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9

This is "too thorough", to put it nicely. One wonders why you just didn't start with $0' + 0 = 0' + 0.$ and conclude $0 = 0'$, using commutativity if you felt necessary. As another general piece of advice, at the end of each proof by contradiction, look back and see if you can't write it as a straightforward "direct proof" that doesn't use contradiction. ...

7

The proof is a bit roundabout. Below I will explain a general way to discover a simpler proof (vs. pulling it out of a hat like magic). The key idea is very simple: one can discover consequences of identities (axioms) by "overlapping" them: looking for a "unified" term that they both apply to. Let's try that here. Suppose that $\,0\,$ and $\,0'$ are both ...

6

It does not just mean that there is always a number that is smaller than the absolute difference between the two given real numbers. Rather it means that there is such a number of the form $1/10^n$, so that digits in the $n$th decimal places in the two numbers differ from each other. PS: For example, two numbers differing in the $5$th place and not before ...

5

Why not $a(\mathbf{0})=a(\mathbf{0}+\mathbf{0})=a(\mathbf{0})+a(\mathbf{0})$; now add the inverse of $a(\mathbf{0})$ to both sides?

5

What you have is nowhere near a proof. The definition of $X$ can be accepted, but it is not conveying any insight transgressing the verbal formulation of the problem. We have to construct a bijective map $$f:\quad {\mathbb N}_{\geq0}\to X,\qquad n\mapsto A_n\ .\tag{1}$$ This map produces for each $n\in{\mathbb N}$ a finite set $A_n\in X$, and each element ...

4

Comments on your proof. Define a set $X=\{A\subseteq\mathbb{N}\mid \text{$A$is finite} \}$. OK We can have a function $g_{n}: \mathbb{N} \rightarrow A_{n}$ << Problem: You have to define $A_n$ Say you wanted to say, $A_n$ is a finite subset of $\mathbb{N}$ defined by: $$A_n = \{n_1,n_2,... | ~ n_i \text{ occurs as exponent in } n = p_1^{n_1} ... 4 Complete the square:$$2x^2-4x+3=(\sqrt 2 x)^2-2\times(\sqrt2 x)\times\frac2{\sqrt2}+\left(\frac2{\sqrt2}\right)^2-\left(\frac2{\sqrt2}\right)^2+3\\=\left(\sqrt 2 x-\left(\frac2{\sqrt2}\right)\right)^2+1$$4 The only vague part is "Let v \in V." This sounds like you're going to prove something is true for all elements of the vector space. But for your purposes, any element v will lead to the contradiction and be enough for your proof. So I might write instead "Choose any v \in V." [But there's a simpler proof that doesn't need that extra variable v ... 4 Here is an approach which does not require knowing anything about the square root function. a^2 < b^2 \implies b^2 - a^2 > 0 \implies (b-a)(b+a) > 0 which implies that b-a and b+a are nonzero and have the same sign. Therefore one of the following is true: b > a and b > -a, so b > |a|, so |b| > |a| b < a and b < ... 4 Suppose \displaystyle \lim_{x \to a}f(x) = c and let \epsilon > 0, then there is \delta >0 such that$$|x-a|<\delta \implies |f(x)-c | < \epsilon$$now note that$$|h| = |(a+h)-a|$$and so$$|h|< \delta \implies |f(a+h)-c | < \epsilon$$which shows that \displaystyle \lim_{h \to 0}f(a+h) = c. It follows that$$\displaystyle \lim_{x ...

4

It isn't true. Take two discs in $\mathbb{R}^2$ of radius 1 with the usual topology (euclidean metric), one centered at the origin, and the other centered at $(\frac{1}{2},0)$ and consider their intersection. This is not a ball in that topology. However, it is an open set [ by the definition of a topology, the intersection of 2 open sets is an open set ].

4

Your idea is along the right track, but not worded well. Note that if any element $g$ in $G$ has even order $2k$, then $g^k=g^{-k}$ is your desired element. Suppose for sake of contradiction that all non-identity elements of $G$ have odd order. Then as you noted, such an element cannot be its own inverse $g=g^{-1}$, else it would have order $2$. Since ...

4

A short proof: $\left(1+\frac{x}{n}\right)^n = e^{n\log\left(1+\dfrac{x}{n}\right)}$ Since $\log(1+x) = x + O(x^2)$ when $x \to 0$, we have $n\log(1 + \frac{x}{n}) = x + O(\frac{x^2}{n})$ when $n\to +\infty$

4

It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$

4

Let $Y = \mathbb{R}$ with usual metric and $X = (0,1) \cup \{2\}$ with metric inherited from the standard metric on $\mathbb{R}$. let $f(x) = x$ on $(0,1)$ and $f(2) = 1$ and take $A = (0,1)$. it is then easy to see that for this particular case your statement doesn't hold

3

Since $A\subseteq\bar{A}$ you know that $f^{-1}(A)\subseteq f^{-1}(\bar{A})$, so $$\overline{f^{-1}(A)}\subseteq f^{-1}(\bar{A})$$ because $f^{-1}(\bar{A})$ is closed by continuity of $f$. Now, if your claim is true, you'd conclude that $\overline{f^{-1}(A)}= f^{-1}(\bar{A})$ for every $A\subseteq Y$. Can you?

3

We have for $t\ge0$ $$x^2=t\iff x^2=(\sqrt t)^2\iff x^2-(\sqrt t)^2=0\\\iff (x-\sqrt t)(x+\sqrt t)=0\iff x=\pm\sqrt t$$

3

Proofs should follow the same rules as any other kind of writing. If the text naturally forms paragraphs then that's how it should be written. Now, usually if you have five pages of text forming a single paragraph that would be poor style, but I suppose it's possible that you could have a five-page-long paragraph if, say, one of the sentences of that ...

3

$D = (-4)^2 - 4*3*2 < 0$, hence the polynomial $2x^2 - 4x + 3$ has no real roots, and so its graph does not intersect the horizontal axis. Thus, the value of this polynomial never change its sign, and, for, instance, when $x = 0$ it has the value $3 > 0$, so it always take positive values.

3

This is a slight simplification of @Sami's solution: $$2x^2-4x+3=2(x^2-2x+1)+1=2(x-1)^2+1>0$$

3

Embed the construction in the complex plane. Let $\omega=\exp\left(\frac{\pi i}{3}\right),B=0,C=1,E=1+v$. Then $A=\omega$ and $D=1+\omega v$, hence $F=B+E-A$ implies: $$F = 1-\omega+v,$$ hence: $$\omega F = \omega -\omega^2 + \omega v = 1+\omega v = D,$$ so $BFD$ is equilateral.

3

Hint: $g \ne h$, so $\exists x \in \mathbb R$ such that $h(x) \ne g(x)$. Use continuity to show that there is an open neighbourhood $N_x$ of $x$ such that $h(y) \ne g(y) \ \forall y \in N_x$

3

To conclude your induction proof, just multiply x both sides : $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ multiply $x$ both sides : \begin{align} \\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\ \end{align} factor ...

2

You start correctly with "both $\overline{LK}$ and $\overline{MN}$ are congruent". (As a style point, in UK usage you would phrase this "The lines $\overline{LK}$ and $\overline{MN}$ are equal in length." It's a bit clearer to keep "congruent" for referring to triangles or other two-or-more-dimensional shapes.) The other property you are given is that ...

2

As others have remarked, you seem to already assume that the set you want to be countable is countable by the way you label the finite sets. However, you can "lexicographically order" the finite sets of natural numbers in many ways, for example the following: If $A$ and $B$ are finite sets of natural numbers, say that $A <B$ if $|A| < |B|$. If $A$ and ...

2

To show $R[A \cup B] = R[A] \cup R[B]$ for a relation $R \subset X \times Y$, $A,B \subset X$, we have to show two inclusions. So take any $y \in R[A \cup B]$; this means that there exists $x \in A \cup B$ such that $(x,y) \in R$. Then either $x \in A$, and then $y \in R[A]$, or $x \in B$ and then $y \in R[B]$. In either case $y \in R[A] \cup R[B]$, as ...

2

The proof is ugly due to the limitation of the Ecludean framework. I try to break it down into 3 diagrams. See the first. As mentioned, $\triangle AEC$ is congruent to $\triangle BDC$. [Proof is therefore skipped.] Then $\alpha = \beta$ [result #1] And BF = AE = BD [result #2] See figure 2. Join EF. Through O, draw MN // AE cutting EF at N. HOK is ...

2

HINT $a(n) = 3 + \dfrac{1}{\color{red}{3+\dfrac{1}{3+\dfrac{1}{3+\cdots }}}} = 3 + \dfrac{1}{\color{red}{a(n-1)}}$

2

Here are three in the order of the bullets above. Note that the quantifiers do cause some issues, especially in #1, but I think these meet the spirit of the question. I am not sure they are accessible to a layman, though. If $A$ is a closed set of real numbers then if $(x_n)$ is a Cauchy sequence in $A$ then $\lim x_n$ is in $A$. Converse: If every Cauchy ...

2

From your comments, I am guessing that when you pick the counter-example ($n=6$ is one instance as you have mentioned) you are having a difficulty in handling the expressions inside the square bracket (the $k$ dependent expressions). So let us pick $n=6$ as our prospective counterexample. To disprove the statement, we need to show that the statement ...

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