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24

If you want to have a "set containment" kind of intuition, you should probably do the opposite: thinking of $P \implies Q$ as $P \subseteq Q$. The intuition is that if you are in some situation where $P$ is true (i.e. "being contained in $P$") then $Q$ is also true for this situation ("contained in $Q$"). In particular, $Q$ may contain more than just $P$, ...


15

Taking @SiongthyeGoh's comment, in the above picture, $AE=1$, $\angle BAE=x$, $\angle BAC=\beta$, and $\angle BAD=\alpha$. It follows that $DE=\sin(x+\alpha)$ and $CE=\sin(x+\beta)$. Applying the cosine theorem to $\triangle DCE$, the given expression is equal to $DC^2$, hence is a constant.


9

Since it seems you are thinking in terms of subsets, It is like saying that if $P\supseteq Q$ then $Q^c\supseteq P^c$ (in the sense that if $P$ is the set of all things we know to be true as our hypothesis, then the entirety of $Q$ is among those things we know as a result to be true. On the other hand if $Q^c$ is the set of all things we know to be true, ...


8

Let $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$ Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$. $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$ $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(...


5

This is wrong. $6$ divides the sum of the digits of $15$, but $6$ does not divide $15$. $6$ divides $12$, but $6$ does not divide the sum of the digits of $12$. What is true is that $3$ divides $n$ if and only if $3$ divides the sum of the digits of $n$. $6$ divides $n$ if and only if $3$ divides the sum of the digits of $n$ and the last digit of $n$ is ...


4

It's overloaded notation, for sure. But I have never seen it mean anything other than the following: Let $f : X \to Y$. Then, for $A \subseteq X$, $f(A) := \{ f(x) : x \in A \}$. The reason there isn't much notice is probably because it's a very common convention. It may even be explicit at some point in some books on Real Analysis. It turns out that ...


4

HINT Use the formulas $\sin^2t=\frac {1-\cos2t} 2$ and $\cos2p+\cos2q=2\cos(p+q)\cos(p-q)$.


3

$$F=\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ $$=1-\{\underbrace{\cos^2(x+\alpha)-\sin^2(x+\beta)}\}-\cos(\alpha-\beta)\{\underbrace{2\sin(x+\alpha)\sin(x+\beta)}\}$$ Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and Werner Formula$(2\sin A\sin B=\cdots),$ $$D=1-\cos(2x+\alpha+\beta)\cos(\...


3

$$\text {Let }\quad L=\sin^2 (x+a)+\sin^2(x+b).$$ $$\text {Let }\quad R=-2\cos (a-b)\sin (x+a) \sin (x+b).$$ $$\text {We have }\quad L=\frac {1}{2}(1-\cos (2x+2a) +\frac {1}{2}(1-\cos (2x+2b).$$ Now with $y=2x+a+b$ and $z=a-b$ we have $$\cos (2x+2a)=\cos (y+z)=\cos y \cos z-\sin y \sin z. $$ $$\cos (2x+2b)=\cos (y-z)=\cos y \cos z +\sin y \sin z .$$ $$\...


3

Just suggesting another approach. If you are familiar with partial fraction, notice that, $$\frac{1}{(2k-1)(2k+1)}=\frac12\left( \frac{1}{2k-1}-\frac{1}{2k+1} \right)$$ Using telescoping sum, $$\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n \frac12\left( \frac{1}{2k-1}-\frac{1}{2k+1} \right)=\frac12\left(1-\frac{1}{2n+1} \right)=\frac{n}{2n+1}$$


3

You have to account for it. However, this is often automatically done, because any statement in the form $$[\forall x\in A, P(X)]\equiv [\forall x,(P(x)\vee\neg x\in A)]$$ is always true when $A=\emptyset$. You must be careful when you make an existence claim, because any satement in the form $$[\exists x\in A,\ P(x)]\equiv [\exists x,\ (x\in A\wedge P(x))]$...


3

The definition of $f$ isn't changing. $f(U)$ is simply standard notation for the image of $U$ under $f$, and similarly for the inverse image $f^{-1}(U)$. Of course $f:X \to Y$ is not the same function as its direct image function, which you could denote $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$ if you really wanted. If category theory interests you,...


3

Everything looks good except for when $p$ is odd, the part where you say it has at most $MN$ edges if the parts have sizes $M$ and $N$ is very good. I would proceed as follows: Clearly $M+N=p$ so we can assume $M\geq N$ and $M=\frac{p+k}{2}$ and $N=\frac{p-k}{2}$, for some non-negative integer $k$. The options for $k$ are $0,2,\dots,p$ if $p$ is even and $...


2

Assume the arithmetic progression, starting with $\frac1{m_1}$, contains $k$ positive terms. Now the (uniform) term difference is $$ d = \frac{1}{m_1} - \frac{1}{m_2} \\ m_2 \geq m_1+1 \implies d \geq \frac{1}{m_1} - \frac{1}{m_1+1}=\frac{1}{m_1(m_1+1)} $$ The entire sequence, which ends in a positive value $\frac{1}{m_k}$, consists of $k-1$ steps of size $...


2

It is important in this problem to understand that the maximun number $k$ is determined by the choice of the two first integers $m_1,m_2$. Let $m_1$ and $m_2=m_1+h$ (where $h\ge 1$); the common difference of the a. p. is $d=\frac{-h}{m_1(m_1+h)}$ so we have $$u_1=\frac{1}{m_1}\\u_2=\frac{1}{m_1+h}\\u_3=\frac{1}{m_1+h}+\frac{-h}{m_1(m_1+h)}=\frac{m_1-h}{m_1(...


2

Since they are in arithmetic progression let us call $1/m_n-1/m_{n+1}=L>0$. We then have that $$\begin{align} \sum_{n=1}^{k-1}\left(\frac{1}{m_n}-\frac{1}{m_{n+1}}\right) &= \sum_{n=1}^{k-1}L \\ &= (k-1)L \\ &= \frac{1}{m_1}-\frac{1}{m_k} \end{align}$$ Which means we just have to show $$k=\frac{\frac{1}{m_1}-\frac{1}{m_k}}{L}+1<m_1+2\qquad (...


2

Some authors use different notation, i.e., for $f\colon A\to B$, the associated direct image function is denoted $f_\to\colon \mathcal P(A)\to \mathcal P(B)$. However, since this association is very common it is standrd to simply write $f$ for the extended set valued function, without warning as you say. Some care must be exercised, since for instance if $...


2

First, note that for any $x$ we have $b(x, x) \geq b(x, x) + b(x, x)$ by the second rule, and so $b(x, x) = 0$ for any $x$. Now let $x\neq y$, so that $b(x, y) = c > 0$. Then by the second and last requirements, $0 = b(x,x) \geq b(x, y) + b(y, x) = 2b(x, y) = 2c > 0$, which is impossible.


2

Generally, yes. Usually, the theorem is easy to prove when the set is empty, so just start off by handling the case where $A$ is empty, then assume $A$ is nonempty so you can pick $x\in A$. EDIT- Though there are some cases where you don't have to handle the two cases. For instance, the definition of $A\subset B$ is $x\in A \implies x\in B$; to prove $A \...


2

You're asking for intuitive sense, and the other answers are great at the logical proofs, but for intuition I like concrete examples. If I have tomatoes, I must have gone to the store. \--------v-------/ \-------------v-------------/ P Q Contrapositive: If I didn't go to the store, I can't have tomatoes. \-----------v-----...


2

if $P$ is true, then $Q$ has to be true. But then suppose that $P$ is true, and $Q$ is false. But, we already said that if $P$ is true, then $Q$ had to be true. Thus, this is an impossibility. Therefore, we can conclude that $P$ is false. The true statement of the contrapositive isn't qualified with a truth value, in the way I've done above, but this is a ...


2

Restrict attention to base elements. A base for the product topology is furnished by elements of the form $B=I_1\times\cdots \times I_N\times \prod_{k>N} {\mathbb R}$ with $N$ finite and each $I_k$ open. Each $f_k^{-1} (I_k)$, $1\leq k\leq N$ is open and their (finite) intersection which equals $f^{-1} (B)$ is thus open.


2

Hint: $$ \frac{a+c}{b+d}-\frac ab=\frac{bc-ad}{b(b+d)} $$ and $$ \frac{a+c}{b+d}-\frac cd=\frac{ad-bc}{d(b+d)} $$


1

It looks like you want a rigorous proof based on axioms. Let $\mathbb{R}^{+} \subset \mathbb{R}$ satisfying: For $a \in \mathbb{R}$, exactly one of the following holds: $a \in \mathbb{R}^{+}$; $-a \in \mathbb{R}^{+}$; or $a = 0$ (The trichotomy law) $a,b \in \mathbb{R}^{+} \Rightarrow a + b \in \mathbb{R}^{+}$ (Closed under addition) $a,b \in \mathbb{R}^{+...


1

Your grasp of these 'proof techniques' is roughly correct, though there are indeed clearer ways of understanding the difference between them. 'Indirect proofs' included not only 'proof by cases' but also 'proof by contradiction' and 'proof by contrapositive'. But I'll focus on the distinction between 'proof by cases' and 'proof without cases'. A 'proof by ...


1

I think this comes down to what it means for one proposition to "contain" another proposition. There are two ways of thinking about this, and both ways of thinking are valid, but they are incompatible. Let's talk about predicates instead of propositions, because predicates work better with "option 2" below. For the sake of example, let's consider the case ...


1

Since for $a\ne0$ zero function does not belong to $C_a$, the claim is only true for $a=0$.$\newcommand{\Ilim}{\operatorname{I-lim}}$ To prove that $C_0$ is a ring, to me the most natural way seems to be use the following facts (which are probably already known to you, if you study I-convergence): If $\Ilim f=x$ and $\Ilim g=y$, then $f+g$ is $I$-...


1

Yes, you do have to do that everytime. It could even be a method to prove that a given set $E$ is empty : Let $x$ be in $E$... and then you get to a contradiction. And you conclude that $E=\emptyset$.


1

Suppose you have two distinct points $x, y$, then you know that $$0=b(x,x) \ge b(x, y) +b(y, x) = 2b(x,y)\ge0.$$ What does this tell you about $x, y$?


1

As was pointed out in the comments, the use of continuous techniques like differentiation when dealing with a discrete problem like this is a little suspect. While it is often a good heuristic, turning it into a proper proof can sometimes be a bit tedious. Carry on Smiling's answer is a great way of answer this problem while staying within the discrete ...



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