Show there is a sequence of rational numbers converging to any irrational number.

tapiat0wa4c
2022-06-03
Answered

Show there is a sequence of rational numbers converging to any irrational number.

You can still ask an expert for help

traforatp5il7

Answered 2022-06-04
Author has **1** answers

The space of real numbers $R$ is the metric completion of $Q$ the space of rational numbers. To construct $R$, start with $Q$, and let $C$ be the space of Cauchy sequences of $Q$. Two elements $({x}_{n})$ and $({y}_{n})$ are equivalent if and only if $({x}_{n}-{y}_{n})$ converges towards $0$, then $R$ is the quotient space for this equivalence relation. You can identify $Q$ with the equivalence classes of constant sequences.

If $[{x}_{n}]$ is the class of the Cauchy sequence $({x}_{n})$, define $({y}_{n}^{i})$ by ${y}_{l}^{i}={x}_{i}$ for every $l$. Then $([{y}_{n}^{i}])$ converges towards $[{x}_{n}]$ and $[{y}_{n}^{i}]$ can be identified with the rational ${x}_{i}$.

If $[{x}_{n}]$ is the class of the Cauchy sequence $({x}_{n})$, define $({y}_{n}^{i})$ by ${y}_{l}^{i}={x}_{i}$ for every $l$. Then $([{y}_{n}^{i}])$ converges towards $[{x}_{n}]$ and $[{y}_{n}^{i}]$ can be identified with the rational ${x}_{i}$.

asked 2022-06-15

System of inequalities - proving $n=k$

Let's say we have the following inequalities:

$n<x+1\le k+1$

and

$k<x+1\le n+1$

How to prove that $n=k$?

Let's say we have the following inequalities:

$n<x+1\le k+1$

and

$k<x+1\le n+1$

How to prove that $n=k$?

asked 2022-08-04

Mr. Kumar lives in the eastern part of singapore. he visitshis aged parents who lives 36 km away .every weekend . he findsthat if he increases the average speed of his vehcile by 12 km /hhe could save 9 minutes of his travelling time .Find the speed atwhich he travels before the increase in speed.

asked 2022-06-13

By using piegonhole principal, prove that for any positive irrational number $r$ and positive real numbers $x,y\in \left(0,1\right)$, $x<y$, there exists a positive integer $n$ such that $x\le nr-\left[nr\right]\le y$, where $\left[nr\right]$ is the integral part of $nr$.

asked 2022-06-08

Reducing the System of linear equations

$\begin{array}{rl}x+2y-3z& =4\\ 3x-y+5z& =2\\ 4x+y+({k}^{2}-14)z& =k+2\end{array}$

I started doing the matrix of the system:

$\left(\begin{array}{cccc}1& 2& -3& 4\\ 3& -1& 5& 2\\ 4& 1& {k}^{2}-14& k+2\end{array}\right)$

But I still don't know how can I reduce this

$\begin{array}{rl}x+2y-3z& =4\\ 3x-y+5z& =2\\ 4x+y+({k}^{2}-14)z& =k+2\end{array}$

I started doing the matrix of the system:

$\left(\begin{array}{cccc}1& 2& -3& 4\\ 3& -1& 5& 2\\ 4& 1& {k}^{2}-14& k+2\end{array}\right)$

But I still don't know how can I reduce this

asked 2022-07-07

Find basis of solutions of this linear system

Supposed to find basis of the subspace of vector space ${\mathbb{R}}^{3}$ of solutions of this linear system of equations:

$y=\{\begin{array}{l}{x}_{1}+2{x}_{2}-{x}_{3}=0\\ 2{x}_{1}+7{x}_{2}-2{x}_{3}=0\\ -{x}_{1}+3{x}_{2}+{x}_{3}=0\end{array}$

I solve this system and I got: ${x}_{1}={x}_{3}$ and ${x}_{2}=0$

$\overrightarrow{x}=\left[\begin{array}{c}{x}_{1}\\ 0\\ {x}_{1}\end{array}\right]={x}_{1}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

Supposed to find basis of the subspace of vector space ${\mathbb{R}}^{3}$ of solutions of this linear system of equations:

$y=\{\begin{array}{l}{x}_{1}+2{x}_{2}-{x}_{3}=0\\ 2{x}_{1}+7{x}_{2}-2{x}_{3}=0\\ -{x}_{1}+3{x}_{2}+{x}_{3}=0\end{array}$

I solve this system and I got: ${x}_{1}={x}_{3}$ and ${x}_{2}=0$

$\overrightarrow{x}=\left[\begin{array}{c}{x}_{1}\\ 0\\ {x}_{1}\end{array}\right]={x}_{1}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

asked 2020-12-25

Simplify the following. Write answers without negative exponnents.
$\left(4{x}^{2}{y}^{-3}\right)}^{-2$

asked 2021-09-06

Perform the indicated operation and express the answer in simplest radical form.