Irrationality of $sqrt{15}$












16












$begingroup$


Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?



Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)



Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).



Therefore $3mid p$ and $3mid q$. Contradiction.






number-theory proof-verification irrational-numbers







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    That's a great proof.
    $endgroup$
    – Yanko
    Oct 25 '17 at 16:18






  • 2




    $begingroup$
    looks correct to me well done
    $endgroup$
    – Isham
    Oct 25 '17 at 16:34






  • 2




    $begingroup$
    Agreed and well explained
    $endgroup$
    – imranfat
    Oct 25 '17 at 16:53






  • 7




    $begingroup$
    Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
    $endgroup$
    – rtybase
    Oct 25 '17 at 18:12
















16












$begingroup$


Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?



Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)



Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).



Therefore $3mid p$ and $3mid q$. Contradiction.






number-theory proof-verification irrational-numbers







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    That's a great proof.
    $endgroup$
    – Yanko
    Oct 25 '17 at 16:18






  • 2




    $begingroup$
    looks correct to me well done
    $endgroup$
    – Isham
    Oct 25 '17 at 16:34






  • 2




    $begingroup$
    Agreed and well explained
    $endgroup$
    – imranfat
    Oct 25 '17 at 16:53






  • 7




    $begingroup$
    Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
    $endgroup$
    – rtybase
    Oct 25 '17 at 18:12














16












16








16


3



$begingroup$


Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?



Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)



Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).



Therefore $3mid p$ and $3mid q$. Contradiction.






number-theory proof-verification irrational-numbers







share|cite|improve this question











$endgroup$




Could someone verify the correctness of this proof for the irrationality of $sqrt{15}$?



Assume $sqrt{15}inmathbf{Q}$, then $sqrt{15}=frac{p}{q}$ with $p,qinmathbf{Z}$ ($qne0$ and $gcd(p,q)=1$).
$implies 15q^2=p^2 implies 15mid p^2 implies 3mid p^2 implies 3mid p$ (Euclid's Lemma)



Now we write $p=3k$ for $kinmathbf{Z}$, then we have $15q^2=9k^2 implies 5q^2=3k^2 implies 3mid 5q^2$. Since $gcd(3,5)=1$ (this lemma: if $amid bc$ and $gcd(a,b)=1$ then $amid c$) this gives $3mid q^2 implies 3mid q$ (Euclid's Lemma).



Therefore $3mid p$ and $3mid q$. Contradiction.






number-theory proof-verification irrational-numbers




number-theory proof-verification irrational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 25 '17 at 16:34









Rodrigo de Azevedo

12.9k41856




12.9k41856










asked Oct 25 '17 at 16:15









Heinz DoofenschmirtzHeinz Doofenschmirtz

602620




602620








  • 7




    $begingroup$
    That's a great proof.
    $endgroup$
    – Yanko
    Oct 25 '17 at 16:18






  • 2




    $begingroup$
    looks correct to me well done
    $endgroup$
    – Isham
    Oct 25 '17 at 16:34






  • 2




    $begingroup$
    Agreed and well explained
    $endgroup$
    – imranfat
    Oct 25 '17 at 16:53






  • 7




    $begingroup$
    Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
    $endgroup$
    – rtybase
    Oct 25 '17 at 18:12














  • 7




    $begingroup$
    That's a great proof.
    $endgroup$
    – Yanko
    Oct 25 '17 at 16:18






  • 2




    $begingroup$
    looks correct to me well done
    $endgroup$
    – Isham
    Oct 25 '17 at 16:34






  • 2




    $begingroup$
    Agreed and well explained
    $endgroup$
    – imranfat
    Oct 25 '17 at 16:53






  • 7




    $begingroup$
    Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
    $endgroup$
    – rtybase
    Oct 25 '17 at 18:12








7




7




$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18




$begingroup$
That's a great proof.
$endgroup$
– Yanko
Oct 25 '17 at 16:18




2




2




$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34




$begingroup$
looks correct to me well done
$endgroup$
– Isham
Oct 25 '17 at 16:34




2




2




$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53




$begingroup$
Agreed and well explained
$endgroup$
– imranfat
Oct 25 '17 at 16:53




7




7




$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12




$begingroup$
Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $gcd(p,q)=1$"
$endgroup$
– rtybase
Oct 25 '17 at 18:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
So you have really proven the following:




Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.







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%2f2489395%2firrationality-of-sqrt15%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












    $begingroup$

    I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
    So you have really proven the following:




    Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
      So you have really proven the following:




      Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
        So you have really proven the following:




        Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.







        share|cite|improve this answer









        $endgroup$



        I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$.
        So you have really proven the following:




        Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $sqrt{n}$ is irrational.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 18:13









        60056005

        36.1k751125




        36.1k751125






























            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%2f2489395%2firrationality-of-sqrt15%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

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Redirect URL with Chrome Remote Debugging Android Devices

            Image
            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more