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5

$a)$ it amounts to solving in $\mathbb{Z}: x^2+y^2=3z^2$. You have that $x^2+y^2 = 0 \pmod 3 \to x = y = 0 \pmod 3$, and you get back the original one using descending method, and this proves $x = y = z = 0$, but this means the first equation $x^2+y^2 = 3$ has no rational solutions.


4

First note that: $x\to \sin(x)$ is locally a bijection on the interval you have given , so it's rational for an infinite number of arguments, in particular whenever it's evaluated on $\arcsin(r)$ for rational $r$ Another way of solving your problem is assuming that $\sin(x)$ and $\cos(x)$ are together rationals and here we will come up with the Pythagorean ...


3

The only nice $x$ in this range for which $\sin x$ is rational are $0$, $\pi/6$, and $\pi/2$. Every other preimage of a rational number is neither rational nor a rational multiple of $\pi$. (In fact I'm almost sure that if $\sin x$ is rational and not in $\{-1,-\frac12,0,\frac12,1\}$, then $x$ is algebraically independent of $\pi$).


3

I don't know if this is what you are looking for, but if we work on the unit circle, $sinx$ is the y-coordinate of any point $(x,y)$. Then $a^2+b^2=1$ , where $a=cosx, b=sinx$ . Then $$ b=\sqrt{1-a^2} \in \mathbb Q $$ iff $(1-a^2)= \frac {p^2}{q^2}$ , so $y$ as the coordinate of a point is rational iff the x coordinate is of the form $ \sqrt { 1-p^2/q^2}$ ...


3

Set $\,x=\sqrt{\dfrac{4-2\sqrt 3\strut}7}$, and eliminate progressively the radicals. You'll obtain the equation: $$49x^4-56x^2+4=0$$ If $x$  is rational, say $x=\dfrac pq$, $p$ and $q$ coprime, we know $p$ is a divisor of $4$, and $q$ a divisor of $49$, whence the statement.


3

What you want to prove is: given an irrational number $b$, then one of the following numbers : $$r^{\frac{1}{b}}\ \ \ \ \ \ r\in \Bbb Q $$ is irrational, this seems to be an attainable result for what we know nowadays about the transcendence of numbers, we have for example the following theorem: Six Exponentials Theorem:Let $(x_1,x_2)$ and ...


2

I am not quite sure where Fubini factors in, but your question can be answered by a slightly stronger statement. Proposition Let $A, B\subset \mathbb{R}$ be Lebesgue measurable sets of positive measure, then the difference set $A - B$ contains an interval. To wit, if $A - B$ contains an interval, it must contain a rational number, which would ...


2

Firstly, you should not say "infinite integers" if you mean "infinitely many integers". If one admits such a thing as an infinite integer, and $n$ is an infinite integer, then $n$ and $n+1$ are infinite integers, but they are not infinitely many, since there are only two of them. This is standard usage in mathematics, regardless of what usages may prevail ...


2

Let $r \in L_2$, i.e. $0 < r < 2$. We want to show that there exist $r_1, r_2 \in L_\sqrt2$ such that $r = r_1 \cdot r_2$. Since $\dfrac 2r > 1$ we can choose $v \in \mathbb N$ such that $$ 1 + \frac 1v < \frac{2}{r} \, . \tag 1 $$ Then define $$ u = \max \{ x \in \mathbb N \mid \frac{x^2}{v^2} < 2 \} \tag 2 $$ so that $$ ...


1

According to the Wikipedia article on Pisot–Vijayaraghavan numbers, this is true. In fact, if $x$ is algebraic (not necessarily rational) then it has to be a Pisot number, and in particular an algebraic integer. A rational integer is of course a bona fide integer.


1

One can prove that $$ Trd (\mathbb{Q}(S)) \aleph_0 + 1 \le \dim (\mathbb{Q}(S)) \le (Trd (\mathbb{Q}(S)) +1) \aleph_0 .$$ In particular $\dim (\mathbb{Q}(S)) = Trd (\mathbb{Q}(S)) \aleph_0$ if $Trd (\mathbb{Q}(S))$ is non-zero. Let $S' \subset S$ be a transcendence basis of $\mathbb{Q}(S)$ then $S'$ is algebraically indepenent over the rationals and ...


1

For $N$ an integer, the general result is that if $x^2+y^2=N$ has rational solutions, then it has at least one integer solution.


1

As shown in this answer, $n$ can be written as the sum of two squares if and only if, in the prime factorization of $n$, each prime that is $\equiv3\pmod4$ appears with even exponent. If $x^z+y^2=3z^2$, then $3$ appears with odd exponent. Thus, there are no rational solutions of $$ \left(\frac xz\right)^2+\left(\frac yz\right)^2=3\tag{1} $$ As noted, ...



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