Prove that if $n$ is not the square of a natural number, then $sqrtn$ is irrational. [duplicate]

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$sqrt a$ is either an integer or an irrational number.






I have this homework problem that I can't seem to be able to figure out:


Prove: If $ninmathbbN$ is not the square of some other $minmathbbN$, then $sqrtn$ must be irrational.


I know that a number being irrational means that it cannot be written in the form $displaystylefracab: a, binmathbbN$ $bneq0$ (in this case, ordinarily it'd be $ainmathbbZ$, $binmathbbZsetminus0$) but how would I go about proving this? Would a proof by contradiction work here?



Thanks!!




elementary-number-theory irrational-numbers




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marked as duplicate by copper.hat, Sasha, sdcvvc, Andrés E. Caicedo, The Chaz 2.0 Aug 31 '12 at 6:16


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    Look at math.stackexchange.com/questions/4467/…
    – copper.hat
    Aug 31 '12 at 4:54














up vote
5
down vote

favorite
2













Possible Duplicate:
$sqrt a$ is either an integer or an irrational number.






I have this homework problem that I can't seem to be able to figure out:


Prove: If $ninmathbbN$ is not the square of some other $minmathbbN$, then $sqrtn$ must be irrational.


I know that a number being irrational means that it cannot be written in the form $displaystylefracab: a, binmathbbN$ $bneq0$ (in this case, ordinarily it'd be $ainmathbbZ$, $binmathbbZsetminus0$) but how would I go about proving this? Would a proof by contradiction work here?



Thanks!!




elementary-number-theory irrational-numbers




share|cite|improve this question













marked as duplicate by copper.hat, Sasha, sdcvvc, Andrés E. Caicedo, The Chaz 2.0 Aug 31 '12 at 6:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    Look at math.stackexchange.com/questions/4467/…
    – copper.hat
    Aug 31 '12 at 4:54












up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2






Possible Duplicate:
$sqrt a$ is either an integer or an irrational number.






I have this homework problem that I can't seem to be able to figure out:


Prove: If $ninmathbbN$ is not the square of some other $minmathbbN$, then $sqrtn$ must be irrational.


I know that a number being irrational means that it cannot be written in the form $displaystylefracab: a, binmathbbN$ $bneq0$ (in this case, ordinarily it'd be $ainmathbbZ$, $binmathbbZsetminus0$) but how would I go about proving this? Would a proof by contradiction work here?



Thanks!!




elementary-number-theory irrational-numbers




share|cite|improve this question














Possible Duplicate:
$sqrt a$ is either an integer or an irrational number.






I have this homework problem that I can't seem to be able to figure out:


Prove: If $ninmathbbN$ is not the square of some other $minmathbbN$, then $sqrtn$ must be irrational.


I know that a number being irrational means that it cannot be written in the form $displaystylefracab: a, binmathbbN$ $bneq0$ (in this case, ordinarily it'd be $ainmathbbZ$, $binmathbbZsetminus0$) but how would I go about proving this? Would a proof by contradiction work here?



Thanks!!





elementary-number-theory irrational-numbers





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share|cite|improve this question




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edited Apr 13 '17 at 12:20









Community♦

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asked Aug 31 '12 at 4:49









roboguy12

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marked as duplicate by copper.hat, Sasha, sdcvvc, Andrés E. Caicedo, The Chaz 2.0 Aug 31 '12 at 6:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by copper.hat, Sasha, sdcvvc, Andrés E. Caicedo, The Chaz 2.0 Aug 31 '12 at 6:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 2




    Look at math.stackexchange.com/questions/4467/…
    – copper.hat
    Aug 31 '12 at 4:54












  • 2




    Look at math.stackexchange.com/questions/4467/…
    – copper.hat
    Aug 31 '12 at 4:54







2




2




Look at math.stackexchange.com/questions/4467/…
– copper.hat
Aug 31 '12 at 4:54




Look at math.stackexchange.com/questions/4467/…
– copper.hat
Aug 31 '12 at 4:54










3 Answers
3






active

oldest

votes

















up vote
12
down vote



accepted










Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $sqrtn$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that



$sqrtn = fracpq$



Then



$n = fracp^2q^2$.



However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that



$n = p^2$



Contradiction since it was assumed that $n neq m^2$ for any $m$.






share|cite|improve this answer





















  • Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
    – roboguy12
    Aug 31 '12 at 4:57






  • 5




    It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
    – Number
    Aug 31 '12 at 5:07











  • Here how can I write q=1
    – codeomnitrix
    Oct 7 '14 at 15:02






  • 1




    @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
    – Jeppe Stig Nielsen
    Apr 24 '16 at 8:03

















up vote
8
down vote













Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $sqrtn=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.






share|cite|improve this answer




























    up vote
    2
    down vote













    This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$



    and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $sqrtn$ is rational, then there exists $tin mathbbN$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.



    In fact, this argument generalizes to showing that if $sqrt[m]n$ is rational, then $n$ is an $m^th$ power.






    share|cite|improve this answer






























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      12
      down vote



      accepted










      Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $sqrtn$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that



      $sqrtn = fracpq$



      Then



      $n = fracp^2q^2$.



      However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that



      $n = p^2$



      Contradiction since it was assumed that $n neq m^2$ for any $m$.






      share|cite|improve this answer





















      • Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
        – roboguy12
        Aug 31 '12 at 4:57






      • 5




        It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
        – Number
        Aug 31 '12 at 5:07











      • Here how can I write q=1
        – codeomnitrix
        Oct 7 '14 at 15:02






      • 1




        @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
        – Jeppe Stig Nielsen
        Apr 24 '16 at 8:03














      up vote
      12
      down vote



      accepted










      Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $sqrtn$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that



      $sqrtn = fracpq$



      Then



      $n = fracp^2q^2$.



      However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that



      $n = p^2$



      Contradiction since it was assumed that $n neq m^2$ for any $m$.






      share|cite|improve this answer





















      • Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
        – roboguy12
        Aug 31 '12 at 4:57






      • 5




        It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
        – Number
        Aug 31 '12 at 5:07











      • Here how can I write q=1
        – codeomnitrix
        Oct 7 '14 at 15:02






      • 1




        @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
        – Jeppe Stig Nielsen
        Apr 24 '16 at 8:03












      up vote
      12
      down vote



      accepted







      up vote
      12
      down vote



      accepted






      Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $sqrtn$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that



      $sqrtn = fracpq$



      Then



      $n = fracp^2q^2$.



      However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that



      $n = p^2$



      Contradiction since it was assumed that $n neq m^2$ for any $m$.






      share|cite|improve this answer













      Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $sqrtn$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that



      $sqrtn = fracpq$



      Then



      $n = fracp^2q^2$.



      However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that



      $n = p^2$



      Contradiction since it was assumed that $n neq m^2$ for any $m$.







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Aug 31 '12 at 4:54









      William

      17k22156




      17k22156











      • Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
        – roboguy12
        Aug 31 '12 at 4:57






      • 5




        It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
        – Number
        Aug 31 '12 at 5:07











      • Here how can I write q=1
        – codeomnitrix
        Oct 7 '14 at 15:02






      • 1




        @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
        – Jeppe Stig Nielsen
        Apr 24 '16 at 8:03
















      • Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
        – roboguy12
        Aug 31 '12 at 4:57






      • 5




        It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
        – Number
        Aug 31 '12 at 5:07











      • Here how can I write q=1
        – codeomnitrix
        Oct 7 '14 at 15:02






      • 1




        @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
        – Jeppe Stig Nielsen
        Apr 24 '16 at 8:03















      Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
      – roboguy12
      Aug 31 '12 at 4:57




      Oh, I see. We basically use the fact that $n$ must be an integer to draw that the denominator must be 1. Got it, thank you very much! I got up until then and couldn't progress further, but this makes it clear!
      – roboguy12
      Aug 31 '12 at 4:57




      5




      5




      It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
      – Number
      Aug 31 '12 at 5:07





      It is important to explicitly mention that the proof uses unique factorization (or Euclid's Lemma) to deduce that $q = 1$. This can fail in rings lacking such properties.
      – Number
      Aug 31 '12 at 5:07













      Here how can I write q=1
      – codeomnitrix
      Oct 7 '14 at 15:02




      Here how can I write q=1
      – codeomnitrix
      Oct 7 '14 at 15:02




      1




      1




      @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
      – Jeppe Stig Nielsen
      Apr 24 '16 at 8:03




      @BillDubuque Yes. This proof seems to use the property that when elements $p$ and $q$ have no common divisor (other than units), then $p^2$ and $q^2$ have no common divisor (other than units). It sounds like a property that holds only in some rings (which ones?).
      – Jeppe Stig Nielsen
      Apr 24 '16 at 8:03










      up vote
      8
      down vote













      Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $sqrtn=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.






      share|cite|improve this answer

























        up vote
        8
        down vote













        Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $sqrtn=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.






        share|cite|improve this answer























          up vote
          8
          down vote










          up vote
          8
          down vote









          Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $sqrtn=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.






          share|cite|improve this answer













          Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $sqrtn=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 31 '12 at 5:40









          Lubin

          40.8k34183




          40.8k34183




















              up vote
              2
              down vote













              This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$



              and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $sqrtn$ is rational, then there exists $tin mathbbN$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.



              In fact, this argument generalizes to showing that if $sqrt[m]n$ is rational, then $n$ is an $m^th$ power.






              share|cite|improve this answer



























                up vote
                2
                down vote













                This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$



                and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $sqrtn$ is rational, then there exists $tin mathbbN$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.



                In fact, this argument generalizes to showing that if $sqrt[m]n$ is rational, then $n$ is an $m^th$ power.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$



                  and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $sqrtn$ is rational, then there exists $tin mathbbN$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.



                  In fact, this argument generalizes to showing that if $sqrt[m]n$ is rational, then $n$ is an $m^th$ power.






                  share|cite|improve this answer















                  This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$



                  and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $sqrtn$ is rational, then there exists $tin mathbbN$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.



                  In fact, this argument generalizes to showing that if $sqrt[m]n$ is rational, then $n$ is an $m^th$ power.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 31 '12 at 6:02


























                  answered Aug 31 '12 at 5:53







                  user5137



















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