# Tag Info

13

Yes. Every rational number $\ge 0$ is the sum of four squares. This is easily derived from Lagrange's Theorem which says that every non-negative integer is the sum of four squares. Note that the formula we get is existential.

10

You have put your finger precisely on the statement that is incorrect. There are two competing conventions with regard to rational exponents. The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write ...

7

Hint: $f(\Bbb Q)$ is countable. New hint given the edit: $f(\Bbb R)$ must be connected, and hence an interval, and hence must contain uncountably many irrationals.

6

A QUITE SIMPLE REMARK $$\begin{cases}\frac{e}{\pi}=t\\e+\pi=s\end{cases}\qquad (*)$$ would imply $$\pi=\frac{s}{t+1}\\e=\frac{st}{t+1}$$ Consequently and least one of $t$ and $s$ in (*) must be trascendental.

5

The issue is that $a^{\frac{1}{n}}$ is multivalued. You could arguably simplify the first calculation into $1 = \sqrt{1} = -1$. Taking different branch cuts is how the "paradox" arises. Essentially, in the context of the reals (or even the complex numbers) $\sqrt{a}$ is one name for two functions, say $\sqrt[+]{a^2} = a$ and $\sqrt[-]{a^2} = -a$. All the ...

5

Definitely not. $\frac{\pi^x}{\zeta(x)}$ is continuous on at least $x\in[3,4)$ (just giving a counterexample here), and non-constant, thus, it'll have rational values for some $x\in[3,4)$ (and so $x$ is not an even integer)

4

For the first one, you can proceed directly: Notice that if the expression is equal to some rational $r$, then $$x + b = r(x + a) \implies x(1 - r) = a - b$$ Now the right side is rational, but the left side is irrational unless..... For the second, I'd suggest proceeding similarly. Write $$x^2 + x + \sqrt 2 = r(y^2 + y + \sqrt 2)$$ and rearrange to ...

4

$\mathbb{R}/\mathbb{Q}$ is a $\mathbb{Q}$-vector space. Choose a $\mathbb{Q}$-basis for $\mathbb{R}/\mathbb{Q}$, and pick a preimage for each basis vector in $\mathbb{R}$. By mapping each basis vector to the chosen preimage, you have successfully constructed a $\mathbb{Q}$-linear map $\mathbb{R}/\mathbb{Q} \to \mathbb{R}$ that splits the quotient map. This ...

4

The existence of 2 rational points on a line $L$ implies the existence of infinitely many. For if a line $L$ has two rational points $p=(x_1,y_2)$, $q=(x_2,y_2)$, then $L$ satisfies an equation with rational coefficients $ax+by=c$, which one can obtain by rewriting the "two-point" form of the equation for $L$ which is $(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)$. ...

4

We need to show that, if any two of $\{x-y,\; x+y,\; x^2-y^2,\; x^2 + y^2\}$ are rational then that at least one of $\{x,\; y,\; x^2,\; y^2,\; xy \}$ is rational. Which means that the answer to the question is no. There are $6$ ways to choose two objects out of $4$. Case 1, 2, and 3: Any two of $\{x-y,\; x+y,\; x^2 - y^2\}$ are rational. Since $(x-y)(x+y) ... 3 I've thought about this question for a while without an answer. The key is to consider the structure of the constructible real numbers. I was actually a bit cavalier with my original definition, "$x$is irrational if$|x-p/q|<q^{-2}$has infinitely many coprime solutions". The problem lies in what is meant by "infinitely many" here. If it means that there ... 3$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$The thing that looks the most suspect in my example above is the 4th equality,$(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... ... 3 No continuous definition of$a^r$can be made for all real$a$and$r$; and likewise, the familiar properties of exponents cannot be extended consistently to all real bases and powers. As a result, there are a number of competing definitions for$a^r$for non-integer values$r$, depending on how much the author wishes to extend these properties, and in what ... 3 I should amend my comment. The infinite tetration$x^{x^{x^{.^{.^.}}}}$is only defined for$e^{-e}<x<e^\frac{1}{e}$Indeed this can be define for$x>1$. For example, if$x=\sqrt 2$the tower is equal to$2$. There are some notes on this topic here. It is easy to convince yourself that this number blows up by calculating the first few values ... 2$(a^r)^s=a^{rs}$can indeed be false for$a<0$, as shown by your example. You can "rescue" this rule by stating instead "$(a^r)^s=a^{rs}=(a^s)^r$, provided all three expressions are defined". (As the product is commutative, you cannot really distinguish$r$and$s$.) 2 At least for the question of a rational function$\Bbb N\to\Bbb Q$, the answer is no. In fact, we can make a stronger claim: If$f:\Bbb N\to\Bbb R$is a rational function, then the range of$f$is either bounded above or bounded below. Proof: Separate the highest order term of$f$, writing it as$f(x)=ax^n+g(x)$where$g\in o(x^n)$for some$n\in\Bbb ...

2

Well, if you've actually proved the biconditional statements you mentioned, then you're done. Alternatively, show that $$\Bbb Q\subseteq\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr),$$ which I leave to you. Then, suppose $z\in\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr).$ Since $z\in\Bbb Q\bigl(\sqrt2\bigr),$ then $z\in\Bbb R.$ From there, we can use the fact that ...

2

Yes, consider $\alpha + \frac{1}{n}$ where $\alpha$ is irrational and $n$ is an integer. $\alpha + \frac{1}{n}$ is also irrational and can be made arbitrarily close to $\alpha$ by choosing $n$ to be sufficiently large.

2

You can show the contrary, i.e. the set of $x \in [0, 1]$ that there are infinitely many $p/q \in \mathbf{Q}$ in reduced form such that $q \geqslant 2$ and $\left| x - p/q \right| < 1 / \left( q \log q \right)^2$ is of measure $0$. Edit: Let $a_i=p_i/q_i \:(p_i<q_i,\:2\leqslant q_i\in\Bbb{N})$ be the enumeration of rational numbers in $[0, 1]$ and ...

2

It is a famous result of Julia Robinson that $(\Bbb{Q}, +, \cdot, 0)$ is undecidable. This implies that $(\Bbb{Q}, +, \cdot, 0)$ does not admit elimination of quantifiers. That the rational numbers are not definable in the first-order theory of the reals follows from this, but also follows from well-known facts about O-minimality of the first-order theory ...

2

Let's use a binary encoding for our way from the top of the tree to a particular fraction. We encode the starting position and each move to the right as $\color{red}1$ while a move to the left is $\color{blue}0$. (This leads to the familiar binary representation of the Calkin-Wilf sequence.) Now, if we move to the right, we move from $\frac ab$ to ...

1

Just to elaborate on the method proposed by user1952009: Let $Q(x)$ be the number of rationals $p/q \in \mathbf{Q} - \left\{0, 1\right\}$ in reduced form with $q \geq 2$ satisfying $$\left| x - \frac{p}{q} \right| < \frac{1}{\left( q \log q \right)^2}$$ and let $\mathbf{Q}(x)$ be the set of such rational numbers. Since $$\int_0^1 Q(x) dx < \infty ... 1 Let x be an arbitrary real. As \mathbb Q is dense in \mathbb R, we can surely find a sequence (q_n)_{n\in\mathbb N} of mutually distinct rationals with |x-q_n| < \frac1{2^{2n}} for all n. Let (p_n)_{n\in\mathbb N} be an enumeration of all rationals (there are certainly infinitely many) not occurring in this sequence. For n\in\mathbb N, ... 1 Hint: Since \bf R is connected and f is non constant hencef(\bf R) is connected and hence uncountable. 1 1. There are p,q\in\mathbb{Z}, with gcd(p,q)=1 such that \frac{x+a}{x+b}=\frac{p}{q}. Then$$(x+a)q=(x+b)p\Leftrightarrow x(q-p)=bp-aq\Leftrightarrow x=\frac{bp-aq}{q-p}\in\mathbb{Q}\Leftrightarrow x=0.$$This is absurd. Therefore p=q and x+a=x+b, so a=b. 2. Let r\in\mathbb{Q} such that \frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r and suppose that ... 1 This is an alternative argument based on Arnaud's suggestion. Without loss of generality, you can assume the rational function f is \frac{p}{q} where n = \deg(p) \ge \deg(q) = m (otherwise trade f in for \frac{1}{f}). But then (using (\frac{p}{q})' = \frac{p'q - pq'}{q^2}) you have:$$ f'(x) = \frac{(n-m)p_nx^{m+n-1} + \mbox{terms of lower ...

1

yes it is true $Q\sqrt2$ and $Q(i)$ are extension of degree $2$ of $Q$ so the degree of $Q(i)\cap Q(\sqrt2)$ is either 1 or $2$, if it is 2 it implies that $Q(i)=Q(\sqrt2)$ thus $\sqrt2=a+bi, a,b\in Q$, by writing $(\sqrt2-a)^2=-b^2$, you obtain $a=0$ unless $\sqrt2\in Q$ a fact which is not true. If $a=0$, $\sqrt2=bi$ thus the $2=-b^2$ this not true also.

1

So that means that there are two irrational numbers surrounding a rational, right? No, it is not the fact that there are only two irrationals surrounding a rational; there are actually infinitely many (for any distance of interest, say $\epsilon > 0$). This is referred to as the set being "dense". So if I take the A.M. of those two irrational ...

1

While I agree with everything in the answer by David, I'll give a different answer here just to put a different emphasis. The fundamental error is to put the rule $(a^r)^s=a^{rs}$ in the box governed by the condition provided that all the expressions used are defined. That is not the right kind of condition for this rule, it requires specific limitations to ...

1

There is a mistake in part (ii): as pointed out in the comments, $0$ is not in $E$. What you wrote in part (i) is true, but more complicated than it needs to be. Here are a few suggestions to simplify your proof: Delete the first sentence. The variable $M$ is not used anywhere in the rest of the proof. Delete the sentence that starts with "Then for some ...

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