New answers tagged

0

At the risk of giving the answer without having the original poster work it through: It's interesting that we are so eager in teaching to "jump" from the field of $\mathbb Q$ to $\mathbb Q \subset \mathbb R$ without pausing with the field of real algebraic numbers in between: $\mathbb Q \subset Algebraic Reals \subset \mathbb R$. Although $\mathbb R$ is an ...


0

As was pointed out in comments, no, neither of those are an algebraic closure of $\mathbb{Q}$. The algebraic closure of $\mathbb{Q}$ would have to contain the root of every polynomial over $\mathbb{Q}$, so $\sqrt{2}$ would have to be in there, and $\mathfrak{i}\sqrt{3}$, and $\sqrt{1+\sqrt{19}}$, and so on. The closure of $\mathbb{Q}$ is called the field of ...


3

Given three points $A=(x_1,y_1), B=(x_2,y_2), C=(x_3,y_3)$ all co-ordinates rational, the 3 line segments joining them have equations with rational coefficients and rational slope. The midpoints P,Q,R of these 3 line segments also have rational coefficients. Now the circum centre of triangle $ABC$ is the intersection of lines passing through P, Q,R and ...


5

$(a, b)$ is the center of that circle. If there are three rational points on this circle, you can find their circumcenter by finding the intersection of the perpendicular bisectors, which can be done by solving a system of linear equations with rational coefficients, so $a$ and $b$ must also be rational.


1

General method: Let $x$ denote the input number Let $|n|$ denote the number of decimal digits in $n$ Split $x$ into the following parts: $\color\red{A}=$ the integer part, i.e., $\lfloor{x}\rfloor$ $\color\green{B}=$ the fraction part's non-periodic prefix $\color\orange{C}=$ the fraction part's periodic postfix Then: ...


0

$.768\overline{786}=\frac{786}{999}-(.786-.768)=\frac{786}{999}-\frac{18}{1000}=\frac{128003}{166500}$


2

Let $x=0.7\color{red} {86}786\overline{786}$ where I fixed what I believe is a typo. Then $1000x=768.768786\overline{786}$ or $999x=786$ If there are $n$ digits in the repeat, you multiply by $10^n$.


0

If $a,b$ are positive integers and $2^{a/b}=7$ then $2^a=7^b$ which make an even number equal to an odd number.


15

Suppose $\log_2 7 = {a\over b}$ for two positive integers $a$ and $b$. $\log_2 7 = { \ln 7 \over \ln 2} = {a \over b}$ Cross multiply, $b \ln 7 = a \ln 2 \implies \ln ( 7^b ) = \ln (2^a)$ Take the $e^{( \ \ )}$ of both sides, $7^b = 2^a$ This is impossible for integers $a$ and $b$ because $7^b$ is always going to be an odd number, while $2^a$ will ...


0

Suppose $a/b$ and $c/d$ are slutions to $x^3+y^3=9$, where the fractions are assumed to be irreducible. I suppose for your problem $a,b,c,d$ are also all positive? In any case, we can then rearrange the equation to$${(ad)}^3+{(bc)}^3=9{(bd)}^3$$ where $a,b,c,d$ are positive integers. Now, look into the prime factorization of $b$ and of $d$. Consider a prime ...


3

This is from Albert Beiler, Recreations In The Theory Of Numbers: The Queen Of Mathematics Entertains. Chapter 25 is called Tilts and Tourneys. This is problem 83, on page 303. The answer is on pages 335-336, and confirms that the desired quantity is rational and positive $x,y$ such that $x^3 + y^3 = 9,$ smallest common denominator, but $(x,y) \neq (2,1)$ ...


1

I am assuming that the question is about Niven's theorem (it is rather poorly phrased) so I give a short proof. If $ \cos(\pi \theta) = \cos( \pi p / q) $, then we have that $ 2\cos(\pi \theta) = \zeta_{2q}^p + \zeta_{2q}^{-p} $ where $ \zeta_{2q} = e^{\pi i/q} $ is a primitive root of $ X^{2q} - 1 = 0 $, and therefore is an algebraic integer. This implies ...


3

The topology doesn't care about exact distance values in the metric; it only cares about what points are close to what others in a broad sense. So for a question where the exact distance values are important, you wouldn't expect the question to say the same under different metrics even if they induce the same topology.


4

At least your integral turns out to have a closed form: $$ \int_{0}^{\pi/2} x \, \frac{\sqrt{\sin x} - \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, \mathrm{d}x = G + \pi \left( \frac{1+2\sqrt{2}}{4} \log 2 - \log (1+\sqrt{2}) \right), $$ where $G$ is the Catalan's constant. So it seems to me that the 'almost rationality' of this integral is just a ...


0

Consider any function $f$ such that $\int_{a}^{b} f(x)dx$ does not have a closed form, now choose $b-a$ small enough that the integral is very close to zero. Eg: $\int_{0}^{1} e^{-x^{100}} dx$ perhaps does not have closed form and is surely very close to zero. Any such function can be modified to make the integral very close to any given rational integer.


1

Integral of $e^{-2x^2}$ from $-2$ to $62$ is about $1/8$ which has a $.003$ error. Also maybe cheating but its possible to get this integral arbitrarily small giving really small errors to something without closed form. This has implication that you could integrate any probability distributions most of which have no close form to 1 plus an arbitrary small ...


3

Well, $$\begin{align}\dfrac{\sqrt{2}+\sqrt{a}}{\sqrt{3}+\sqrt{b}} ~=~& \dfrac{(\sqrt{2}+\sqrt{a})(\sqrt{3}-\sqrt{b})}{(\sqrt{3}+\sqrt{b})(\sqrt{3}-\sqrt{b})} & \textsf{except when }b=3 \\[2ex] =~&\dfrac{\sqrt{6}+\sqrt{3a}-\sqrt{2b}-\sqrt{ab}}{3-b} \end{align}$$ Then, when might this be rational? ...


0

I would say no consider only two points which form an ellipse then by the erdos anning theorem all points of mutual rational distance (ellipse) which includes the other point. all points then have rational coordinates. this is poorly worded take the first two points as foci. take the radius such that the ellipse hits the third point.


4

From your comment below, you apparently take a point with irrational coordinates to mean a point with at least one irrational coordinate. Suppose the point $(x,y)$ is a rational distance from all of $a,b,c$. Then the square of its distance from $a$ is $x^2+y^2$ and the square of its distance from $b$ is $(x-1)^2+y^2$. These must both be rational and hence ...


3

The claim is false without additional assumptions on $\frac mn$. For example if $\frac mn=\frac1{10}$ then your next guess is $10.05$ which is further away from $\sqrt 2$. The claim is true, however, if we impose a condition like $\frac mn>1$. Write $x=\frac mn$. You want to show that $\frac x2 +\frac1x$ is closer to $\sqrt 2$ than $x$ is. In other ...


-1

You have indicated how to obtain next approximation starting from given present value by drawing a tangent of graph of $ y =x^2 $ when $ y=2 $ and reading off the new improved value by the iterative method of Newton. Let $$ f= x^2 -2;\, f^{\prime} = 2 x \tag{1} $$ Successive iteration values are suffixed : $$ x_{N +1} = x_{N} - \frac{f}{f \prime} = x_N - ...


7

Divide both sides by $ab$. We get $\frac{1}{a}>\frac{1}{b}$.


3

$\frac{m}{n}$ approximately equals $\sqrt 2$ Suppose $\epsilon$ is our the error in our estimate. i.e. $\frac{m}{n} + \epsilon = \sqrt 2$ or $(\frac{m}{n} + \epsilon)^2 = 2$ $\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon + \epsilon^2 = 2$ Now we need to solve for $\epsilon.$ If $\epsilon$ is small then then $\epsilon^2$ is very small. Which is a little ...


0

You can use the iterative method: $a_{n+1}=\frac{a_n+\frac{2}{a_n}}{2}$. Thereby $a_{n+1}$ is more accurate than $a_n$. If $a_n=\frac{m}{n}$: $a_{n+1}=\frac{\frac{m}{n}+\frac{2}{\frac{m}{n}}}{2}$ This is equivalent to: $a_{n+1}={\frac{m}{2n} + \frac{n}{m}}$


1

Hint 1 : Putting $x=\frac{m}{n}$, try to show that if $x$ is an approximation to $\sqrt{2}$ then $f(x)=\frac{x}{2}+\frac{1}{x}$ is a better approximation to $\sqrt{2}$. Hint 2: $\frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}}=f'(c)$ for some $c$ between $x$ and $\sqrt{2}$ (this is the mean value theorem) If you don't know about derivatives, you can also compute that ...


0

I can't comment, but I'm afraid that the answer of George is incorrect. Indeed the equation has at least 2 solutions, namely $[0:1:0]$ and $[-1:-1:1]$


1

Unless I missed something in your question, your field $S$ is obtained by adjoining to $\mathbf Q$ all the square roots of all the positive rational numbers, or equivalently, of all the natural integers, or equivalently, of all the positive rational primes $p$. As a compositum of quadratic fields, $S$ is thus the maximal real subfield contained in $T$ := ...


1

Your solution does not work. First, $m$ is always different from $0$ because $H(\mathbb Q)$ contains the cyclic subgroup $\{(1,0),(-1,0)\}$. Second, what does it mean that "$\mathbb Q(\sqrt{A})^*$ has no linearly independent elements"? Because actually $\mathbb Q(\sqrt{A})^*$ contains a lot of linearly independent elements! To prove the claim, one can ...


4

One of the most remarkable integrals for me is the Borwein sequence. They do have closed form, but somehow I feel they should be mentioned here. More interesting integrals.


1

You have done most of the work, you just need to prove transitivity. If $a R b$ and $b R c$, $\frac{a}{b}=2^{m_1}$ and $\frac{b}{c}=2^{m_2}$, multiply them together, you have $$\frac{a}{c}=2^{m_1+m_2}$$ Since $m_1+m_2 \in \mathbb{Z}$, $\frac{a}{c} \in H$, i.e. $a Rc$.


0

General method: Let $x$ denote the input number Let $|n|$ denote the number of decimal digits in $n$ Split $x$ into the following parts: $\color\red{A}=$ the integer part, i.e., $\lfloor{x}\rfloor$ $\color\green{B}=$ the fraction part's non-periodic prefix $\color\orange{C}=$ the fraction part's periodic postfix Then: ...


1

$$0.412\overline{8754}=\frac{4128754-412}{9999000}$$ Edit: the numerator is the difference of the number build by the preperiod followed by the period, in our case $4128754$, and the preperiod, here $412$. For the denominator, write down as much nines as the period is long, here $9999$, followed by as much zeros as the preperiod is long, in our case $000$.


3

Say the largest element is $r$, It is obvious that $\frac{2+r}{2}$ is still en element of $S$, which is a rational number. What's more $\frac{2+r}{2} > r $, that is a contradiction.


0

Can you use that the rationals are dense? If yes, the answer is immediate: assume $S$ has a maximum, say $r_0$ and therefore, $r_0<2$. But since the rationals are dense in the real numbers, you can find $r_1\in\mathbb Q$, such that $r_0<r_1<2$. It follows that $r_1\in S$, but that contradicts the assumption that $r_0$ is the maximum of $S$.



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