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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 isX = \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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