Is this a valid definition of the rationals?












0












$begingroup$


$$mathbb{Q}=left{sum_{n=1}^k f(n)mid k,ninmathbb{N}land ftext{ is a finite composition of $+$, $-$, $div$, $times$}right}$$



Reasoning:



Any real number can be described by a (sometimes infinite) sum of rational numbers. If such a sum is taken to be $q=sum_{n=1}^k f(n)$, then every real number which is not rational can be approximated to arbitrary precision by increasingly large $k$. If $f(n)$ is composed solely of elementary arithmetic operations ($+,-,div,times$), then $q$ remains rational for all $k<infty$.



If the quotient of any two rational numbers is also rational, then it follows that for elementary functions $f$ and $g$, the quotient of the summations
$sum_{n=1}^k f(n)$ and $sum_{n=1}^k g(n)$ is always rational even as $k$ tends towards infinity.



Intuitively, it would seem that the quotient of any two such summations is always rational even if $k=infty$. However, this is not the case, as there are many infinite sums satisfying the above conditions which are irrational.



Therefore, the sum is rational iff the upper bound $k$ is finite.



Becuase any real number may be represented as a summation, it follows that any rational number can be represented as a summation.



Thus, every rational number can be represented as a finite sum of elementary functions.






algebra-precalculus elementary-number-theory summation rational-numbers







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 16:40






  • 2




    $begingroup$
    How do you define "a finite composition of +, −, ÷, ×"?
    $endgroup$
    – BPP
    Dec 18 '18 at 16:42










  • $begingroup$
    @Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
    $endgroup$
    – R. Burton
    Dec 18 '18 at 16:44






  • 2




    $begingroup$
    "Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
    $endgroup$
    – Noah Schweber
    Dec 18 '18 at 16:50






  • 3




    $begingroup$
    This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 16:53
















0












$begingroup$


$$mathbb{Q}=left{sum_{n=1}^k f(n)mid k,ninmathbb{N}land ftext{ is a finite composition of $+$, $-$, $div$, $times$}right}$$



Reasoning:



Any real number can be described by a (sometimes infinite) sum of rational numbers. If such a sum is taken to be $q=sum_{n=1}^k f(n)$, then every real number which is not rational can be approximated to arbitrary precision by increasingly large $k$. If $f(n)$ is composed solely of elementary arithmetic operations ($+,-,div,times$), then $q$ remains rational for all $k<infty$.



If the quotient of any two rational numbers is also rational, then it follows that for elementary functions $f$ and $g$, the quotient of the summations
$sum_{n=1}^k f(n)$ and $sum_{n=1}^k g(n)$ is always rational even as $k$ tends towards infinity.



Intuitively, it would seem that the quotient of any two such summations is always rational even if $k=infty$. However, this is not the case, as there are many infinite sums satisfying the above conditions which are irrational.



Therefore, the sum is rational iff the upper bound $k$ is finite.



Becuase any real number may be represented as a summation, it follows that any rational number can be represented as a summation.



Thus, every rational number can be represented as a finite sum of elementary functions.






algebra-precalculus elementary-number-theory summation rational-numbers







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 16:40






  • 2




    $begingroup$
    How do you define "a finite composition of +, −, ÷, ×"?
    $endgroup$
    – BPP
    Dec 18 '18 at 16:42










  • $begingroup$
    @Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
    $endgroup$
    – R. Burton
    Dec 18 '18 at 16:44






  • 2




    $begingroup$
    "Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
    $endgroup$
    – Noah Schweber
    Dec 18 '18 at 16:50






  • 3




    $begingroup$
    This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 16:53














0












0








0





$begingroup$


$$mathbb{Q}=left{sum_{n=1}^k f(n)mid k,ninmathbb{N}land ftext{ is a finite composition of $+$, $-$, $div$, $times$}right}$$



Reasoning:



Any real number can be described by a (sometimes infinite) sum of rational numbers. If such a sum is taken to be $q=sum_{n=1}^k f(n)$, then every real number which is not rational can be approximated to arbitrary precision by increasingly large $k$. If $f(n)$ is composed solely of elementary arithmetic operations ($+,-,div,times$), then $q$ remains rational for all $k<infty$.



If the quotient of any two rational numbers is also rational, then it follows that for elementary functions $f$ and $g$, the quotient of the summations
$sum_{n=1}^k f(n)$ and $sum_{n=1}^k g(n)$ is always rational even as $k$ tends towards infinity.



Intuitively, it would seem that the quotient of any two such summations is always rational even if $k=infty$. However, this is not the case, as there are many infinite sums satisfying the above conditions which are irrational.



Therefore, the sum is rational iff the upper bound $k$ is finite.



Becuase any real number may be represented as a summation, it follows that any rational number can be represented as a summation.



Thus, every rational number can be represented as a finite sum of elementary functions.






algebra-precalculus elementary-number-theory summation rational-numbers







share|cite|improve this question









$endgroup$




$$mathbb{Q}=left{sum_{n=1}^k f(n)mid k,ninmathbb{N}land ftext{ is a finite composition of $+$, $-$, $div$, $times$}right}$$



Reasoning:



Any real number can be described by a (sometimes infinite) sum of rational numbers. If such a sum is taken to be $q=sum_{n=1}^k f(n)$, then every real number which is not rational can be approximated to arbitrary precision by increasingly large $k$. If $f(n)$ is composed solely of elementary arithmetic operations ($+,-,div,times$), then $q$ remains rational for all $k<infty$.



If the quotient of any two rational numbers is also rational, then it follows that for elementary functions $f$ and $g$, the quotient of the summations
$sum_{n=1}^k f(n)$ and $sum_{n=1}^k g(n)$ is always rational even as $k$ tends towards infinity.



Intuitively, it would seem that the quotient of any two such summations is always rational even if $k=infty$. However, this is not the case, as there are many infinite sums satisfying the above conditions which are irrational.



Therefore, the sum is rational iff the upper bound $k$ is finite.



Becuase any real number may be represented as a summation, it follows that any rational number can be represented as a summation.



Thus, every rational number can be represented as a finite sum of elementary functions.






algebra-precalculus elementary-number-theory summation rational-numbers




algebra-precalculus elementary-number-theory summation rational-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 16:39









R. BurtonR. Burton

687110




687110








  • 2




    $begingroup$
    Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 16:40






  • 2




    $begingroup$
    How do you define "a finite composition of +, −, ÷, ×"?
    $endgroup$
    – BPP
    Dec 18 '18 at 16:42










  • $begingroup$
    @Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
    $endgroup$
    – R. Burton
    Dec 18 '18 at 16:44






  • 2




    $begingroup$
    "Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
    $endgroup$
    – Noah Schweber
    Dec 18 '18 at 16:50






  • 3




    $begingroup$
    This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 16:53














  • 2




    $begingroup$
    Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 16:40






  • 2




    $begingroup$
    How do you define "a finite composition of +, −, ÷, ×"?
    $endgroup$
    – BPP
    Dec 18 '18 at 16:42










  • $begingroup$
    @Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
    $endgroup$
    – R. Burton
    Dec 18 '18 at 16:44






  • 2




    $begingroup$
    "Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
    $endgroup$
    – Noah Schweber
    Dec 18 '18 at 16:50






  • 3




    $begingroup$
    This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 16:53








2




2




$begingroup$
Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
$endgroup$
– Arthur
Dec 18 '18 at 16:40




$begingroup$
Note that $sum_{n=1}^k$ is itself a finite composition of $+$, so it is redundant. It's enough with just $f(1)$.
$endgroup$
– Arthur
Dec 18 '18 at 16:40




2




2




$begingroup$
How do you define "a finite composition of +, −, ÷, ×"?
$endgroup$
– BPP
Dec 18 '18 at 16:42




$begingroup$
How do you define "a finite composition of +, −, ÷, ×"?
$endgroup$
– BPP
Dec 18 '18 at 16:42












$begingroup$
@Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
$endgroup$
– R. Burton
Dec 18 '18 at 16:44




$begingroup$
@Arthur it is redundant if $k$ is finite, yes. The only reason it was important to specify was because of the 'natural' extension to the reals - that being the case where $k$ is infinite.
$endgroup$
– R. Burton
Dec 18 '18 at 16:44




2




2




$begingroup$
"Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
$endgroup$
– Noah Schweber
Dec 18 '18 at 16:50




$begingroup$
"Therefore, the sum is rational iff the upper bound k is finite." Nope - consider $sum_{i=1}^infty 2^{-i}$.
$endgroup$
– Noah Schweber
Dec 18 '18 at 16:50




3




3




$begingroup$
This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
$endgroup$
– Don Thousand
Dec 18 '18 at 16:53




$begingroup$
This whole post seems like it's a more muddy construction of the rationals that requires a lot of casework and cleanup. I'm not sure I see the purpose.
$endgroup$
– Don Thousand
Dec 18 '18 at 16:53










1 Answer
1






active

oldest

votes


















4












$begingroup$

If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:




$mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,times,-,$ and $div$.




One direction is easy to prove: A positive rational ${aover b}$ can always be represented as $(1+...+1)over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $mathbb{Q}$ is contained in any such set. Conversely, clearly $1inmathbb{Q}$ and $mathbb{Q}$ is closed under $+,times,-,$ and $div$, so we're done.



Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$sum_{i=1}^infty{1div 2times ...times 2mbox{ ($i$ times)}},$$ more clearly written as $sum_{i=1}^infty 2^{-i}$, which is of course just $1$.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
    $endgroup$
    – Andrés E. Caicedo
    Dec 18 '18 at 16:58












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045379%2fis-this-a-valid-definition-of-the-rationals%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:




$mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,times,-,$ and $div$.




One direction is easy to prove: A positive rational ${aover b}$ can always be represented as $(1+...+1)over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $mathbb{Q}$ is contained in any such set. Conversely, clearly $1inmathbb{Q}$ and $mathbb{Q}$ is closed under $+,times,-,$ and $div$, so we're done.



Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$sum_{i=1}^infty{1div 2times ...times 2mbox{ ($i$ times)}},$$ more clearly written as $sum_{i=1}^infty 2^{-i}$, which is of course just $1$.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
    $endgroup$
    – Andrés E. Caicedo
    Dec 18 '18 at 16:58
















4












$begingroup$

If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:




$mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,times,-,$ and $div$.




One direction is easy to prove: A positive rational ${aover b}$ can always be represented as $(1+...+1)over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $mathbb{Q}$ is contained in any such set. Conversely, clearly $1inmathbb{Q}$ and $mathbb{Q}$ is closed under $+,times,-,$ and $div$, so we're done.



Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$sum_{i=1}^infty{1div 2times ...times 2mbox{ ($i$ times)}},$$ more clearly written as $sum_{i=1}^infty 2^{-i}$, which is of course just $1$.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
    $endgroup$
    – Andrés E. Caicedo
    Dec 18 '18 at 16:58














4












4








4





$begingroup$

If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:




$mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,times,-,$ and $div$.




One direction is easy to prove: A positive rational ${aover b}$ can always be represented as $(1+...+1)over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $mathbb{Q}$ is contained in any such set. Conversely, clearly $1inmathbb{Q}$ and $mathbb{Q}$ is closed under $+,times,-,$ and $div$, so we're done.



Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$sum_{i=1}^infty{1div 2times ...times 2mbox{ ($i$ times)}},$$ more clearly written as $sum_{i=1}^infty 2^{-i}$, which is of course just $1$.






share|cite|improve this answer









$endgroup$



If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:




$mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,times,-,$ and $div$.




One direction is easy to prove: A positive rational ${aover b}$ can always be represented as $(1+...+1)over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $mathbb{Q}$ is contained in any such set. Conversely, clearly $1inmathbb{Q}$ and $mathbb{Q}$ is closed under $+,times,-,$ and $div$, so we're done.



Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$sum_{i=1}^infty{1div 2times ...times 2mbox{ ($i$ times)}},$$ more clearly written as $sum_{i=1}^infty 2^{-i}$, which is of course just $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 16:55









Noah SchweberNoah Schweber

127k10151293




127k10151293








  • 3




    $begingroup$
    I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
    $endgroup$
    – Andrés E. Caicedo
    Dec 18 '18 at 16:58














  • 3




    $begingroup$
    I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
    $endgroup$
    – Andrés E. Caicedo
    Dec 18 '18 at 16:58








3




3




$begingroup$
I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
$endgroup$
– Andrés E. Caicedo
Dec 18 '18 at 16:58




$begingroup$
I want to emphasize that the wording suggested here is cleaner than what was posted in the question, which is not rigorous and difficult to parse. Better to use words and a careful description (as in this answer) than a lot of symbols.
$endgroup$
– Andrés E. Caicedo
Dec 18 '18 at 16:58


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045379%2fis-this-a-valid-definition-of-the-rationals%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Brian Clough

Image
Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
Read more

Cáceres

Image
Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
Read more