1\cdots a_n)^{1/n}$.


2

This only requires basic inequation manipulation ($n \in \mathbb{N}$): $${n \over n+ 1} < {n + 1\over n+ 2}$$ $$\iff n(n+2) < (n + 1)^2$$ $$\iff n^2+2n < n^2 + 2n + 1$$ $$\iff 0 < 1$$ Done.


2

you don't need that result(actually usually that result is proved USING the fact that maximal planer graphs have only triangular regions!) To prove your result, observe that it isn't true for $n=1$ and $n=2$ so let $n\geq 3$. Assume to the contrary that there is a maximal planar graph $G=(V,E)$ embedded in the plane with a region that is not a triangle. ...


2

A homeomorphism by definition is a bicontinous BIJECTION. Meaning they must have the same cardinalities. $\mathbb{R}$ has cardinality $2^{\aleph_0}$ while $\mathbb{Q}$ has cardinality $\aleph_0<2^{\aleph_0}$.


1

Hint: By Taylor expansion about zero, $$ \mu(\theta_n)=\mu(0)+\mu'(0)(\theta_n-0)+o(\theta_n-0)\\ \iff\sqrt{n}(\mu(\theta_n)-\mu(0))=\sqrt{n}\mu'(0)\theta_n+\sqrt{n}o(\theta_n)\\ \iff \sqrt{n}(\mu(\theta_n)-\mu(0))=\sqrt{n}\theta_n\mu'(0)+\sqrt{n}\theta_n o(1)\\ $$ Furthermore, $\frac{\mu'(0)}{\sigma(0)}=o(1)$ is just a constant. Given the other stuff you ...


1

Take a region which is not a triangle. What is your definition of region? Can you add an edge? Hope that helps,


1

The condition $(d,a)=1$ is part of the hypothesis. So the first two sentences of the proof are superfluous, and should be deleted. Once that is done, the proof is good. The "By Bezout" part is too informal. You should say that by Bezout, there are integers $s$ and $t$ such that $\dots$. The proof is, as you observed, essentially the same as the proof of ...


1

Note that $$ \frac{n}{n+1} = 1 - \frac{1}{n+1},$$ $$ \frac{n+1}{n+2} = 1 - \frac{1}{n+2},$$ so actually you just need to show that $$\frac{1}{n+1} > \frac{1}{n+2}.$$


1

We want to consider the properties that are invariant under isomorphism and see that there is one that the two groups do not share. For example, $D_4\times\mathbb{Z}_3$ has an element of order $3$ in its center. The elements of order $3$ in $S_4$ are the $3$-cycles, none of which are in the center. For another example, $D_4\times \mathbb{Z}_3$ has an ...


1

Uniqueness part (in words) For all A1,A2 in P(U), if { [for all B(B in P(U) implies A1 union B = A1] and [for all B(B in P(U) implies A2 union B = A2] } then A1 = A2. Now to symbolize this is direct.


1

Let $S$ be the reference set. One can indeed consider $X$, $Y$, $U$ and $V$ as elements of the Boolean ring $\mathcal{P}(S)$ of subsets of $S$, where the addition is the symmetric difference and the product is intersection. Then the union of two subsets $A$ and $B$ is $A + B + AB$, the complement of $A$ is $1 + A$, the set difference $A \ B$ is $A(1+B) = A ...


1

If you have already on hand a theorem that uniform convergence implies convergence of the integral, like Baby Rudin 7.16, then the proof you've been given is perfectly rigorous. The triangle comment establishes that the value of the integral is independent of $n$. In more detail, the integral from $0$ to $1/n$ is ${1\over 2}(1/n)n={1\over 2}$ because ...


1

An alternative solution: The number of ways to arrange the $16$ students in a line and then split it into $4$ equal segments: $$16!=20922789888000$$ The number of ways to do it such that each segment contains a graduate student: ...


1

The specifics of your actual questions (about how $A_3=A_3\cap A_2\cap A_1$ and such) seem to be adequately answered above. Here instead, I provide an alternative solution to the stated problem which may be more intuitive than the book's approach. Let us approach this via direct counting. Temporarily assume that the four groups are considered distinct ...



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