Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < varepsilon$












0












$begingroup$


I'm having difficulties making progress in proving:
$$forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$



To clarify, $r$ is a real number and $q$ is a rational number.



Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ exists n in Z^{+}$ such that $na>b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.



I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!






real-analysis inequality proof-writing real-numbers rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Dirichlet's approximation theorem should be helpful.
    $endgroup$
    – rtybase
    Jan 12 at 10:40






  • 1




    $begingroup$
    You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
    $endgroup$
    – Wojowu
    Jan 12 at 10:42
















0












$begingroup$


I'm having difficulties making progress in proving:
$$forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$



To clarify, $r$ is a real number and $q$ is a rational number.



Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ exists n in Z^{+}$ such that $na>b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.



I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!






real-analysis inequality proof-writing real-numbers rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Dirichlet's approximation theorem should be helpful.
    $endgroup$
    – rtybase
    Jan 12 at 10:40






  • 1




    $begingroup$
    You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
    $endgroup$
    – Wojowu
    Jan 12 at 10:42














0












0








0





$begingroup$


I'm having difficulties making progress in proving:
$$forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$



To clarify, $r$ is a real number and $q$ is a rational number.



Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ exists n in Z^{+}$ such that $na>b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.



I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!






real-analysis inequality proof-writing real-numbers rational-numbers







share|cite|improve this question











$endgroup$




I'm having difficulties making progress in proving:
$$forall varepsilon > 0, exists q in Q text{ where } 0 < |r - q| < varepsilon $$



To clarify, $r$ is a real number and $q$ is a rational number.



Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ exists n in Z^{+}$ such that $na>b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.



I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!






real-analysis inequality proof-writing real-numbers rational-numbers




real-analysis inequality proof-writing real-numbers rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 10:45









rtybase

10.7k21533




10.7k21533










asked Jan 12 at 10:33









Zen'zZen'z

243




243












  • $begingroup$
    Dirichlet's approximation theorem should be helpful.
    $endgroup$
    – rtybase
    Jan 12 at 10:40






  • 1




    $begingroup$
    You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
    $endgroup$
    – Wojowu
    Jan 12 at 10:42


















  • $begingroup$
    Dirichlet's approximation theorem should be helpful.
    $endgroup$
    – rtybase
    Jan 12 at 10:40






  • 1




    $begingroup$
    You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
    $endgroup$
    – Wojowu
    Jan 12 at 10:42
















$begingroup$
Dirichlet's approximation theorem should be helpful.
$endgroup$
– rtybase
Jan 12 at 10:40




$begingroup$
Dirichlet's approximation theorem should be helpful.
$endgroup$
– rtybase
Jan 12 at 10:40




1




1




$begingroup$
You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
$endgroup$
– Wojowu
Jan 12 at 10:42




$begingroup$
You know that there is a rational between any two distinct reals. Consider $r$ and $r+epsilon$.
$endgroup$
– Wojowu
Jan 12 at 10:42










2 Answers
2






active

oldest

votes


















1












$begingroup$

Choose some positive integer $n$ such that $n>frac{1}{varepsilon}$. Define $q$ as $frac{[nr]+1}{n}$. Then, $|r-q|=frac{1-{nr}}{n}in (0;varepsilon)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
    $endgroup$
    – Zen'z
    Jan 13 at 18:27










  • $begingroup$
    For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
    $endgroup$
    – richrow
    Jan 13 at 20:05










  • $begingroup$
    For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
    $endgroup$
    – richrow
    Jan 13 at 20:07












  • $begingroup$
    So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
    $endgroup$
    – Zen'z
    Jan 13 at 20:10












  • $begingroup$
    No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
    $endgroup$
    – richrow
    Jan 13 at 21:40



















0












$begingroup$

No need to any theorem. Let $rnotinBbb Q$ and $$x=10^kcdot r$$for some value of $kin Bbb R$ then define $$ntriangleqBiglfloor10^kcdot rBigrfloor$$which leads to $$nle 10^kcdot r<n+1$$or $${nover 10^k}-{1over 10^k}<r<{n+1over 10^k}={nover 10^k}+{1over 10^k}$$finally $$0<|r-{nover 10^k}|<{1over 10^k}<epsilon$$ and $q={nover 10^k}$ when $k>-log epsilon$



For the case $rin Bbb Q$, define $q=r+{1over 10^k}$ for large enough $kin Bbb N$.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    Choose some positive integer $n$ such that $n>frac{1}{varepsilon}$. Define $q$ as $frac{[nr]+1}{n}$. Then, $|r-q|=frac{1-{nr}}{n}in (0;varepsilon)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
      $endgroup$
      – Zen'z
      Jan 13 at 18:27










    • $begingroup$
      For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
      $endgroup$
      – richrow
      Jan 13 at 20:05










    • $begingroup$
      For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
      $endgroup$
      – richrow
      Jan 13 at 20:07












    • $begingroup$
      So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
      $endgroup$
      – Zen'z
      Jan 13 at 20:10












    • $begingroup$
      No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
      $endgroup$
      – richrow
      Jan 13 at 21:40
















    1












    $begingroup$

    Choose some positive integer $n$ such that $n>frac{1}{varepsilon}$. Define $q$ as $frac{[nr]+1}{n}$. Then, $|r-q|=frac{1-{nr}}{n}in (0;varepsilon)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
      $endgroup$
      – Zen'z
      Jan 13 at 18:27










    • $begingroup$
      For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
      $endgroup$
      – richrow
      Jan 13 at 20:05










    • $begingroup$
      For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
      $endgroup$
      – richrow
      Jan 13 at 20:07












    • $begingroup$
      So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
      $endgroup$
      – Zen'z
      Jan 13 at 20:10












    • $begingroup$
      No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
      $endgroup$
      – richrow
      Jan 13 at 21:40














    1












    1








    1





    $begingroup$

    Choose some positive integer $n$ such that $n>frac{1}{varepsilon}$. Define $q$ as $frac{[nr]+1}{n}$. Then, $|r-q|=frac{1-{nr}}{n}in (0;varepsilon)$.






    share|cite|improve this answer









    $endgroup$



    Choose some positive integer $n$ such that $n>frac{1}{varepsilon}$. Define $q$ as $frac{[nr]+1}{n}$. Then, $|r-q|=frac{1-{nr}}{n}in (0;varepsilon)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 12 at 11:01









    richrowrichrow

    1615




    1615












    • $begingroup$
      Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
      $endgroup$
      – Zen'z
      Jan 13 at 18:27










    • $begingroup$
      For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
      $endgroup$
      – richrow
      Jan 13 at 20:05










    • $begingroup$
      For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
      $endgroup$
      – richrow
      Jan 13 at 20:07












    • $begingroup$
      So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
      $endgroup$
      – Zen'z
      Jan 13 at 20:10












    • $begingroup$
      No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
      $endgroup$
      – richrow
      Jan 13 at 21:40


















    • $begingroup$
      Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
      $endgroup$
      – Zen'z
      Jan 13 at 18:27










    • $begingroup$
      For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
      $endgroup$
      – richrow
      Jan 13 at 20:05










    • $begingroup$
      For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
      $endgroup$
      – richrow
      Jan 13 at 20:07












    • $begingroup$
      So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
      $endgroup$
      – Zen'z
      Jan 13 at 20:10












    • $begingroup$
      No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
      $endgroup$
      – richrow
      Jan 13 at 21:40
















    $begingroup$
    Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
    $endgroup$
    – Zen'z
    Jan 13 at 18:27




    $begingroup$
    Could you clarify the notation of [nr] in defining q as $frac{[nr]+1}{n} $ and the notation of {nr} in defining $|r - q| = frac{1-{nr}}{n} $? Thanks!
    $endgroup$
    – Zen'z
    Jan 13 at 18:27












    $begingroup$
    For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
    $endgroup$
    – richrow
    Jan 13 at 20:05




    $begingroup$
    For real $x$ we define the floor function of $x$ - $[x]$ as integer $n$ such that $nleq x<n+1$ (it's easy to see that such integer exists and unique).
    $endgroup$
    – richrow
    Jan 13 at 20:05












    $begingroup$
    For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
    $endgroup$
    – richrow
    Jan 13 at 20:07






    $begingroup$
    For more information about floor and ceiling functions see here en.m.wikipedia.org/wiki/Floor_and_ceiling_functions or here mathworld.wolfram.com/FloorFunction.html.
    $endgroup$
    – richrow
    Jan 13 at 20:07














    $begingroup$
    So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
    $endgroup$
    – Zen'z
    Jan 13 at 20:10






    $begingroup$
    So the [nr] defined in q and the {nr} in |r - q| both evaluates to integers by definition of the floor function? Also, I don't get how you got to |r - q| = $frac{1 - [nr]}{n}$. Could you help me understand the steps involved? Thanks again!
    $endgroup$
    – Zen'z
    Jan 13 at 20:10














    $begingroup$
    No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
    $endgroup$
    – richrow
    Jan 13 at 21:40




    $begingroup$
    No, ${x}$ is known as fractional part of $x$ and it's defined as ${x}:=x-[x]$. Hence, from definition we always have $0leq {x}<1$.
    $endgroup$
    – richrow
    Jan 13 at 21:40











    0












    $begingroup$

    No need to any theorem. Let $rnotinBbb Q$ and $$x=10^kcdot r$$for some value of $kin Bbb R$ then define $$ntriangleqBiglfloor10^kcdot rBigrfloor$$which leads to $$nle 10^kcdot r<n+1$$or $${nover 10^k}-{1over 10^k}<r<{n+1over 10^k}={nover 10^k}+{1over 10^k}$$finally $$0<|r-{nover 10^k}|<{1over 10^k}<epsilon$$ and $q={nover 10^k}$ when $k>-log epsilon$



    For the case $rin Bbb Q$, define $q=r+{1over 10^k}$ for large enough $kin Bbb N$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No need to any theorem. Let $rnotinBbb Q$ and $$x=10^kcdot r$$for some value of $kin Bbb R$ then define $$ntriangleqBiglfloor10^kcdot rBigrfloor$$which leads to $$nle 10^kcdot r<n+1$$or $${nover 10^k}-{1over 10^k}<r<{n+1over 10^k}={nover 10^k}+{1over 10^k}$$finally $$0<|r-{nover 10^k}|<{1over 10^k}<epsilon$$ and $q={nover 10^k}$ when $k>-log epsilon$



      For the case $rin Bbb Q$, define $q=r+{1over 10^k}$ for large enough $kin Bbb N$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No need to any theorem. Let $rnotinBbb Q$ and $$x=10^kcdot r$$for some value of $kin Bbb R$ then define $$ntriangleqBiglfloor10^kcdot rBigrfloor$$which leads to $$nle 10^kcdot r<n+1$$or $${nover 10^k}-{1over 10^k}<r<{n+1over 10^k}={nover 10^k}+{1over 10^k}$$finally $$0<|r-{nover 10^k}|<{1over 10^k}<epsilon$$ and $q={nover 10^k}$ when $k>-log epsilon$



        For the case $rin Bbb Q$, define $q=r+{1over 10^k}$ for large enough $kin Bbb N$.






        share|cite|improve this answer









        $endgroup$



        No need to any theorem. Let $rnotinBbb Q$ and $$x=10^kcdot r$$for some value of $kin Bbb R$ then define $$ntriangleqBiglfloor10^kcdot rBigrfloor$$which leads to $$nle 10^kcdot r<n+1$$or $${nover 10^k}-{1over 10^k}<r<{n+1over 10^k}={nover 10^k}+{1over 10^k}$$finally $$0<|r-{nover 10^k}|<{1over 10^k}<epsilon$$ and $q={nover 10^k}$ when $k>-log epsilon$



        For the case $rin Bbb Q$, define $q=r+{1over 10^k}$ for large enough $kin Bbb N$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 14:16









        Mostafa AyazMostafa Ayaz

        15.3k3939




        15.3k3939






























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