Prime Factorization of $m^2$












0












$begingroup$


If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.



I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:



Proof:
let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}cdots p_k^{e_k}$



Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}cdots p_k^{2e_k}$



These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?






number-theory proof-verification prime-factorization







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:01






  • 1




    $begingroup$
    One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
    $endgroup$
    – Bill Dubuque
    Jul 27 '17 at 2:02












  • $begingroup$
    Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:02












  • $begingroup$
    @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
    $endgroup$
    – OLE
    Jul 27 '17 at 2:04










  • $begingroup$
    @BillDubuque is their a nicer more stronger way of proving this?
    $endgroup$
    – OLE
    Jul 27 '17 at 2:06
















0












$begingroup$


If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.



I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:



Proof:
let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}cdots p_k^{e_k}$



Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}cdots p_k^{2e_k}$



These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?






number-theory proof-verification prime-factorization







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:01






  • 1




    $begingroup$
    One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
    $endgroup$
    – Bill Dubuque
    Jul 27 '17 at 2:02












  • $begingroup$
    Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:02












  • $begingroup$
    @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
    $endgroup$
    – OLE
    Jul 27 '17 at 2:04










  • $begingroup$
    @BillDubuque is their a nicer more stronger way of proving this?
    $endgroup$
    – OLE
    Jul 27 '17 at 2:06














0












0








0





$begingroup$


If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.



I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:



Proof:
let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}cdots p_k^{e_k}$



Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}cdots p_k^{2e_k}$



These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?






number-theory proof-verification prime-factorization







share|cite|improve this question











$endgroup$




If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.



I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:



Proof:
let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}cdots p_k^{e_k}$



Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}cdots p_k^{2e_k}$



These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?






number-theory proof-verification prime-factorization




number-theory proof-verification prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 27 '17 at 2:07









Bill Dubuque

213k29196654




213k29196654










asked Jul 27 '17 at 1:59









OLEOLE

463419




463419












  • $begingroup$
    I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:01






  • 1




    $begingroup$
    One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
    $endgroup$
    – Bill Dubuque
    Jul 27 '17 at 2:02












  • $begingroup$
    Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:02












  • $begingroup$
    @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
    $endgroup$
    – OLE
    Jul 27 '17 at 2:04










  • $begingroup$
    @BillDubuque is their a nicer more stronger way of proving this?
    $endgroup$
    – OLE
    Jul 27 '17 at 2:06


















  • $begingroup$
    I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:01






  • 1




    $begingroup$
    One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
    $endgroup$
    – Bill Dubuque
    Jul 27 '17 at 2:02












  • $begingroup$
    Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
    $endgroup$
    – Gregory Grant
    Jul 27 '17 at 2:02












  • $begingroup$
    @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
    $endgroup$
    – OLE
    Jul 27 '17 at 2:04










  • $begingroup$
    @BillDubuque is their a nicer more stronger way of proving this?
    $endgroup$
    – OLE
    Jul 27 '17 at 2:06
















$begingroup$
I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
$endgroup$
– Gregory Grant
Jul 27 '17 at 2:01




$begingroup$
I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good.
$endgroup$
– Gregory Grant
Jul 27 '17 at 2:01




1




1




$begingroup$
One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
$endgroup$
– Bill Dubuque
Jul 27 '17 at 2:02






$begingroup$
One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent).
$endgroup$
– Bill Dubuque
Jul 27 '17 at 2:02














$begingroup$
Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
$endgroup$
– Gregory Grant
Jul 27 '17 at 2:02






$begingroup$
Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even.
$endgroup$
– Gregory Grant
Jul 27 '17 at 2:02














$begingroup$
@GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
$endgroup$
– OLE
Jul 27 '17 at 2:04




$begingroup$
@GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either.
$endgroup$
– OLE
Jul 27 '17 at 2:04












$begingroup$
@BillDubuque is their a nicer more stronger way of proving this?
$endgroup$
– OLE
Jul 27 '17 at 2:06




$begingroup$
@BillDubuque is their a nicer more stronger way of proving this?
$endgroup$
– OLE
Jul 27 '17 at 2:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

It's in fact an if and only if :




Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.




Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = prod p_i^{frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.



Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = prod p_i^{r_i}$ suggests the prime factorization for $k = prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.



Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p mid frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.



Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=frac m{p^k}$, $p^{k+1} mid m$, or that $p^{2k+2} mid m^2$, a contradiction by uniqueness of prime factorization.



On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.



EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.






share|cite|improve this answer











$endgroup$














    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%2f2373053%2fprime-factorization-of-m2%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









    1












    $begingroup$

    It's in fact an if and only if :




    Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.




    Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = prod p_i^{frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.



    Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = prod p_i^{r_i}$ suggests the prime factorization for $k = prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.



    Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p mid frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.



    Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=frac m{p^k}$, $p^{k+1} mid m$, or that $p^{2k+2} mid m^2$, a contradiction by uniqueness of prime factorization.



    On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.



    EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      It's in fact an if and only if :




      Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.




      Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = prod p_i^{frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.



      Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = prod p_i^{r_i}$ suggests the prime factorization for $k = prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.



      Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p mid frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.



      Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=frac m{p^k}$, $p^{k+1} mid m$, or that $p^{2k+2} mid m^2$, a contradiction by uniqueness of prime factorization.



      On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.



      EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        It's in fact an if and only if :




        Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.




        Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = prod p_i^{frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.



        Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = prod p_i^{r_i}$ suggests the prime factorization for $k = prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.



        Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p mid frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.



        Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=frac m{p^k}$, $p^{k+1} mid m$, or that $p^{2k+2} mid m^2$, a contradiction by uniqueness of prime factorization.



        On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.



        EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.






        share|cite|improve this answer











        $endgroup$



        It's in fact an if and only if :




        Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.




        Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = prod p_i^{frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.



        Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = prod p_i^{r_i}$ suggests the prime factorization for $k = prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.



        Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p mid frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.



        Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=frac m{p^k}$, $p^{k+1} mid m$, or that $p^{2k+2} mid m^2$, a contradiction by uniqueness of prime factorization.



        On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.



        EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 16:36

























        answered Jul 27 '17 at 2:59









        астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

        40.1k33577




        40.1k33577






























            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%2f2373053%2fprime-factorization-of-m2%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

            Måne

            Image
            För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
            Read more

            Storängen

            Image
            För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
            Read more

            VLT Carioca

            Image
            VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
            Read more