Verification for a Proof By Contradiction concerning a 5-root being irrational The Next CEO of...

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Verification for a Proof By Contradiction concerning a 5-root being irrational



The Next CEO of Stack OverflowProof that the set of irrational numbers is dense in realsThe square root of a prime number is irrationalUnderstanding the proof of “$sqrt{2}$ is irrational” by contradiction.Why do we consider that $p$ & $q$ are co-primes when proving square root of a prime number is irrational?Proof by contradictionPlease critique my proof that $sqrt{12}$ is irrationalProof that, for every positive integer $n$, $log_7(n)$ is either integer or irrationalIs this a valid proof for $sqrt5$ being irrational?Is this proof of $sqrt{3}$ being an irrational number correct?Proof by contradiction that an expression is irrational












1












$begingroup$


I was asked to prove that for a non negative $m$ number, that is not a 5-th power of an integer, its 5-root, $sqrt[5]m$ is not a rational number.
Thanks for any feedback.



Proof By Contradiction:



Suppose Not,



Assume that for a non-negative $m$ number, that is not a 5-th power of an integer, its 5-root $sqrt[5]m$ is a rational number.



Being a rational number we can say:$$sqrt[5]m=frac{p}{q}quad text{for} ;p,qinBbb{Z};text{and}:qneq0$$
Then by algebra,$$m=left(frac{p}{q}right)^{5}Rightarrow m=frac{p^{5}}{q^{5}}$$
Consider when $frac{p}{q}=n$, $n inBbb Z$,



This is a contradiction with the fact that $m$ is not a 5-th power of an integer.



Hence, $sqrt[5]{m}$ is not a rational number.






real-analysis proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your attempt stops making sense to me after $m=p^5/q^5$.
    $endgroup$
    – learner
    Mar 16 at 5:27






  • 1




    $begingroup$
    How can you be sure that $frac{p}{q}$ will be an integer $n$?
    $endgroup$
    – Sujit Bhattacharyya
    Mar 16 at 5:30






  • 1




    $begingroup$
    It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
    $endgroup$
    – Robert Shore
    Mar 16 at 5:31
















1












$begingroup$


I was asked to prove that for a non negative $m$ number, that is not a 5-th power of an integer, its 5-root, $sqrt[5]m$ is not a rational number.
Thanks for any feedback.



Proof By Contradiction:



Suppose Not,



Assume that for a non-negative $m$ number, that is not a 5-th power of an integer, its 5-root $sqrt[5]m$ is a rational number.



Being a rational number we can say:$$sqrt[5]m=frac{p}{q}quad text{for} ;p,qinBbb{Z};text{and}:qneq0$$
Then by algebra,$$m=left(frac{p}{q}right)^{5}Rightarrow m=frac{p^{5}}{q^{5}}$$
Consider when $frac{p}{q}=n$, $n inBbb Z$,



This is a contradiction with the fact that $m$ is not a 5-th power of an integer.



Hence, $sqrt[5]{m}$ is not a rational number.






real-analysis proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your attempt stops making sense to me after $m=p^5/q^5$.
    $endgroup$
    – learner
    Mar 16 at 5:27






  • 1




    $begingroup$
    How can you be sure that $frac{p}{q}$ will be an integer $n$?
    $endgroup$
    – Sujit Bhattacharyya
    Mar 16 at 5:30






  • 1




    $begingroup$
    It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
    $endgroup$
    – Robert Shore
    Mar 16 at 5:31














1












1








1





$begingroup$


I was asked to prove that for a non negative $m$ number, that is not a 5-th power of an integer, its 5-root, $sqrt[5]m$ is not a rational number.
Thanks for any feedback.



Proof By Contradiction:



Suppose Not,



Assume that for a non-negative $m$ number, that is not a 5-th power of an integer, its 5-root $sqrt[5]m$ is a rational number.



Being a rational number we can say:$$sqrt[5]m=frac{p}{q}quad text{for} ;p,qinBbb{Z};text{and}:qneq0$$
Then by algebra,$$m=left(frac{p}{q}right)^{5}Rightarrow m=frac{p^{5}}{q^{5}}$$
Consider when $frac{p}{q}=n$, $n inBbb Z$,



This is a contradiction with the fact that $m$ is not a 5-th power of an integer.



Hence, $sqrt[5]{m}$ is not a rational number.






real-analysis proof-verification







share|cite|improve this question









$endgroup$




I was asked to prove that for a non negative $m$ number, that is not a 5-th power of an integer, its 5-root, $sqrt[5]m$ is not a rational number.
Thanks for any feedback.



Proof By Contradiction:



Suppose Not,



Assume that for a non-negative $m$ number, that is not a 5-th power of an integer, its 5-root $sqrt[5]m$ is a rational number.



Being a rational number we can say:$$sqrt[5]m=frac{p}{q}quad text{for} ;p,qinBbb{Z};text{and}:qneq0$$
Then by algebra,$$m=left(frac{p}{q}right)^{5}Rightarrow m=frac{p^{5}}{q^{5}}$$
Consider when $frac{p}{q}=n$, $n inBbb Z$,



This is a contradiction with the fact that $m$ is not a 5-th power of an integer.



Hence, $sqrt[5]{m}$ is not a rational number.






real-analysis proof-verification




real-analysis proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 16 at 5:18









MACMAC

283




283








  • 1




    $begingroup$
    Your attempt stops making sense to me after $m=p^5/q^5$.
    $endgroup$
    – learner
    Mar 16 at 5:27






  • 1




    $begingroup$
    How can you be sure that $frac{p}{q}$ will be an integer $n$?
    $endgroup$
    – Sujit Bhattacharyya
    Mar 16 at 5:30






  • 1




    $begingroup$
    It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
    $endgroup$
    – Robert Shore
    Mar 16 at 5:31














  • 1




    $begingroup$
    Your attempt stops making sense to me after $m=p^5/q^5$.
    $endgroup$
    – learner
    Mar 16 at 5:27






  • 1




    $begingroup$
    How can you be sure that $frac{p}{q}$ will be an integer $n$?
    $endgroup$
    – Sujit Bhattacharyya
    Mar 16 at 5:30






  • 1




    $begingroup$
    It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
    $endgroup$
    – Robert Shore
    Mar 16 at 5:31








1




1




$begingroup$
Your attempt stops making sense to me after $m=p^5/q^5$.
$endgroup$
– learner
Mar 16 at 5:27




$begingroup$
Your attempt stops making sense to me after $m=p^5/q^5$.
$endgroup$
– learner
Mar 16 at 5:27




1




1




$begingroup$
How can you be sure that $frac{p}{q}$ will be an integer $n$?
$endgroup$
– Sujit Bhattacharyya
Mar 16 at 5:30




$begingroup$
How can you be sure that $frac{p}{q}$ will be an integer $n$?
$endgroup$
– Sujit Bhattacharyya
Mar 16 at 5:30




1




1




$begingroup$
It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
$endgroup$
– Robert Shore
Mar 16 at 5:31




$begingroup$
It's not quite that easy. For all you know, $mq^5$ could be the fifth power of an integer even though $m$ is not. You need to use the prime factorization of $m$, just as is done to prove $sqrt{2}$ is irrational.
$endgroup$
– Robert Shore
Mar 16 at 5:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

Well, you are on the right track, but you cannot consider when $p/q = n in mathbb Z$.



HINT



I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:



$q^5 cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.



Can you continue?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
    $endgroup$
    – MAC
    Mar 20 at 1:23








  • 1




    $begingroup$
    @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
    $endgroup$
    – Yujie Zha
    Mar 20 at 1:26





















2












$begingroup$

You may assume (wlog) that $qgt 0$, so since $p^5/q^5=minBbb Z$, this implies $q^5mid p^5$. You can furthermore assume (wlog) that $gcd(p,q)=1$. This would imply that $qmid p$, so $1=gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$




In fact, this argument can be generalized to $sqrt[n]m$ where $m$ is not a perfect $n^text{th}$ power






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Well, you are on the right track, but you cannot consider when $p/q = n in mathbb Z$.



    HINT



    I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:



    $q^5 cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.



    Can you continue?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
      $endgroup$
      – MAC
      Mar 20 at 1:23








    • 1




      $begingroup$
      @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
      $endgroup$
      – Yujie Zha
      Mar 20 at 1:26


















    1












    $begingroup$

    Well, you are on the right track, but you cannot consider when $p/q = n in mathbb Z$.



    HINT



    I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:



    $q^5 cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.



    Can you continue?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
      $endgroup$
      – MAC
      Mar 20 at 1:23








    • 1




      $begingroup$
      @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
      $endgroup$
      – Yujie Zha
      Mar 20 at 1:26
















    1












    1








    1





    $begingroup$

    Well, you are on the right track, but you cannot consider when $p/q = n in mathbb Z$.



    HINT



    I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:



    $q^5 cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.



    Can you continue?






    share|cite|improve this answer









    $endgroup$



    Well, you are on the right track, but you cannot consider when $p/q = n in mathbb Z$.



    HINT



    I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:



    $q^5 cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.



    Can you continue?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 16 at 5:36









    Yujie ZhaYujie Zha

    6,98611729




    6,98611729












    • $begingroup$
      Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
      $endgroup$
      – MAC
      Mar 20 at 1:23








    • 1




      $begingroup$
      @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
      $endgroup$
      – Yujie Zha
      Mar 20 at 1:26




















    • $begingroup$
      Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
      $endgroup$
      – MAC
      Mar 20 at 1:23








    • 1




      $begingroup$
      @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
      $endgroup$
      – Yujie Zha
      Mar 20 at 1:26


















    $begingroup$
    Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
    $endgroup$
    – MAC
    Mar 20 at 1:23






    $begingroup$
    Are you saying that from the fact m exists in Z+ and m>0, p,q >0,and that we can define p,q as coprime with each other. From this we have two cases: where p=1, or q=1?
    $endgroup$
    – MAC
    Mar 20 at 1:23






    1




    1




    $begingroup$
    @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
    $endgroup$
    – Yujie Zha
    Mar 20 at 1:26






    $begingroup$
    @MAC No, I mean since p/q is a rational number, then q,p are either coprime; or q = 1; or p = 1. For each scenario, you could check for your equation, and you’ll find that only when q=1 will make you equation solvable, but in that case, m is 5-th power of an integer
    $endgroup$
    – Yujie Zha
    Mar 20 at 1:26













    2












    $begingroup$

    You may assume (wlog) that $qgt 0$, so since $p^5/q^5=minBbb Z$, this implies $q^5mid p^5$. You can furthermore assume (wlog) that $gcd(p,q)=1$. This would imply that $qmid p$, so $1=gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$




    In fact, this argument can be generalized to $sqrt[n]m$ where $m$ is not a perfect $n^text{th}$ power






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You may assume (wlog) that $qgt 0$, so since $p^5/q^5=minBbb Z$, this implies $q^5mid p^5$. You can furthermore assume (wlog) that $gcd(p,q)=1$. This would imply that $qmid p$, so $1=gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$




      In fact, this argument can be generalized to $sqrt[n]m$ where $m$ is not a perfect $n^text{th}$ power






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You may assume (wlog) that $qgt 0$, so since $p^5/q^5=minBbb Z$, this implies $q^5mid p^5$. You can furthermore assume (wlog) that $gcd(p,q)=1$. This would imply that $qmid p$, so $1=gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$




        In fact, this argument can be generalized to $sqrt[n]m$ where $m$ is not a perfect $n^text{th}$ power






        share|cite|improve this answer









        $endgroup$



        You may assume (wlog) that $qgt 0$, so since $p^5/q^5=minBbb Z$, this implies $q^5mid p^5$. You can furthermore assume (wlog) that $gcd(p,q)=1$. This would imply that $qmid p$, so $1=gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$




        In fact, this argument can be generalized to $sqrt[n]m$ where $m$ is not a perfect $n^text{th}$ power







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 16 at 5:37









        learnerlearner

        1,110417




        1,110417






























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            六本木駅

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            六本木駅 7番出入口（東京ミッドタウン横） （2008年7月） ろっぽんぎ Roppongi （六本木ヒルズ・東京ミッドタウン前*） 東京都港区六本木六丁目1-25（東京メトロ） 北緯35度39分46秒東経139度43分53秒 座標: 北緯35度39分46秒 東経139度43分53秒 東京都港区赤坂九丁目7-39（東京都交通局） 北緯35度39分48秒東経139度43分55.6秒 所属事業者 東京地下鉄（東京メトロ） 東京都交通局（都営地下鉄） 電報略号 ロホ（東京メトロ） 六（東京都交通局、駅名略称） 駅構造 地下駅 ホーム 計4面4線 （日比谷線 2面2線 大江戸線 上下二層2面2線） 乗降人員 -統計年度- （東京メトロ）134,902人/日 （東京都交通局）100,587人/日 -2017年- 開業年月日 1964年（昭和39年）3月25日 乗入路線 2 路線 所属路線 ● 東京メトロ日比谷線 駅番号 ○ H 04 キロ程 16.1km（北千住起点） ◄ H 03 広尾 (1.7km) (1.5km) 神谷町 H 05 ► 所属路線 ● 都営地下鉄大江戸線 駅番号 ○ E 23 キロ程 23.2km（都庁前起点） ◄ E 22 麻布十番 (1.1km) (1.3km) 青山一丁目 E 24 ► 備考 * 『東京ミッドタウン前』表記は大江戸線のみ テンプレートを表示 六本木駅 （ろっぽんぎえき）は、東京都港区にある、東京地下鉄（東京メトロ）・東京都交通局（都営地下鉄）の駅である。所在地は東京メトロが六本木六丁目、東京都交通局が赤坂九丁目となっている。 東京メトロの日比谷線と、都営地下鉄の大江戸線が乗り入れている。駅番号は日比谷線が H 04 、大江戸線が E 23 である。 目次 1 歴史 2 駅構造 2.1 東京メトロ 2.1.1 のりば 2.1.2 改札内設備 2.2 都営地下鉄 2.2.1 のりば 2.2.2 改札内設備 3 利用状況 3.1 年度...
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            Integral that is continuous and looks like it converges to a geometric seriesTesting if a geometric series converges by taking limit to infinitySummation of arithmetic-geometric series of higher orderGeometric series with polynomial exponentHow to Recognize a Geometric SeriesShowing an integral equality with series over the integersDiscontinuity of a series of continuous functionsReasons why a Series ConvergesSum of infinite geometric series with two terms in summationUsing geometric series for computing IntegralsLimit of geometric series sum when $r = 1$

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            Why did Rep. Omar conclude her criticism of US troops with the phrase "NotTodaySatan"? Contradiction proof for inequality of P and NP? What is meant by electrons moving to lower potential energy? Negative Resistance Zonal Statistics is returning null values in ArcGIS Desktop? Why do distances seem to matter in the Foundation world? Injection into a proper class and choice without regularity A faster way to compute the largest prime factor "Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English? Do I need to watch Ant-Man and the Wasp and Captain Marvel before watching Avengers: Endgame? How bug prioritization works in agile projects vs non agile Is this a typo in Section 1.8.1 Mathematics for Computer Science? Drawing a german abacus as in the books of Adam Ries How to have a sharp product image? Creating a chemical industry from a medieval tech level without petroleum Partitioning va...
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            Joseph Lister

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            Joseph Lister Fødd 5. april 1827 Upton House, Newham Død 10. februar 1912 Kent Nasjonalitet Det sameinte kongeriket Storbritannia og Irland Område kirurgi Yrke lege, universitetslærar, kirurg Institusjonar King’s College London University of Edinburgh University of Glasgow Alma mater University College London Medlem Royal Society Kungliga Vetenskapsakademien Det ungarske vitskapsakademiet Det greske filologiske selskap i Konstantinopel grunnlagt 1861 American Academy of Arts and Sciences Kungliga Vetenskaps- och Vitterhetssamhället i Göteborg Deutsche Akademie der Naturforscher Leopoldina Académie nationale de médecine Royal Academy of Medicine of Belgium Joseph Lister (fødd 5. april 1827 i Upton i Essex, død 10. februar 1912 i Walmer) var ein britisk kirurg. Han er kjend for å ha oppdaga at karbolsyre påverkar bakteriane. Ved å rense sår og å leggje på desifiserte banda...
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