8

My favourite, and the cleanest way in my oppinion is the following classical solution: Let $f(x)= \lfloor x\rfloor+\lfloor x+\frac{1}{2}\rfloor-\lfloor 2x \rfloor$. Then, it is trivial to see that $f(x+\frac{1}{2})=f(x)$. $f(x)=0 \forall x \in [0, \frac{1}{2})$. From here it follows immediatelly that $f \equiv 0$. P.S. By using $f(x)= \lfloor x\rfloor+\...


8

Note that the assumption $|A| \le |B|$ means that there is one-one function $f: A \to B$. Define a function $$F: \mathcal{P}(A) \to \mathcal{P}(B)$$ by $F(X)= f[X]:= \{f(x): x \in X\}$ for $X \in \mathcal{P}(A)$. Then $F$ is one-one. To see this, assume that $X \ne Y$. Then without loss of generality, we may assume that $X \not\subseteq Y$ and thus there is $...


7

There's no intuitionism/constructivism issue going on here at all. First of all, as a general matter of practice you need to keep separate the notions of proof by contradiction and proof by negation. The latter is intuitionistically acceptable! Specifically, let's say we interpret "$\neg p$" as shorthand for "$p\rightarrow\perp$" (this ...


4

If $|B|<|A|$, then there is an injection from $B$ to $A$. You’re assuming that $|A|\le|B|$, so there is also an injection from $A$ to $B$. The Cantor-Schröder-Bernstein theorem then says that $|A|=|B|$, contradicting the assumption that $|B|<|A|$. Thus, $|B|\not<|A|$. That’s all there is to it.


3

$$\cos \beta = \frac{|AB|}{|BC|} = \frac{\sqrt 5}{5},\quad \sin \beta = \frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}$$ Given $|AB|: |BC| = \sqrt5:5$, we assume $$|AB| = \sqrt5k, |BC| = 5k $$ where $k$ is a constant. Now using $$\frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}, \quad |AH| = 2k$$ $$|AH| + |BH| + |HC| = |AH| + |BC| = 7k = 7\implies k = 1.$$ So, $$|AH| = 2, |...


3

I'll answer the paragraphs first, then the question. Paragraphs It is right ... condition of our question. Of course, that's correct. Not all $a,b,c$ will satisfy this equation. The reason why the statement is useful is that some $a,b,c$ do satisfy the statement (extreme example : $a = 10^{10} , b = 10^{10}+1 , c=2$)! If not ... can be infinite. Yes, ...


3

The picture should really look like this: We have $\sin\beta=2/\sqrt5$ and $\tan\beta=CA/AB=2$. Then $HB,HA,HC$ are in geometric progression with ratio $2$; since their sum is $7$ it can easily be worked out that the hypotenuse $BC$ is $5$. Then $AB/BC=\cos\beta$ gives $AB=\sqrt5$, $CA=2\sqrt5$ and $A(\triangle ABC)=5$.


3

Lemma: [Smoothing of convex functions] Given a convex function $f(x)$, and variables $ a \leq b$, if $ \epsilon < b-a$, then $ f(a) + f(b) \geq f( a + \epsilon ) + f(b- \epsilon)$. (This is "well-known", so I'm not going to prove it. If you're stuck, explain what you've tried.) Corollary: Given a convex function $f(x)$, and variables $ a \leq b \...


3

I do not have access to Lee's book, but the notation $(X,\mathcal{E},\varphi)$ suggests that the characteristic maps of the cells are part of the structure. In that case we do not need AC. In fact, for each cell $e = e^n_\alpha$ we have an associated characteristic map $\varphi_e : D^n \to X$. Then take $d_e = \varphi_e(0)$. If the characteristic maps do not ...


2

Alternative solution: WLOG, assume $a \ge b \ge c = 1$. Denote $a^x = u, b^x = v$. Then $u \ge v \ge 1$. We have \begin{align*} f'(x) &= A \ln a + B\ln b \\ &= A\ln \frac{a}{b} + (A + B) \ln b \end{align*} where \begin{align*} A &= {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+u \right) ^{2}}} -{\frac {u}{ \left( u+v \right) ^{2}}},\\ B &={\frac {...


2

HINT Let $a = \overline{AH}$, $b = \overline{CH}$ and $c = \overline{HB}$. Then we have that \begin{align*} \begin{cases} a = \tan(\beta)c\\\\ a = \tan(\pi/2 - \beta)b \end{cases} \Rightarrow a(1 + \tan(\beta) + \cot(\beta)) = 7 \end{align*} Can you take it from here?


2

A set is well ordered if every nonempty subset has a least element. Your set is $$X = \mathbb N/2 = \{\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}, \frac{5}{2}, ...\}.$$ You want to show that every nonempty subset $A$ of $X$ has a least element. Already, you know that $\mathbb N$ is well ordered. Isn't there a natural association between subsets of $\...


2

Let $x \in X$ and $\epsilon >0$. There exist $y \in \bigcup_n V_n$ such that $\|x-y\| <\epsilon$. Now $\|P_n x-x\| \leq \|P_n x-P_ny\|+\|P_ny-y\|+\|y-x\|\leq (M+1)\epsilon +\|P_n y-y\|$ where $M=\sup_n \|P_n\|$. Can you finish?


2

The line of proof works, though the proof has a typo as quoted, and it may not be written in the easiest way to follow. The discriminant is given as: $D = (b^2 + a^2 - c^2)x - 4a^2b^2$ Here is the typo, and the discriminant can not depend on $x$. The expression above is wrong (though what's used afterwards appears to be based on the correct expression). ...


2

$\cos\beta=\dfrac{\sqrt5}{5}$ and $\sin\beta=\dfrac{2\sqrt5}{5}.$ Let $\overline{AH}=x,$ then you can obtain values $$\overline{AB}= \dfrac{\sqrt{5}x}{2},\qquad \overline{BC}= \dfrac{5x}{2},\qquad \overline{AC}=\sqrt{5}x$$ and $$\overline{AH}=x,\qquad \overline{BH}=\dfrac{x}{2},\qquad \overline{CH}=2x$$ using simple trigonometry. Now, compute the value of $x$...


1

For $x > 0$: $[2x]$ is the number of positive integers $\leq 2x$. Count them by separating the even and odd numbers. For general $x$: add a sufficiently large integer to it.


1

Did you know that $\mathbb{N}$ is well ordered? Assuming this, let $A=\left\{\dfrac{a}{2} \mid a\in\mathbb{N}\right\}$. Take $B\subseteq A$ such that $B\neq\emptyset$. Define $C=\{ 2z\mid z\in B\}$. Clearly $C\neq\emptyset$ since $B\neq\emptyset$. Moreover, if we take $z\in B$ then there exist $a\in\mathbb{N}$ such that $z=\dfrac{a}{2}$. Then $2z=a$. Thus $C\...


1

You can refer to the definition of a well ordered set to prove that any subset of a well ordered set is well ordered. A well ordered set is one that any subset has a smallest element. If you take a subset of a well ordered set, all of its subsets are subsets of the larger set, so they have a smallest element. This shows the subset is well ordered.


1

You are over-complicating things. $\cos:\mathbb{R}\to[-1,1]$ is not injective but it is surjective. To show that it is surjective, use the intermediate value theorem. You know that $\cos(\pi)=-1$ and $\cos(0)=1$, so by the intermediate value theorem, for any $y\in[-1,1]$, there is an $x\in[0,\pi]$ with $\cos(x)=y$. Since $|\cos(x)|\leq 1$, then it is ...


1

I should point out that my first pass at this problem had an algebraic flaw that caused me to believe that the domain of $f$ had to be $(-1,1).$ It was only after seeing zwim's answer, and re-checking my math that I saw and corrected my error. First see comments following the question. This is how I would approach it: Let $\displaystyle f(x) = x/\sqrt{x^2 +...


1

A shorter proof might be: Let $x$ be a natural number, but $x\ne 1$. Then $x\in A_{x-1}$ since $(x-1+1)\cdot1=x.$ $x =1$ is not in any $A_n$ since $y\in A_n \rightarrow y>n$. Hence $\cup A_n = \mathbb{N}\backslash\{1\}$. Assume now that $x\in \mathbb{N}$. Then $x$ is not an element of $A_{x}$ since we had $y\in A_n \rightarrow y>n$. Hence $\cap A_n =\...


1

HINT Start with length of $ HB =1$. Chase the sides using Pythagoras thm. A property of right triangles can be used ( square of altitude equals product of base segments ; $HC=HA^2/HB$). $$\text{ScaleFactor=}\frac{7}{4+2+1}=1$$ $$\text{Area = ScaleFactor}^2 \cdot \frac12 \cdot (4+1) \cdot 2 =5. $$


1

This is a proof without homothety. Let $AB'$ intersect $l$ at point $P$. $l$ is the perpendicular bisector of $AA'$ and $BB'$. $\angle APX=\angle B'PY$ Hence, $\angle A'PX=\angle APX=\angle B'PY=\angle BPY$ and thereafter points $A'$, $P$ and $B$ are collinear.


1

The formula is a rewritten form of the fundamental Stirling number of the second kind recurrence relation $$S(n, k) = S(n-1, k-1) + kS(n-1, k).$$ Maybe some extra notation would help. Let $f_m(n) = S(n, n-m)$. For each fixed $m$, you want to show $f_m(n)$ is a polynomial in $n$ of some to-be-determined degree. The rewritten recurrence relation is then $$f_m(...


1

I tried to make a bit more for this interesting problem. With $$f(x)=x^{\frac{x}{x+1}}-\left(\frac{x}{x+1}\right)^{x}+\ln\left(x\right)$$ what you want to show is that, for $x \geq 14$ $$f(x) < x < f(x)+\frac 1 e$$ that is to say that $$\lim_{x\to \infty } \, (x-f(x))=\frac 1 e-\epsilon \quad \text{and} \quad \lim_{x\to \infty } \, (f(x)-x+\frac 1 e)=\...


1

If $|B|<|A|$, then $|B|\le|A|$ and $|B|\neq|A|$ (according to textbook). Assume $|A|\le|B|$, then by Cantor Bernstein Theorem if $|B|\le |A|$ and $|A|\le|B|$ then $|B|=|A|$; however, $|B|\neq|A|$. This is a contradition. Hence $|A|\not\le|B|$. Hence if $|B|\lt|A|$ then $|A|\not\le|B|$. Therefore by contraposition, if $|A|\le|B|$ then $|B|\not\lt|A|$.


1

fleablood’s comment hits the nail on the head: you can’t just jump from $|A| < |B|$ to $\neg (|B| < |A|)$ because this assumes Cantor-Bernstein. There is no inherent contradiction between $|A| < |B|$ and $|B| < |A|$ that follows trivially from their definitions; the OP is subtly assuming there is a contradiction because the notation is ...


1

We are looking for solutions to $$ \exp(y\log x) = \exp(x\log y) \, . $$ Taking logs of both sides and rearranging, this equation becomes $$ \frac{x}{\log x}=\frac{y}{\log y} \, . $$ Consider the graph of $f(z)=z/\log z$: Its unusual shape can be explained by a few observations: For $0<z<1$, $\log z$ is negative, so $f(z)$ is negative. As $z\to1$, $\...


1

Let $x$ be an element of $S$. Since $S$ is finite, there exist integers $i, p >0$ such that $x^i = x^{i+p}$. It follows that, for all $k \geqslant i$, $x^k = x^{k+p}$. In particular, if $k$ is a multiple of $p$, say $k = qp$, one has $(x^k)^2 = x^{2k} = x^{k+qp} = x^k$ and thus $x^k$ is idempotent.


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