Can any improvements be made to my proof that “$sqrt{3} $ is irrational”?












5












$begingroup$


Suppose, for contradiction that $sqrt{3}$ is rational. Then there exists $a,b in mathbb{Z}$ such that $$frac{a}{b}= sqrt{3},$$



where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.



Now, consider $b$ to be odd, then a is odd. Then for $m,n in mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\ 4m^2+4m+1=12n^2+12n+3\
2(2m^2+2m)=2(6n^2+6n+1)\2(m^2+m)=2(3n^2+3n)+1.$$



The LHS is even since $m^2+m in mathbb{Z}$ and the RHS is odd since $ 3n^2+3nin mathbb{Z}$. This is a contradiction, and we therefore conclude that $sqrt{3}$ is irrational.






proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    "Suppose, for contradiction that $sqrt{3}$ is rational."
    $endgroup$
    – John
    Dec 20 '18 at 18:34






  • 2




    $begingroup$
    What a silly error. Thanks.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:37






  • 1




    $begingroup$
    Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
    $endgroup$
    – fleablood
    Dec 20 '18 at 19:29
















5












$begingroup$


Suppose, for contradiction that $sqrt{3}$ is rational. Then there exists $a,b in mathbb{Z}$ such that $$frac{a}{b}= sqrt{3},$$



where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.



Now, consider $b$ to be odd, then a is odd. Then for $m,n in mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\ 4m^2+4m+1=12n^2+12n+3\
2(2m^2+2m)=2(6n^2+6n+1)\2(m^2+m)=2(3n^2+3n)+1.$$



The LHS is even since $m^2+m in mathbb{Z}$ and the RHS is odd since $ 3n^2+3nin mathbb{Z}$. This is a contradiction, and we therefore conclude that $sqrt{3}$ is irrational.






proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    "Suppose, for contradiction that $sqrt{3}$ is rational."
    $endgroup$
    – John
    Dec 20 '18 at 18:34






  • 2




    $begingroup$
    What a silly error. Thanks.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:37






  • 1




    $begingroup$
    Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
    $endgroup$
    – fleablood
    Dec 20 '18 at 19:29














5












5








5


0



$begingroup$


Suppose, for contradiction that $sqrt{3}$ is rational. Then there exists $a,b in mathbb{Z}$ such that $$frac{a}{b}= sqrt{3},$$



where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.



Now, consider $b$ to be odd, then a is odd. Then for $m,n in mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\ 4m^2+4m+1=12n^2+12n+3\
2(2m^2+2m)=2(6n^2+6n+1)\2(m^2+m)=2(3n^2+3n)+1.$$



The LHS is even since $m^2+m in mathbb{Z}$ and the RHS is odd since $ 3n^2+3nin mathbb{Z}$. This is a contradiction, and we therefore conclude that $sqrt{3}$ is irrational.






proof-verification proof-writing







share|cite|improve this question











$endgroup$




Suppose, for contradiction that $sqrt{3}$ is rational. Then there exists $a,b in mathbb{Z}$ such that $$frac{a}{b}= sqrt{3},$$



where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.



Now, consider $b$ to be odd, then a is odd. Then for $m,n in mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\ 4m^2+4m+1=12n^2+12n+3\
2(2m^2+2m)=2(6n^2+6n+1)\2(m^2+m)=2(3n^2+3n)+1.$$



The LHS is even since $m^2+m in mathbb{Z}$ and the RHS is odd since $ 3n^2+3nin mathbb{Z}$. This is a contradiction, and we therefore conclude that $sqrt{3}$ is irrational.






proof-verification proof-writing




proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 18:37

























asked Dec 20 '18 at 18:32







user503154















  • 7




    $begingroup$
    "Suppose, for contradiction that $sqrt{3}$ is rational."
    $endgroup$
    – John
    Dec 20 '18 at 18:34






  • 2




    $begingroup$
    What a silly error. Thanks.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:37






  • 1




    $begingroup$
    Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
    $endgroup$
    – fleablood
    Dec 20 '18 at 19:29














  • 7




    $begingroup$
    "Suppose, for contradiction that $sqrt{3}$ is rational."
    $endgroup$
    – John
    Dec 20 '18 at 18:34






  • 2




    $begingroup$
    What a silly error. Thanks.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:37






  • 1




    $begingroup$
    Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
    $endgroup$
    – fleablood
    Dec 20 '18 at 19:29








7




7




$begingroup$
"Suppose, for contradiction that $sqrt{3}$ is rational."
$endgroup$
– John
Dec 20 '18 at 18:34




$begingroup$
"Suppose, for contradiction that $sqrt{3}$ is rational."
$endgroup$
– John
Dec 20 '18 at 18:34




2




2




$begingroup$
What a silly error. Thanks.
$endgroup$
– user503154
Dec 20 '18 at 18:37




$begingroup$
What a silly error. Thanks.
$endgroup$
– user503154
Dec 20 '18 at 18:37




1




1




$begingroup$
Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
$endgroup$
– fleablood
Dec 20 '18 at 19:29




$begingroup$
Pretty slick. As a corollary (or a lemma toward), no square is of the form $4k+3$. Rather cute.
$endgroup$
– fleablood
Dec 20 '18 at 19:29










2 Answers
2






active

oldest

votes


















3












$begingroup$

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 equiv 1 mod 4,$$
and
begin{align*}
b^2 &equiv 1mod 4\
3b^2 &equiv 3mod 4,
end{align*}
contradiction.



Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you. That is a better method.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:53



















0












$begingroup$

Other proof



we know that
$$1<sqrt{3}<2$$



assume $a=bsqrt{3}in Bbb N$



with $b>1$.



Let $$A={cin Bbb N, c>1 : csqrt{3}in Bbb N}$$



$$Aneq emptyset $$
let $m=min A$.



then



$$alpha=m(sqrt{3}-1)in A text{ and } alpha<min A$$
which is a contradiction, thus $sqrt{3}notin Bbb Q$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:12








  • 1




    $begingroup$
    Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:23













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%2f3047840%2fcan-any-improvements-be-made-to-my-proof-that-sqrt3-is-irrational%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown
























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 equiv 1 mod 4,$$
and
begin{align*}
b^2 &equiv 1mod 4\
3b^2 &equiv 3mod 4,
end{align*}
contradiction.



Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you. That is a better method.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:53
















3












$begingroup$

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 equiv 1 mod 4,$$
and
begin{align*}
b^2 &equiv 1mod 4\
3b^2 &equiv 3mod 4,
end{align*}
contradiction.



Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you. That is a better method.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:53














3












3








3





$begingroup$

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 equiv 1 mod 4,$$
and
begin{align*}
b^2 &equiv 1mod 4\
3b^2 &equiv 3mod 4,
end{align*}
contradiction.



Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.






share|cite|improve this answer









$endgroup$



Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 equiv 1 mod 4,$$
and
begin{align*}
b^2 &equiv 1mod 4\
3b^2 &equiv 3mod 4,
end{align*}
contradiction.



Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 18:39









Alex KruckmanAlex Kruckman

28.2k32658




28.2k32658








  • 2




    $begingroup$
    Thank you. That is a better method.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:53














  • 2




    $begingroup$
    Thank you. That is a better method.
    $endgroup$
    – user503154
    Dec 20 '18 at 18:53








2




2




$begingroup$
Thank you. That is a better method.
$endgroup$
– user503154
Dec 20 '18 at 18:53




$begingroup$
Thank you. That is a better method.
$endgroup$
– user503154
Dec 20 '18 at 18:53











0












$begingroup$

Other proof



we know that
$$1<sqrt{3}<2$$



assume $a=bsqrt{3}in Bbb N$



with $b>1$.



Let $$A={cin Bbb N, c>1 : csqrt{3}in Bbb N}$$



$$Aneq emptyset $$
let $m=min A$.



then



$$alpha=m(sqrt{3}-1)in A text{ and } alpha<min A$$
which is a contradiction, thus $sqrt{3}notin Bbb Q$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:12








  • 1




    $begingroup$
    Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:23


















0












$begingroup$

Other proof



we know that
$$1<sqrt{3}<2$$



assume $a=bsqrt{3}in Bbb N$



with $b>1$.



Let $$A={cin Bbb N, c>1 : csqrt{3}in Bbb N}$$



$$Aneq emptyset $$
let $m=min A$.



then



$$alpha=m(sqrt{3}-1)in A text{ and } alpha<min A$$
which is a contradiction, thus $sqrt{3}notin Bbb Q$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:12








  • 1




    $begingroup$
    Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:23
















0












0








0





$begingroup$

Other proof



we know that
$$1<sqrt{3}<2$$



assume $a=bsqrt{3}in Bbb N$



with $b>1$.



Let $$A={cin Bbb N, c>1 : csqrt{3}in Bbb N}$$



$$Aneq emptyset $$
let $m=min A$.



then



$$alpha=m(sqrt{3}-1)in A text{ and } alpha<min A$$
which is a contradiction, thus $sqrt{3}notin Bbb Q$.






share|cite|improve this answer











$endgroup$



Other proof



we know that
$$1<sqrt{3}<2$$



assume $a=bsqrt{3}in Bbb N$



with $b>1$.



Let $$A={cin Bbb N, c>1 : csqrt{3}in Bbb N}$$



$$Aneq emptyset $$
let $m=min A$.



then



$$alpha=m(sqrt{3}-1)in A text{ and } alpha<min A$$
which is a contradiction, thus $sqrt{3}notin Bbb Q$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 19:09

























answered Dec 20 '18 at 19:00









hamam_Abdallahhamam_Abdallah

38.2k21634




38.2k21634








  • 1




    $begingroup$
    More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:12








  • 1




    $begingroup$
    Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:23
















  • 1




    $begingroup$
    More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:12








  • 1




    $begingroup$
    Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
    $endgroup$
    – Bill Dubuque
    Dec 20 '18 at 19:23










1




1




$begingroup$
More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
$endgroup$
– Bill Dubuque
Dec 20 '18 at 19:12






$begingroup$
More conceptually: the denominator set $A$ contains coprimes $b$ and $a$ (by $a^2/b^2=3,Rightarrow,,a/b = 3b/a),$ thus since $A$ is closed under subtractiion it is closed under gcd, so it contains $1 = gcd(a,b),,$ i.e. $,sqrt 3,$ can be written with denominator $1$, i.e. $,sqrt 3in Bbb Z,,$ contradiction. See my posts on denominator descent for more on this viewpoint.
$endgroup$
– Bill Dubuque
Dec 20 '18 at 19:12






1




1




$begingroup$
Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
$endgroup$
– Bill Dubuque
Dec 20 '18 at 19:23






$begingroup$
Alternatively by $,a^2/b = 3b,Rightarrow, bmid a^2,$ so $,b=1,,$ by $,a,b,$ coprime and Euclid's lemma (generally. the least denominator of a fraction divides every denominator, a special case of the fact that ideals are principal in $,Bbb Z) $
$endgroup$
– Bill Dubuque
Dec 20 '18 at 19:23




















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%2f3047840%2fcan-any-improvements-be-made-to-my-proof-that-sqrt3-is-irrational%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more