Apostol Proof for Finite Decimal Approximations to Real Numbers












1












$begingroup$


I'm self-learning real analysis, and I am trying to understand a part in the proof for the following theorem in Mathematical Analysis by Apostol:




Let $xgeq 0$. Then for every integer $n geq 1$, there exists a finite decimal $r_{n} = a_{0}.a_{1}a_{2}cdots a_{n}$ such that $r_{n} leq x < r_{n} + frac{1}{10^{n}}$.




Here is the given proof:




let S be the set of all nonnegative integers that are less than or equal to $x$. Since $0 in S$, $S$ is not empty. Since $x$ is an upper bound of $S$, $S$ has a supremum $a_0=sup(S)$. Since $a_0le x$, $a_0 in S$, and so $a_0$ is a nonnegative integer. Then $a_0=lfloor x rfloor$ is the greatest integer in $x$. Then $a_0le x < a_0 + 1$.



Let $a_1=lfloor 10x-10a_0rfloor$ be the greatest integer in $10x-10a_0$. Then $0le 10x-10a_0=color{red}{10(x-a_0)<10}$. Then $0le a_1 le 9$. Since $a_1=lfloor 10x-10a_0rfloor$, then $a_1le 10x-10a_0<a_1+1$. Then $a_1$ is the largest integer that satisfies $a_1+10a_0le10x<a_1+1+10a_0$ or $a_0+a_1/10le x<a_0+a_1/10+1/10$.



In general, choosing $a_1$, $a_2$, ..., $a_{n-1}$ with $0le a_ile9$, let $a_n$ be the largest integer that satisfies $a_0 + a_1/10 + cdots + a_n/10^n le x < a_0 + a_1/10 + cdots + a_n/10^n + 1/10^n$. Then $0le a_n le 9$. Then $r_n le x < r_n + 1/10^n$ with $r_n=a_0.a_1a_2cdots a_n$. $blacksquare$




The part I don't understand is in red. Why is $10(x-a_{0})<10$? I thought $x-a_{0}$ is nonnegative since $a_{0}leq x$. Is it because $x$ is not an integer?






real-analysis proof-explanation decimal-expansion







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  • 1




    $begingroup$
    $x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
    $endgroup$
    – fleablood
    Jan 1 at 9:18
















1












$begingroup$


I'm self-learning real analysis, and I am trying to understand a part in the proof for the following theorem in Mathematical Analysis by Apostol:




Let $xgeq 0$. Then for every integer $n geq 1$, there exists a finite decimal $r_{n} = a_{0}.a_{1}a_{2}cdots a_{n}$ such that $r_{n} leq x < r_{n} + frac{1}{10^{n}}$.




Here is the given proof:




let S be the set of all nonnegative integers that are less than or equal to $x$. Since $0 in S$, $S$ is not empty. Since $x$ is an upper bound of $S$, $S$ has a supremum $a_0=sup(S)$. Since $a_0le x$, $a_0 in S$, and so $a_0$ is a nonnegative integer. Then $a_0=lfloor x rfloor$ is the greatest integer in $x$. Then $a_0le x < a_0 + 1$.



Let $a_1=lfloor 10x-10a_0rfloor$ be the greatest integer in $10x-10a_0$. Then $0le 10x-10a_0=color{red}{10(x-a_0)<10}$. Then $0le a_1 le 9$. Since $a_1=lfloor 10x-10a_0rfloor$, then $a_1le 10x-10a_0<a_1+1$. Then $a_1$ is the largest integer that satisfies $a_1+10a_0le10x<a_1+1+10a_0$ or $a_0+a_1/10le x<a_0+a_1/10+1/10$.



In general, choosing $a_1$, $a_2$, ..., $a_{n-1}$ with $0le a_ile9$, let $a_n$ be the largest integer that satisfies $a_0 + a_1/10 + cdots + a_n/10^n le x < a_0 + a_1/10 + cdots + a_n/10^n + 1/10^n$. Then $0le a_n le 9$. Then $r_n le x < r_n + 1/10^n$ with $r_n=a_0.a_1a_2cdots a_n$. $blacksquare$




The part I don't understand is in red. Why is $10(x-a_{0})<10$? I thought $x-a_{0}$ is nonnegative since $a_{0}leq x$. Is it because $x$ is not an integer?






real-analysis proof-explanation decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
    $endgroup$
    – fleablood
    Jan 1 at 9:18














1












1








1





$begingroup$


I'm self-learning real analysis, and I am trying to understand a part in the proof for the following theorem in Mathematical Analysis by Apostol:




Let $xgeq 0$. Then for every integer $n geq 1$, there exists a finite decimal $r_{n} = a_{0}.a_{1}a_{2}cdots a_{n}$ such that $r_{n} leq x < r_{n} + frac{1}{10^{n}}$.




Here is the given proof:




let S be the set of all nonnegative integers that are less than or equal to $x$. Since $0 in S$, $S$ is not empty. Since $x$ is an upper bound of $S$, $S$ has a supremum $a_0=sup(S)$. Since $a_0le x$, $a_0 in S$, and so $a_0$ is a nonnegative integer. Then $a_0=lfloor x rfloor$ is the greatest integer in $x$. Then $a_0le x < a_0 + 1$.



Let $a_1=lfloor 10x-10a_0rfloor$ be the greatest integer in $10x-10a_0$. Then $0le 10x-10a_0=color{red}{10(x-a_0)<10}$. Then $0le a_1 le 9$. Since $a_1=lfloor 10x-10a_0rfloor$, then $a_1le 10x-10a_0<a_1+1$. Then $a_1$ is the largest integer that satisfies $a_1+10a_0le10x<a_1+1+10a_0$ or $a_0+a_1/10le x<a_0+a_1/10+1/10$.



In general, choosing $a_1$, $a_2$, ..., $a_{n-1}$ with $0le a_ile9$, let $a_n$ be the largest integer that satisfies $a_0 + a_1/10 + cdots + a_n/10^n le x < a_0 + a_1/10 + cdots + a_n/10^n + 1/10^n$. Then $0le a_n le 9$. Then $r_n le x < r_n + 1/10^n$ with $r_n=a_0.a_1a_2cdots a_n$. $blacksquare$




The part I don't understand is in red. Why is $10(x-a_{0})<10$? I thought $x-a_{0}$ is nonnegative since $a_{0}leq x$. Is it because $x$ is not an integer?






real-analysis proof-explanation decimal-expansion







share|cite|improve this question











$endgroup$




I'm self-learning real analysis, and I am trying to understand a part in the proof for the following theorem in Mathematical Analysis by Apostol:




Let $xgeq 0$. Then for every integer $n geq 1$, there exists a finite decimal $r_{n} = a_{0}.a_{1}a_{2}cdots a_{n}$ such that $r_{n} leq x < r_{n} + frac{1}{10^{n}}$.




Here is the given proof:




let S be the set of all nonnegative integers that are less than or equal to $x$. Since $0 in S$, $S$ is not empty. Since $x$ is an upper bound of $S$, $S$ has a supremum $a_0=sup(S)$. Since $a_0le x$, $a_0 in S$, and so $a_0$ is a nonnegative integer. Then $a_0=lfloor x rfloor$ is the greatest integer in $x$. Then $a_0le x < a_0 + 1$.



Let $a_1=lfloor 10x-10a_0rfloor$ be the greatest integer in $10x-10a_0$. Then $0le 10x-10a_0=color{red}{10(x-a_0)<10}$. Then $0le a_1 le 9$. Since $a_1=lfloor 10x-10a_0rfloor$, then $a_1le 10x-10a_0<a_1+1$. Then $a_1$ is the largest integer that satisfies $a_1+10a_0le10x<a_1+1+10a_0$ or $a_0+a_1/10le x<a_0+a_1/10+1/10$.



In general, choosing $a_1$, $a_2$, ..., $a_{n-1}$ with $0le a_ile9$, let $a_n$ be the largest integer that satisfies $a_0 + a_1/10 + cdots + a_n/10^n le x < a_0 + a_1/10 + cdots + a_n/10^n + 1/10^n$. Then $0le a_n le 9$. Then $r_n le x < r_n + 1/10^n$ with $r_n=a_0.a_1a_2cdots a_n$. $blacksquare$




The part I don't understand is in red. Why is $10(x-a_{0})<10$? I thought $x-a_{0}$ is nonnegative since $a_{0}leq x$. Is it because $x$ is not an integer?






real-analysis proof-explanation decimal-expansion




real-analysis proof-explanation decimal-expansion






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share|cite|improve this question




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edited Dec 31 '18 at 3:44









Dan Uznanski

7,03321528




7,03321528










asked Dec 31 '18 at 2:15









numericalorangenumericalorange

1,817312




1,817312








  • 1




    $begingroup$
    $x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
    $endgroup$
    – fleablood
    Jan 1 at 9:18














  • 1




    $begingroup$
    $x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
    $endgroup$
    – fleablood
    Jan 1 at 9:18








1




1




$begingroup$
$x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
$endgroup$
– fleablood
Jan 1 at 9:18




$begingroup$
$x-a_0$ is nonnegative. But it's also less than $1$. So multiplying it by 10 will result in something less than.
$endgroup$
– fleablood
Jan 1 at 9:18










2 Answers
2






active

oldest

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3












$begingroup$

Since $a_0$ is defined to be $a_0 = left[xright]$ in the proof, this means by the definition of "greatest integer" that $a_0 leq x lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 lt 10$, which factors to $10left(x - a_0right) lt 10$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $a_0le x <a_0+1$ so



    $0le (x-a_0) < 1$ so



    $10*0le 10(x-a_0)<1*10$ so



    $0le 10 (x-a_0)<10$.






    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









      3












      $begingroup$

      Since $a_0$ is defined to be $a_0 = left[xright]$ in the proof, this means by the definition of "greatest integer" that $a_0 leq x lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 lt 10$, which factors to $10left(x - a_0right) lt 10$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Since $a_0$ is defined to be $a_0 = left[xright]$ in the proof, this means by the definition of "greatest integer" that $a_0 leq x lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 lt 10$, which factors to $10left(x - a_0right) lt 10$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Since $a_0$ is defined to be $a_0 = left[xright]$ in the proof, this means by the definition of "greatest integer" that $a_0 leq x lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 lt 10$, which factors to $10left(x - a_0right) lt 10$.






          share|cite|improve this answer











          $endgroup$



          Since $a_0$ is defined to be $a_0 = left[xright]$ in the proof, this means by the definition of "greatest integer" that $a_0 leq x lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 lt 10$, which factors to $10left(x - a_0right) lt 10$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 31 '18 at 3:30

























          answered Dec 31 '18 at 2:24









          John OmielanJohn Omielan

          4,1692215




          4,1692215























              1












              $begingroup$

              $a_0le x <a_0+1$ so



              $0le (x-a_0) < 1$ so



              $10*0le 10(x-a_0)<1*10$ so



              $0le 10 (x-a_0)<10$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $a_0le x <a_0+1$ so



                $0le (x-a_0) < 1$ so



                $10*0le 10(x-a_0)<1*10$ so



                $0le 10 (x-a_0)<10$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $a_0le x <a_0+1$ so



                  $0le (x-a_0) < 1$ so



                  $10*0le 10(x-a_0)<1*10$ so



                  $0le 10 (x-a_0)<10$.






                  share|cite|improve this answer









                  $endgroup$



                  $a_0le x <a_0+1$ so



                  $0le (x-a_0) < 1$ so



                  $10*0le 10(x-a_0)<1*10$ so



                  $0le 10 (x-a_0)<10$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 9:13









                  fleabloodfleablood

                  73.4k22791




                  73.4k22791






























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