Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.












3














Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.



My Attempt:



Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$



According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.



So, how do I proceed from here?






algebra-precalculus polynomials







share|cite|improve this question






















  • Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
    – Guy
    Apr 29 '17 at 14:59










  • I see no reason to downvote.
    – ajotatxe
    Apr 29 '17 at 15:00










  • @Guy, Please elaborate more.
    – blue_eyed_...
    Apr 29 '17 at 15:01










  • @ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
    – mathreadler
    Apr 29 '17 at 15:03










  • Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
    – Guy
    Apr 29 '17 at 15:03
















3














Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.



My Attempt:



Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$



According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.



So, how do I proceed from here?






algebra-precalculus polynomials







share|cite|improve this question






















  • Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
    – Guy
    Apr 29 '17 at 14:59










  • I see no reason to downvote.
    – ajotatxe
    Apr 29 '17 at 15:00










  • @Guy, Please elaborate more.
    – blue_eyed_...
    Apr 29 '17 at 15:01










  • @ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
    – mathreadler
    Apr 29 '17 at 15:03










  • Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
    – Guy
    Apr 29 '17 at 15:03














3












3








3







Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.



My Attempt:



Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$



According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.



So, how do I proceed from here?






algebra-precalculus polynomials







share|cite|improve this question













Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.



My Attempt:



Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$



According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.



So, how do I proceed from here?






algebra-precalculus polynomials




algebra-precalculus polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 29 '17 at 14:58









blue_eyed_...

3,23821645




3,23821645












  • Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
    – Guy
    Apr 29 '17 at 14:59










  • I see no reason to downvote.
    – ajotatxe
    Apr 29 '17 at 15:00










  • @Guy, Please elaborate more.
    – blue_eyed_...
    Apr 29 '17 at 15:01










  • @ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
    – mathreadler
    Apr 29 '17 at 15:03










  • Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
    – Guy
    Apr 29 '17 at 15:03


















  • Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
    – Guy
    Apr 29 '17 at 14:59










  • I see no reason to downvote.
    – ajotatxe
    Apr 29 '17 at 15:00










  • @Guy, Please elaborate more.
    – blue_eyed_...
    Apr 29 '17 at 15:01










  • @ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
    – mathreadler
    Apr 29 '17 at 15:03










  • Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
    – Guy
    Apr 29 '17 at 15:03
















Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59




Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59












I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00




I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00












@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01




@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01












@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03




@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03












Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03




Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03










2 Answers
2






active

oldest

votes


















1














We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.



So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...



In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.



So for $n=1:$



$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.



Assume true for $n$. Now we must prove true $n+1:$



Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$






share|cite|improve this answer





























    0














    By induction over $n$:



    If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
    That is $f$ has no roots in $F$.



    Suppose $n>0:$



    Suppose the claim is true for every polynomial of degree $n-1$.



    If $f$ has not roots in $F$, we're done.



    Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.



    By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.



    It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.



    Hence, $f$ has at most $n$ different roots in $F$






    share|cite|improve this answer





















    • what is $F$ , what have you assumed it?
      – blue_eyed_...
      Apr 29 '17 at 15:50










    • @AlbertEinstein $F$ is a field.
      – Itay4
      Apr 29 '17 at 15:52











    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%2f2257711%2fprove-that-every-equation-of-degree-n-in-x-has-n-roots-and-no-more%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









    1














    We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.



    So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...



    In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.



    So for $n=1:$



    $f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.



    Assume true for $n$. Now we must prove true $n+1:$



    Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$






    share|cite|improve this answer


























      1














      We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.



      So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...



      In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.



      So for $n=1:$



      $f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.



      Assume true for $n$. Now we must prove true $n+1:$



      Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$






      share|cite|improve this answer
























        1












        1








        1






        We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.



        So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...



        In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.



        So for $n=1:$



        $f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.



        Assume true for $n$. Now we must prove true $n+1:$



        Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$






        share|cite|improve this answer












        We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.



        So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...



        In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.



        So for $n=1:$



        $f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.



        Assume true for $n$. Now we must prove true $n+1:$



        Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 29 '17 at 15:08









        mrnovice

        5,3781326




        5,3781326























            0














            By induction over $n$:



            If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
            That is $f$ has no roots in $F$.



            Suppose $n>0:$



            Suppose the claim is true for every polynomial of degree $n-1$.



            If $f$ has not roots in $F$, we're done.



            Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.



            By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.



            It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.



            Hence, $f$ has at most $n$ different roots in $F$






            share|cite|improve this answer





















            • what is $F$ , what have you assumed it?
              – blue_eyed_...
              Apr 29 '17 at 15:50










            • @AlbertEinstein $F$ is a field.
              – Itay4
              Apr 29 '17 at 15:52
















            0














            By induction over $n$:



            If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
            That is $f$ has no roots in $F$.



            Suppose $n>0:$



            Suppose the claim is true for every polynomial of degree $n-1$.



            If $f$ has not roots in $F$, we're done.



            Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.



            By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.



            It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.



            Hence, $f$ has at most $n$ different roots in $F$






            share|cite|improve this answer





















            • what is $F$ , what have you assumed it?
              – blue_eyed_...
              Apr 29 '17 at 15:50










            • @AlbertEinstein $F$ is a field.
              – Itay4
              Apr 29 '17 at 15:52














            0












            0








            0






            By induction over $n$:



            If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
            That is $f$ has no roots in $F$.



            Suppose $n>0:$



            Suppose the claim is true for every polynomial of degree $n-1$.



            If $f$ has not roots in $F$, we're done.



            Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.



            By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.



            It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.



            Hence, $f$ has at most $n$ different roots in $F$






            share|cite|improve this answer












            By induction over $n$:



            If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
            That is $f$ has no roots in $F$.



            Suppose $n>0:$



            Suppose the claim is true for every polynomial of degree $n-1$.



            If $f$ has not roots in $F$, we're done.



            Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.



            By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.



            It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.



            Hence, $f$ has at most $n$ different roots in $F$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 29 '17 at 15:08









            Itay4

            1,257722




            1,257722












            • what is $F$ , what have you assumed it?
              – blue_eyed_...
              Apr 29 '17 at 15:50










            • @AlbertEinstein $F$ is a field.
              – Itay4
              Apr 29 '17 at 15:52


















            • what is $F$ , what have you assumed it?
              – blue_eyed_...
              Apr 29 '17 at 15:50










            • @AlbertEinstein $F$ is a field.
              – Itay4
              Apr 29 '17 at 15:52
















            what is $F$ , what have you assumed it?
            – blue_eyed_...
            Apr 29 '17 at 15:50




            what is $F$ , what have you assumed it?
            – blue_eyed_...
            Apr 29 '17 at 15:50












            @AlbertEinstein $F$ is a field.
            – Itay4
            Apr 29 '17 at 15:52




            @AlbertEinstein $F$ is a field.
            – Itay4
            Apr 29 '17 at 15:52


















            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f2257711%2fprove-that-every-equation-of-degree-n-in-x-has-n-roots-and-no-more%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