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You get $(bx)(sy)=ar$ alright, but rewriting it as $(\frac{b}{1}\cdot \frac{a}{b})(\frac{s}{1} \cdot \frac{r}{s})=ar$ is the wrong way to proceed. Instead use the associative and commutative law on the left-hand side to get $$(bs)(xy)=ar$$ In other words, if $x$ is a number such that $bx=a$ and $y$ is a number such that $sy=r$, then $xy$ will be a number ...

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Question 1 You need to be slightly more careful about what you can and can't do. Def. 1 doesn't allow you to conclude that $\frac{a}{a}=1$, only that $\frac{a}{a}=\frac{1}{1}$, and so your proof doesn't work. You won't be able to derive the definition of a fraction from the definition of multiplication. (Indeed, if you have no rule telling you when things ...

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For your first problem, notice that when you say: if $ba'=ab'$, then $\frac{a}{b}\times\frac{b'}{a'}=1$ you are in fact using definition $2$. For the second problem, the author is simply telling you that you have a canonical map $\mathbb{Z}\to\mathbb{Q}$ when you look at $\mathbb{Q}$ as the ring of fractions of $\mathbb{Z}$. For the third ...

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The equation is symmetric in $x$ and $y,$ suggesting solutions of the form $x=y.$ Plugging that in, we find $$2x^3-6x^2=0$$ This gives the solution $x=y=0.$ Simplifying by dividing by $2x^2$ gives $x-3=0,$ leading to another solution $x=y=3.$

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Which are the solutions over the integers? $(x+y)^3=\underbrace{x^3+y^3}_{6xy}+3xy(x+y)=3xy(x+y+2)\iff6|(x+y)$, since, on one hand, $3$ is a prime, and, on the other hand, x and y being of opposite parity would lead to contradiction. Hence, we have $x=2a+r$, and $y=2b-r$, with $r\in\{0,1\}$, and, at the same time, $x=3A+R$, and $y$$=3B-R, also with ... 3 If x and y are rational, then so is y/x=\alpha. Then x^3+y^3=6xy becomes$$ (\alpha^3+1)x^3-6\alpha x^2=0\tag{1} $$and then x=0 or x=\dfrac{6\alpha}{\alpha^3+1}. Thus, for any rational \alpha, we have the rational solutions$$ \left(\frac{6\alpha}{\alpha^3+1},\frac{6\alpha^2}{\alpha^3+1}\right)\tag{2} $$Since ... 0 Wolfram Alpha says that there are no rational solutions except the one you noted, x=y=3 although. It seems that it chose to skip the trivial x=y=0 though. The link has some irrational solutions too, if you need them. 5 Lemma 1: For any real number x, there exists a sequence of rational numbers (x_n) such that \lim (x_n) = x. Proof: take e.g. x_n to be the decimal expansion of x truncated to n digits. Lemma 2: For any real number x, there exists a sequence of irrational numbers (y_n) such that \lim (y_n) = x. Proof: The idea is to start with rationals ... 3 For (1), observe how \mathbb{R} is usually constructed - as the set of all finite limit points of sequences in \mathbb{Q}. So not only is (1) wrong, you have that for every x \in \mathbb{R} there is a sequence (a_n) contained in \mathbb{Q} such that \lim_{n\to\infty}a_n = x. For (2) - if that was true, then only sequences of rational numbers ... 24 Both should be no. For (2), just divide by \pi of your first example. 1 Here's a hint. Let U be an open set on \mathcal{T}. Let x be any element of \mathcal{T}. Is it true that we can find some set V which is open in the topology of \mathcal{T}' where x \in V \subseteq U? 0 Ok this question is tricky! As you pointed out that because \mathbb{Q} is dense in \mathbb{R} we will not find any open Set that contains just one rational number. As we will not find any open set that will contain just one irrational number. "I see a set of measure zero to have gaps in it that stop it from having a full interval." This is basically ... 1 We want to show that$$a + b\sqrt 2 = c + d\sqrt 3 \tag 1$$with a, b, c, d rational implies b=d=0 From (1)$$ 2 b^2 = (c + d\sqrt 3 -a)^2 = (c-a)^2 + 2 (c-a) d \sqrt 3 + 3 d^2$$or$$ 2 (c-a) d \sqrt{3} = 2 b^2 - (c-a)^2 - 3 d^2 $$since \sqrt{3} is irrational we need both sides to be identically zero. If d \ne 0 then$$ a = c, ... 0 You're basically trying to prove that if$a+b\sqrt2=c+d\sqrt3$, for rational$a,b,c,d$, then$b=d=0$. Try placing the square roots on one side, then square both sides. Rearrange, and try to prove that$b$or$d$must be equal to 0. 3 For the proof you can just give some explicit construction that works. E.g., note that if$x < y$, then$x < x + \frac{1}{3}(y - x) < x + \frac{2}{3}(y - x) < y$. Then argue that if$x, y \in\mathbb{Q}$, then the two new numbers are rationals, too. 0 You can just use the fact that rationals numbers are dense. So given$a$and$b$, there is a rational number in$(a,b)$and there is a rational number in both of$(a,x)$and$(x,b)$, where$x$is any number in$(a,b)$. 1 let$a$and$b$be distinct rational numbers. Let$a<b$without loss of generality. $$a<\frac{a+\frac{a+b}{2}}{2}<\frac{a+b}{2}<b$$ As a logical expression this could look like$(a,b\in \mathbb Q\wedge a<b)\rightarrow \exists p,q\in \mathbb Q|a<p<q<b $0 If you restrict to finite decimal expansions, then every number lives in the ring$\mathbb{Z}[1/10]$. I'm not sure it's a disaster, but it's definitely not the real numbers... 2 You are right. The computation is just too inefficient. The best known attack on ECDLP, the pollard rho attack, would be useless against elliptic curves over the rationals. Consider this, if you were to do the computations over a finite field of say 512-bits, you will only have to deal with 512-bit intermediate values along the way. Considering the same ... 2 Hint$ $Any rational root of$\,x^3-p\,$is an integer, by the Rational Root Test. Alternatively$\, a^3 = pb^3\,$contradicts the uniqueness of prime factorizations, since the prime$\,p\,$occurs to power a multiple of$\,3\,$on the lhs, but a nonmultiple$\,1\!+\!3n\,$on rhs, i.e.$\,0\not\equiv 1\pmod 3.\,$This is a generalization of the ... 3 The main point is: The cube root of a natural number is rational iff it is infact an integer. More generally, any rational root of a monic polynomial with integer coefficients (such as$X^3-n$) is in fact integer. So if$\sqrt[3] n$is rational then$n$is a cube (and cannot be prime). 9 Suppose$\sqrt[3]{P} = \dfrac{a}{b}$where$a$and$b$have no common factors (i.e. the fraction is in reduced form). Then you have $$b^3 P = a^3.$$ Both sides must be divisible by$a$(if they're both equal to$a^3$). We already know that$a$does not divide$b$(when we assumed the fraction is reduced). So then$a$must divide$P$. EDIT: and if$a = ...

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$\Bbb Q[\sqrt 2]$ is a field and a countable dense linear order. All countable dense linear orders without endpoints are order isomorphic. That doesn't say there is a nice way to define the isomorphism by a formula.

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Yes. It's a much celebrated theorem of Julia Robinson. Julia Robinson, Definability and Decision Problems in Arithmetic. The Journal of Symbolic Logic, Vol. 14, No. 2 (Jun., 1949) , pp. 98-114

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