New answers tagged

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 ...


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.


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 ...


0

[Please don't upvote - just adding answer so question not stuck in unanswered state forever] As per the numerous comments, your proof is correct.


0

As Henning Makholm pointed out in his comment, your proof that there is always a smaller rational number should point to the existence of $\frac{-p + 1}{q} < \frac{-p}{q}$. Otherwise, yes: you have have supposed that your conclusion was false and shown that that supposition led to a contradiction. Therefore your conclusion must be true.


0

Well, Algebra meets Calculus and they disagree on some points, literally (unless well guided). Algebra says, "I have a polynomial $x^n=a$ with $n$ different complex roots. ($a\neq 0$, $n$ integer). And for positive real numbers I can have a function $\sqrt[n]a$ that is positive and solves the equation" Convenience says: "Oh, so I can write ...


0

The easiest way of seeing this is to remember that compactness is equivalent to sequential compactness in metric spaces. Now, as $[a,b]$ is not a complete metric space, find a Cauchy sequence without a limit inside $[a,b]$ (say a sequence whose limit would be $a+\frac{\sqrt2(b-a)}2$). Now you have a sequence in the interval without a convergent subsequence, ...


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 ...


0

As @lulu states in the comments of the question, the Lindermann-Weierstrass Theorem is a theorem that goes along these lines: If I have $\sin(n)$ where $n$ is algebraic e.g. it is a solution to a polynomial with rational coefficients, then $\sin(n)$ is transcendental. And, furthermore, transcendental numbers are not rational. However, I can tell you that ...


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 ...


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 ...


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.


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 ...


0

I think that the fourth equality introduces the error, since $a^{rs}=(a^r)^s$ may not hold if considering complex numbers. For example, \begin{align} -1=e^{i\pi}=e^{2i\pi \cdot \frac{1}{2}} \neq (e^{2i\pi})^{\frac{1}{2}}=1. \end{align}


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) ...


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 $s$.)


-1

Rational exponents: $a^{m/n} = (\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}$ provided $m$ and $n$ are integers with no common factors and $\sqrt[n]{a}$ is a real number. (NOTE: This is a terrible definition, and you could argue that it prevents any exponent from being well-defined.) The laws of rational exponents hold provided all the expressions used are defined. ...


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

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 ...


0

You have four questions. The answer to the first is yes. The answer to the second is yes. However, although given irrationals $x<y$, there is a rational $r$ with $x<r<y$, it need not be the arithmetic mean ${1\over2}(x+y)$. The arithmetic mean can easily be irrational: suppose, for example, $x=\sqrt{3}-\sqrt{2}$ and $y=\sqrt{3}+\sqrt{2}$. ...


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 ...


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")... ...


0

HINT: Use the fact that Between any two rational numbers, we have infinite irrational numbers and similarly, Between any two irrational numbers, we have infinite rational numbers.


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.


0

Your solution for n=3 includes an implicit change of variables: $$ \left(a,b,c,d\right)=\left(x+y,x-y,u+v,u-v\right) $$ $$ r = \left(2x/2u\right)\left(x^2+3y^2\right)/\left(u^2+3v^2\right)$$ at which point the substitution $$ \left(x,y,u,v\right)=\left(3ps^3t,9qt^4,9qst^3,ps^4\right)$$ yields the desired result of $$r=p/q$$ A similar two-step substitution ...


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 $(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)$. ...


0

If nice is taken to mean analytic function there sure is. There exists a result from complex analysis that states that if $\lim_{n\to\infty} c_n=\infty$ and $a_n$ is a sequence of complex number then there's an analytic functionn $f$ such than $f(c_n) = a_n$. Now let $c_n = n$ and $a_n$ be an enumeration of $\mathbb Q$ then this result guarantees the ...


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

Edit: I'm now disavowing this answer, but keeping it here to maintain the comments. Consider my newer answer elsewhere for, I think, a better treatment. I would currently argue that it's the 2nd equality in the example, $(-1)^1 = (-1)^\frac{2}{2}$, that contains the initial error. It's illegitimate to write $(-1)^\frac{2}{2}$ because that's an undefined ...


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 ...


4

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 ...


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 ...


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 ...


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 ...


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.


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.


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 ...


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 ...


0

Hint: Show that $\Bbb Q$ is dense in $\Bbb R$ by showing that any basic open set contains a rational point. The basic open sets are sets of the form: $\{x-\epsilon<y<x+\epsilon\}$ as $\epsilon$ and $x$ range over all real numbers. Note that this is the same as saying that a basic open set is an interval of the form $(a,b)$ for $a<b \in \Bbb R$.


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 ...


15

Let $x=3^{1/2}$ and $y=\log_{3}(4)$. Then $x^y=2$. The proof that $x$ is irrational is familiar. For $y$, suppose $y=p/q$ where $p$ and $q$ are positive integers. Then $3^{p/q}=4$, so $3^p=4^q$. This is impossible, since $4^q$ is even and $3^p$ is odd.



Top 50 recent answers are included