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0 votes
1 answer
26 views

On pairs of integers satisfying an inequality

Let us consider the following set $A:=\{(r,t)| r \in \mathbb N \cup \{0\}, t \in \mathbb Z, t \leq r-5\}$ My question is the follow : Does there exists any pair $(r,t)$ belonging to $A$ such that it ...
HARRY's user avatar
  • 961
3 votes
1 answer
89 views

Irrationality of an "Euler-like" number

Let $(a_n)_{n=0}^{\infty}$ be a sequence of zeroes and ones such that $a_n=1$ for infinitely many $n$. Let $\displaystyle x:=\sum_{n=0}^{\infty} \frac{a_n}{n!} .$ Is $x$ irrational? I believe it is, ...
jenda358's user avatar
  • 511
0 votes
0 answers
44 views

Is this proof missing a "push-up" step? Every non-empty set of real number bounded from below has a greatest lower bound.

This is from BBFSK, Vol-I pages 142 and 143. The reference to $\S{4.1}$ means for whole numbers $a_i\lt{g}$ and $g\gt{1}$ $$r_n=\sum _{i=0}^{n} a_i g^{-i}.$$ The expression in the sentence following (...
Steven Thomas Hatton's user avatar
1 vote
1 answer
62 views

Can we guarantee that there exists an $\epsilon' > 0$ such that holds for this inequality?

I am currently trying to prove the multiplicative limit law: let $(a_n)^{\infty}_{n=m}, (b_n)^{\infty}_{n=m}$ be convergent sequences of real numbers, and $X, Y$ be the real numbers $X = \lim_{n\to \...
javajav's user avatar
  • 137
1 vote
2 answers
126 views

Please explain the proof of the validity of the Cauchy convergence criterion for real number sequences.

My question pertains to BBFSK, Vol I, Pages 143 and 144. The following appears in the context of developing the real numbers as limits of sequences of rational numbers. It is also easy to prove ...
Steven Thomas Hatton's user avatar
1 vote
0 answers
24 views

Archimedean ordering and a greatest $r_n<x\in{M}$ shows every number less than a lower bound is also a lower bound?

This question pertains to BBFSK Vol I, page 143. The topic is the definition of the greatest lower bound of a non-empty set $M\subset{\mathbb{R}}$ which is bounded from below. For $1<g\in\mathbb{...
Steven Thomas Hatton's user avatar
2 votes
4 answers
57 views

Bound a natural by two consecutive powers

I'm working in the following problems: Given two naturals $m$ and $n$, there exist a natural $d$ such that $$m^{d}\leq n \leq m^{d+1}.$$ Afterwards I need to show that: If one chooses an arbitrary $...
Keith's user avatar
  • 1,167
1 vote
1 answer
257 views

well-ordering principle for natural numbers from the definition of real numbers.

Define the set of real numbers $\mathbb{R}$ by means of the following axiom: There exists a totally ordered field $(\mathbb{R},+,\cdot,\leq)$ which is Dedekind complete. We also assume that $a\leq b$ ...
user avatar
6 votes
2 answers
3k views

can I find the closest rational to any given real, if I assume that the denominator is not larger than some fixed n

Given $n\in\mathbb{N}$ and $r\in\mathbb{R}$(or $r\in\mathbb{Q}$), is that possible to find a rational $\frac{a}{b}$such that, $b<n$ and $\frac{a}{b}$ is the "closest" rational to r? With "closest"...
Unavailable's user avatar
6 votes
5 answers
5k views

What is a natural number? [duplicate]

According to page 25 of the book A First Course in Real Analysis, an inductive set is a set of real numbers such that $0$ is in the set and for every real number $x$ in the set, $x + 1$ is also in the ...
Timothy's user avatar
  • 803
0 votes
2 answers
112 views

Find least positive integer $n$ which satisfies the condition

Find least positive integer $n$ which satisfies the condition $| {0.5}^{\frac{1}{n}}-1|<10^{-3} $ $0.9^{n}<10^{-3}$ What I have tried is I have split the first inequality took log on both ...
user avatar
1 vote
1 answer
1k views

Application of the Archimedean Property

Prove if that $0<a<b$ where $a,b \in \mathbb{R}$ then there exist some $n \in \mathbb{N}$ such that $\frac{1}{n} < a$ and $b < n $ The question states to use the Archimedean Property; If $...
J.Jones's user avatar
  • 227
2 votes
1 answer
6k views

Given a number ε > 0, prove there exists a natural number $N$ such that 1/N < ε

I believe there are three cases. I think I have figured out the first one. Case 1: Let $ε$ = 1 and N > 1. Take $ N$=2. Then 1/2 < $ε$ ≡ 1/2 < 1. Case 2: Let 0<$ε$<1. I believe that I ...
Mathgirl's user avatar
  • 527
0 votes
3 answers
594 views

Square root of non-negative number [closed]

How can one be sure that for every real number $x\in\mathbb R$ with $x\geq0$ there is a unique number $a\geq 0$ such that $a^2 = x$?
user avatar
0 votes
2 answers
155 views

Intuitive justification of associative law

It has been previously asked how one can see that multiplication of real numbers is associative. The answer given there is this: Let's use the following analogy for the multiplication case; suppose ...
user avatar
2 votes
1 answer
234 views

were Irrational numbers discovered at Archimedes's age?

Archimedes axiom states a property of real numbers, while the real numbers include all the rational numbers and all the irrational numbers. I wonder were Irrational numbers discovered at Archimedes's ...
iMath's user avatar
  • 2,267
11 votes
2 answers
978 views

Is there a mathematical statement that is linking integer limits to real limits?

I saw a question asking for the limit $$\lim_{n \to \infty}\frac{\tan(n)}{n}.$$ At first I thought that the limit assumed $n$ to be a real number. So I gave the advice to use $\pi/2+2\pi k$ and $2\...
MrYouMath's user avatar
  • 15.9k
1 vote
4 answers
1k views

If $b-a>1$ then there is a $k\in \mathbb{Z}$ such that $a<k<b$

Given $a, b \in \mathbb{R}$, such that $b-a>1$, there is at least one $k\in \mathbb{Z}$ such that $a<k<b$. My attempt: Consider $E:=(a,b)\cap \mathbb{N}$. We need to show that $E$ is not ...
sequence's user avatar
  • 9,708
7 votes
2 answers
4k views

Let $x$ be a real number. Prove the existence of a unique integer $a$ such that $a \leq x < a+1$

Let $x\in \mathbb{R}$ , Using the Well-Ordering Property of $\mathbb{N}$ and the Archimedean Property of $\mathbb{R}$, show that there exist a unique $a \in \mathbb{Z}$ such that $a \leq x < a+1$ ...
some1fromhell's user avatar
1 vote
0 answers
38 views

how to find continued fractions with terms less than or equal to -2?

How can one find a continued fraction with all the terms less than or equal to -2? meaning that $x=[a_0,a_1,...,a_n]$ with $a_i\leq -2$.
7779052's user avatar
  • 963
9 votes
0 answers
138 views

If $n^x\in\Bbb Z,$ for every $n\in\Bbb Z^+,$ then $x\in\Bbb Z$ [duplicate]

Let $x$ is a real number such that $n^x\in\Bbb Z,$ for every positive integer $n.$ Prove that $x$ is an integer. I got that problem here and it looks difficult, I tried writing $x$ as $\lfloor x\...
CIJ's user avatar
  • 3,457
4 votes
2 answers
587 views

Cantor Sets in perfect sets in the Real numbers

My thesis is related with the Cantor sets. I was reading a lot of papers, blogs, etc, in order to look for the mean properties of these sets. In one blog a read a proposition. ''Every perfect set ...
Jonathan Ortiz's user avatar
1 vote
1 answer
49 views

Can we find a relation between the three integrers $m$, $j$ and $k$?

Let $r>4$ and $n>1$ positive integers and let $α$ be a positive real number. Let us define the following three positive integers: $$ \begin{align*} m &= \lfloor r^{(n+1)^2} \alpha \rfloor \...
DER's user avatar
  • 3,021
0 votes
2 answers
94 views

Find the number of digits of the number $k$ in function of $r$ and $n$

Let $α∈(0,1)$ be an irrational number with infinitely digits after the decimal point. Let $r>4$ and $n>1$ be positive integers. Let $$k=⌊r^{n²}α⌋$$ where $⌊.⌋$ is the floor function. My ...
DER's user avatar
  • 3,021
4 votes
6 answers
389 views

Why is $0^0$ undefined when $x^x=1$ as $x$ approaches $0$?

This question was born in another post available here. I believe $0^0=1$, because $x^x$ is continuous as $x$ approaches $0$. Consider $\lim_{x \to 0}x^x$. Let $$f(x_n)=\bigg(\frac{1}{x}\bigg)^{\...
atomteori's user avatar
  • 618
1 vote
2 answers
113 views

Proving any $x \in [0,1]$ belongs to infinitely many $S^{k}_{n}:=[\frac{k-1}{2^n},\frac{k}{2^n}]$

I am trying to prove the following: Define $S^{k}_{n}:=[\frac{k-1}{2^n},\frac{k}{2^n}]$ i) Given any $x \in [0,1]$, then $x$ belongs to infinitely many $S^{k}_{n}$ ii) Any $x \in [0,1]$ also belongs ...
Mel's user avatar
  • 293