# Tag Info

1

If the number is rational, using multiplication formulas for $\cos$ and $\sin$ you can express the original parametrization in terms of $\cos$ and $\sin$ of multiples of a single argument. Then you can use a rational parametrization of the circle to replace those $\cos$ and $\sin$ by rational functions of the parameters, and then you can eliminate the ...

2

This is about 2-adic restrictions. First, odd squares of integers are $1 \pmod 8.$ Integer squares can only be $0,1,4 \pmod 8$ in any case. Therefore the sum of three integer squares cannot be $7 \pmod 8.$ Next, if the sum of three squares is divisible by $4,$ so $x^2 + y^2 + z^2 = k$ with $k \equiv 0 \pmod 4,$ then $x,y,z$ must be even so we can divide ...

1

Hint to get you started: If all of $a,b,c$ are multiples of $7$ then so is $n$ and we can divide all variables by $7$. Hence we may assume that not all of $a,b,c$ are multiples of $7$. A square can only be $\equiv 0,1,2,4\pmod 7$. For three of these to add to $0$, we must have $0+0+0$ (which we just excluded) or $1+2+4$.

2

Here's a way to generate lots of primitive solutions, though not all of them. $A = 4r^4 - 4p^3r$ $B = 8pr^3 + p^4$ $C = 20p^3r^3 - p^6 + 8r^6$ Then $A^3 + B^3 = C^2$

1

Consider a set of the form $(a,b)\cap\mathbb{Q}$; it's useful to consider $\mathbb{Q}$ as a subspace of its completion $\mathbb{R}$), because this tells us that $(c,d)\cap\mathbb{Q}$ is open in $\mathbb{Q}$, for any choice of $c,d\in\bar{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$. Suppose now $a$ is irrational. Then consider a decreasing sequence $(a_n)$ ...

1

To show non-compactness, we shall just formalise your convergence idea. Let $a_n$ be a strictly decreasing sequence tending to $\sqrt{2}$ (for example take alternating terms of the truncated continued fraction) and let $b_n$ be a strictly increasing sequence tending to $\sqrt{3}$. Consider the sets $U_n=(a_n,b_n)\cap \mathbb{Q}$ (open by the same argument ...

2

Another approach: Observe that the topology on $\mathbb Q$ is the same as the subspace topology on $\mathbb Q\subseteq\mathbb R$ as a subspace of $\mathbb R$ where $\mathbb R$ is equipped with its usual topology. You can write: $$E=\mathbb Q\cap\left([-\sqrt3,-\sqrt2]\cup[\sqrt2,\sqrt3]\right)$$ Here $F:=[-\sqrt3,-\sqrt2]\cup[\sqrt2,\sqrt3]$ is closed in $\... 2$x\in\mathbb Q\iff \exists a,b\in\mathbb Z: b\ne0,\ x=\frac ab$Let$r=\frac pq$,$s=\frac tu$, where$p,q,t,u\in\mathbb Z$and$u,q\ne0$. Then, If$s=0$,$\frac rs$is not defined. Assuming,$s\ne0$,thus,$t\ne0$.$\frac rs=\frac {pu}{qt}$, where$pu,qt\in\mathbb Z$, as$t,q\ne0$;$qt\ne0$. Thus,$\frac rs\in\mathbb Q$.$r-s=\frac {pu-tq}{qu}$, where$pu-...

0

Rudin doesn't give a #*@$whether there is a rational square root of 2 or not. What he's trying to show is that you can divide all the rational numbers into two sets that exhaust the rationals; that one set can have every element larger than every element in the other yet it is possible to have no limits to either set; you can take infinitely many larger ... 0 Here is the classic Put the natural numbers on the horizontal and vertical axes of a grid. Filling the grid, take integer on the horizontal axis, and divide it by the integer on the vertical axis. The grid is filled with the positive rational numbers. Yes many numbers are duplicated, but every rational is represented. Now draw a serpentine line that ... 0$ f(\pm\frac{a}{b}) = \pm2^a 3^b $is an injection$ \mathbb{Q} \to \mathbb{Z} $. 3 The numerator and the denominator are the "parts" of a fraction. It comes from the terminology of simple fractions. The fraction$\dfrac{2}{3}$is two parts out of three parts. 3 If we make a rectangular grid with integer coordinates, it's possible to assign a unique angle to any rational number, using the definition$\tan \phi=y/x$for$\phi \in (-\pi/2, \pi/2)$. For positive rationals it would look something like this: It's intuitively clear to me that the lines corresponding to the rational numbers can't fill all the space ... 3 The rational numbers can be represented diagrammatically as a 'tree' (for example Farey Sequence or Sternâ€“Brocot tree). Such a tree, continued infinitely, is a fractal. The outer edge of the fractal will always be rough, whereas the real numbers (if similarly represented) would be smooth. 3 This is a somewhat more "geometric" argument, somewhat analogous to Cantor's. Consider any sequence of distinct real numbers$x_n$. I will construct a real number$y$that is not in the sequence, as the limit of a sequence$y_n$. Start by choosing$y_1 \ne x_1$. At stage$n$, we will have a real number$y_n$that is not$x_1, \ldots, x_n$. Let$a_1, \...

12

Here is a geometric difference between the rationals and the irrationals: the length of the rationals equals zero, whereas the length of the irrationals equals infinity. Here's why the rationals have zero length. Start with an enumeration of the rationals: $$p_1,p_2,p_3,...$$ Pick your favorite tiny positive number $\epsilon>0$. For each $k=1,2,3,...... 16 The rationals can be mapped into the lattice points$(n, m)$, which are an infinite set of isolated points in the plane. The irrationals, by any of the standard ways of mapping two reals into one (such as zipping the digits of a pair of reals) fill the whole plane (with quibbles about two representations or points missed being disregarded). 4 In general, my favorite way to think about this is to think about a countable set (rationals) as the set of all finite strings of$0$'s and$1$'s and an uncountable set (reals) as the set of all infinite sequences of$0$'s and$1$'s. I think this makes it "geometrically" obvious, at least for me. EDIT 1: Any countable set is in bijection with$\mathbb N$, ... 0 @HennoBrandsma gave you a sketch of one formal proof. Here's another way: If we use the Cantor-Shreoder-Bernstien (CSB) theorem, a direct formal proof becomes easier. The CSB theorem states a bijection exists between 2 well defined nonempty sets A and B iff there exists injective functions f and g where$f: A \rightarrow B$and$g:B \rightarrow A$. Now let'... 3 The idea is indeed correct, but needs a little work to formalise. You note that for every real$r$there is a sequence$(a_n(r))_n$of points that are all distinct and say (strictlty) increasing, so that$a(r)_n \rightarrow r$as$n \rightarrow \infty$. You could choose the finite decimal approximations of$r$(when we write$r$as an infinite decimal ... 0 EDIT I accidentally tried solving the wrong problem. I did not know about the$\{\cdot\}$notation for fractional part. This is an attempt to show that$x^2+x=1$has no solutions for$x\in\mathbb{Q}$. Here's my attempt: Assume there are$p,q \in {\mathbb Z}$so that they are relative prime. $$\left(\frac{p}{q}\right)^2 + \left(\frac{p}{q}\right) = 1\... 5 Suppose that there exists such a positive rational number. We have$$x^2-\lfloor x^2\rfloor+x-\lfloor x\rfloor =1,$$i.e.$$x^2+x=\lfloor x^2\rfloor +\lfloor x\rfloor +1$$We can set x:=p/q where p,q are positive integer with \gcd(p,q)=1, then$$x^2+x=\frac{p}{q}\left(\frac pq+1\right)=m\tag1$$where m\in\mathbb Z. Then,$$(1)\implies mq^2=p(p+q)\... 7 Note that if$\{x^2\} + \{x\} = 1$, then$x^2 + x$is an integer. Solve the equation$x^2 + x = n$for an arbitrary$n$, and see that if it has rational solutions, then those rationals must be integers, which means that$\{x^2\} + \{x\} = 0$. To see that$x^2 + x$must be an integer, note that for any$y$we may use the floor function to write$y = \lfloor ...

0

Suppose this was not true. Then, we can express $\frac{2y}x$ in the form $\frac pq$, where $p,q\in\mathbb Z,q\ne0$. Moreover, as $x$ is rational, and x is non-zero, we can express $x$ as $\frac ab$ where $a,b\in\mathbb Z$ and $a\ne0,b\ne0$. Thus, $$\frac{2y}{x}=\frac{2yb}{a}=\frac{p}{q}$$ $$y=\frac{ap}{2bq}$$ Doing so, we have represented $y$ as a ratio of ...

1

Hint: product and quotient of rationals are rational. So $x/2$ is a rational and if ...

1

$$\frac 1n=0.abcabcabc......\Rightarrow\frac 1n = \frac{abc}{10^3}(1+\frac{1}{10^3}+....+\frac{1}{10^{3n}}+....)=\frac{abc}{999}$$ It follows $$abc=\frac{999}{n}=\frac{3^3\cdot 37}{n}$$ $3^3\cdot 37$ has $(3+1)(1+1)=8$ factors from which we have eight solutions for $n$. The "trivial" ones are $1,3,9$. The other five are $27,37,111,333,999$ whose decimal ...

0

It should at least be a factor of $999$. So, $$999 = 3^3 \cdot 37$$ Your choices then are $1, 3, 9, 27, 37, 111, 333,$ and $999$. The choices $n=1, 3,$ and $9$ have repetends of length $1$. All others have repetend of length $3$. As for why this works, let's say that $k = 0.abcabcabc ...$ is a repeating decimal with a pattern of three digits $a,b,c$ (...

4

Hint: Let $n\in \mathbb N$ such that $n > x$. Take $y=n-x$. Make sure to justify that such an $n$ exists and that $y$ is irrational.

1

Even better - for every real $x$, there is a subsequence of $(a_n)$ converging to $x$. Indeed, start by letting $a_{n_1}$ be any element of the sequence in $(x-1,x+1)$. Then, given $a_{n_1},\ldots,a_{n_k}$, since there are infinitely many rationals in each interval, there must exist $n_{k+1}>n_k$ such that $a_{n_{k+1}}\in(x-\frac1{k+1},x+\frac1{k+1})$. ...

2

The argument given by Elio Joseph can be refined in order to prove that for any given real number, there always exists a subsequence which converges to it. For that, let $x \in \mathbb{R}$ be a real number. Consider the interval $[x-1,x+1]$. It contains infinitely many rationals, but in particular it contains one rational, which is some $a_{n_1}$ of our ...

1

Yes there is. Consider the interval $I:=[0,1]$. Since it contains infinitely many rationals, your sequence will have value in $I$ forever. So you can extract the sub-sequece such that all the values $(a_{\varphi(n)})$ of the new sequence are in $I$ : $$\forall n \quad a_{\varphi(n)}\in I.$$ Since $I$ is a compact and $(a_{\varphi(n)})$ is a sequence ...

1

A generic line intersects a conic in $2$ points. If the conic and the line are rational (given by rational coefficients), then the set consisting of those two points is invariant by $\Bbb Q$-automorphisms, so either each point is rational, either they are defined over a quadratic extension of $\Bbb Q$ and the automorphism of that extension swaps the two ...

0

Use Euclid's Algorithm to compute $\gcd(a,b)$, then divide the numerator and denominator by it. Voila! Reduced fraction! Here is a link: https://en.wikipedia.org/wiki/Euclidean_algorithm

5

You have two cases. If the number of decimals is finite, let's say $x=a.a_1 a_2 ... a_n$. Then obviously $$x = \frac{a a_1 a_2...a_n}{10^n}$$ and then you simplify this fraction by looking for common divisors. If the number of decimals is infinite, then it must have a period $p$, that is $$x=a. a_1 a_2 ... a_p a_1 a_2 ... a_p ...$$ You have 10^p x-x = a ...

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