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1

No. Your statements are correct, but it is incorrect to think of the real numbers like this. Because the rationals and irrationals are dense in $\mathbb{R}$, for any two distinct numbers in either set you can find another in between them (in either set, as you mentioned), so you can't possibly have a notion of 'adjacent'-ness of rationals and irrationals. ...


0

For the top four, you need to put the fractions over common denominators. To do that, factor the denominators and collect all the unique terms. It works similar to the naturals, were you would do $\frac 18 + \frac 16=\frac 1{2^3}+\frac 1{2\cdot 3}=\frac 3{2^3\cdot 3}+\frac 4{2^3\cdot 3}=\frac 7{24}$ For the top right, you should recognize $x-2=-(2-x)$ ...


0

You need to clarify that $x \neq q$ when you are quantifying over all $x$, and that fractions are always given in lowest terms. If you do that, then it is true. Lemma: If $x = m/n \neq q$, where $q = r/s$, then $|x - q| \geq 1/ns$. Corollary: If $\delta < 1/ns$ for some positive integer $n$, and $|x - q| < \delta$ for $x$ rational, then $x = m'/n'$ ...


1

Let $f_n:[0,1]\to\Bbb{R}$ be defined, for $x\ne r_n$, by $f_n(x)=\dfrac{1}{\sqrt{\vert x-r_n\vert}}$. Clearly, for every $n$ we have $$ \Vert f_n \Vert_1=\int_0^{r_n}\frac{dx}{\sqrt{r_n-x}}+ \int_{r_n}^1\frac{dx}{\sqrt{x-r_n}}=2(\sqrt{r_n}+\sqrt{1-r_n})\leq 2\sqrt{2} $$ Now the function $f:[0,1]\to [0,+\infty]$ defined by $f=\sum_{n\geq}\frac{1}{n^2}f_n$, ...


0

For any continuous probability distribution on $\mathbb{R}$, the probability of picking a rational number is zero. In other words, if $X$ is a random variable which has a density, $$ P(\mathbb{Q}) = 0 \text{.} $$ The reason is that for continuous probability distributions, its probability measure $\mathbb{P}$ is absolutely continuous compared to the ...


0

Choose an $N$ large enough such that $\frac{1}{N}<i$. Certainly this is possible since $lim(\frac{1}{N}) = 0$. Now let x = $min(\frac{1}{N}, 1-\frac{1}{N})$. And so: $$0<x<i<1-x<1$$


0

Take $a_0$ = 0/1 and $b_0$ = 1/1. Those enclose your number $i$. Now for each next step $n+1$, take the "mediant" fraction of $a_n$ and $b_n$ gained by adding their nominators and denominators and making a new fraction from the sum. Call it $m_n$. Now if $m_n$ > i, let $a_{n+1}$ = $a_n$ < i < $b_{n+1}$ = $m_n$, otherwise let $a_{n+1}$ = $m_n$ < ...


1

Assuming your closed form follows some general formula, all such irrational numbers are a function of rational numbers. As such, you can enumerate them. You can then use Cantor's diagonal argument to construct an irrational number that does not belong to the original set, hence there are irrational numbers than cannot be written using your scheme.


19

The answer is no. The number of expressions which can be formed out of finite combinations of rational numbers is countable, whereas the number of irrational numbers is uncountable, so it cannot be possible.


33

A little extra detail, you have included a function, square root. Suppose we add in a finite vocabulary of functions that can also be used, logarithm base $e,$ exponential, trigonometry, inverse trig, hyperbolic trig, your favorite list of less "elementary" functions suh as hypergeometric, Lambert W, whatever. Any expression is still a finite string ...


11

No, by a counting argument that's worth remembering: There are $\mathfrak{c}=2^{\aleph_0}$ real numbers, but the set of finite sets of rational numbers is countable. The proof on this last statement is relatively straightforward: since we can map rational numbers to whole numbers, we can map finite sets of rationals to finite sets of whole numbers. But ...


0

Hint $\,\ 0\ne \alpha = \sqrt b\! -\!\sqrt d\in\Bbb Q\,\Rightarrow\ \alpha' = \sqrt b\!+\!\sqrt d = \dfrac{\ b-d}{\sqrt b \!-\! \sqrt d}\in\Bbb Q\,\Rightarrow\,\alpha\!+\!\alpha' = 2\sqrt b\in \Bbb Q\,\Rightarrow\Leftarrow$


3

Set $t=a-c$. Then you have $t+\sqrt{b}=\sqrt{d}$. Therefore $$ t^2+2t\sqrt{b}+b=d. $$ If $t\ne 0$, this means…


2

$$\frac x2=\frac{2\sqrt2}{\sqrt2+1}$$ Applying Componendo & Dividendo, $$\frac{x+2}{x-2}=\frac{2\sqrt2+(\sqrt2+1)}{2\sqrt2-(\sqrt2+1)}=\frac{3\sqrt2+1}{\sqrt2-1}=\frac{(3\sqrt2+1)(\sqrt2+1)}{2-1}$$ $$=7+4\sqrt2$$ Similarly, $$\frac x{2\sqrt2}=\frac2{\sqrt2+1}$$ Apply Componendo & Dividendo again


3

Here's another attempt at doing the whole thing, without the possibly confusing notation. Let's define a kind of object that we'll call a "prn". A "prn" is defined as an ordered pair $(a, b)$ (with some additional structure that we'll define below), and to make it clear that it's a "prn" and not just an ordered pair, we'll denote $(a \star b)$. We can ...


0

The main motivation of defining the rationals this way is of course the attempt to make sense of division, since none of the integers except $1$ have a multiplicative inverse. So we start with definition 2, but the symbols used are misleading, since they suggest that division is already taking place, which is not the case. Remember, until now we only know ...


1

Suppose $x \in \mathbb{Q}$ than $x=\dfrac{p}{q}$, where $p,q \in \mathbb{Z}$ and $x^3=\dfrac{p\cdot p \cdot p}{q \cdot q \cdot q}=\dfrac{p^3}{q^3}$. Since the product of two integers is a integer, We conclude $x^3 \in \mathbb{Q}$


0

Rewrite the equation as $y^2-x^2=5$, which gives $(y+x)(y-x)=5$. For the set $\mathbb Q$ you can replace the equation as $(y+x)(y-x)=5z^2$, and solve that for integers. Then $\frac xz$ and $\frac yz$ are the required solutions. Suppose we put $5z^2=720$, then we can represent this as, say, two numbers $8*90$, and $z=12$. This leads to $y=49, x=41, ...


0

If $x$ and $y$ are to be rational, then the factorization $(y-x)(y+x)=5$ implies $y-x=r$ and $y+x=5/r$, where $r$ can be any (nonzero) rational number, in which case we find $$x={5-r^2\over2r}\quad\text{and}\quad y={5+r^2\over2r}$$ If $x$ and $y$ are to be integers, then the factorization requires that $r\in\{\pm1,\pm5\}$, so the only solutions are ...


3

For solutions over $\Bbb Z$ see comments by others. Finding solutions over $\Bbb Q$ is equivalent to solving $$x^2+5z^2=y^2$$ over $\Bbb Z$, with $x,y,z$ coprime. Without loss of (much) generality assume that $x,y,z$ are positive. Two points first: $x,z$ cannot both be even as then $y$ is even, contradicting coprimality; $x,z$ cannot both be odd as then ...


-1

Why do you need computer/software help for this small problem? You are looking for two squares differing by 5. Expand $(x+1)^2$. See how much it differs from $x^2$. This would give you an idea about the difference about consecutive squares.


0

There is such a set $M$. For example, let $M$ be the set of points $(\cos\theta,\sin\theta)$ such that $\tan(\theta/4)$ is rational. Since $\cos x$ and $\sin x$ are rational functions of $\tan(x/2)$, it follows that for any two points $(\cos\theta,\sin\theta)$ and $(\cos \phi,\sin\phi)$ in $M$, we have $\cos(\theta/2)$, $\sin(\theta/2)$, $\cos(\phi/2)$, ...


0

Hint $\ (\sqrt a\!+\!\sqrt b\, + \sqrt c)(\sqrt a\! +\!\sqrt b\, - \sqrt c)\, =\, a\!+\!b\!-\!c+2\sqrt{ab}\ \ (=\, 2\sqrt{ab}\ \ {\rm if}\ \ a+b=c)$


1

Hint: I think a much more elegant way to go about this problem would be to capitalize on the convenient algebraic properties of the golden ratio. Start by converting all the roots of 5 to expressions involbing $\phi$ instead: e.g., $\phi=\frac{\sqrt{5}+1}{2}\implies \sqrt{5}+2=2\phi+1$, and so forth.


1

The next step would be to combine the square roots, $$\sqrt{\sqrt5 - 2} * \sqrt{\sqrt5 + 2} = \sqrt{(\sqrt5 - 2)(\sqrt5 + 2)}$$ Hope that helps.


2

Squaring the fraction gives $$\frac{2\sqrt5+2\sqrt{\underbrace{(\sqrt{5}+2)(\sqrt5-2)}_{=1}}}{\sqrt5+1}=\frac{2\sqrt5+2}{\sqrt5+1}=2$$


1

$$N^2 = \frac{(\sqrt5 + 2) + (\sqrt 5 - 2) + 2\sqrt{(\sqrt 5 + 2)(\sqrt 5 - 2)}}{\sqrt 5 + 1} =\\=\frac{2\sqrt 5 + 2\sqrt{5-4}}{\sqrt 5 + 1} =...$$ Can you see where this is going?


6

What I would do is multiply by the first term plus the conjugate of the last two terms. I have highlighted the important parts of the following expression to make it easier to understand. ...


1

Your "long calculation" was obviously wrong, since the denominator should be $$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$


1

$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$ $$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$ ...



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