Terence Tao Exercise 5.4.3: Integer part of $x$ proof.












5












$begingroup$


I am reading Terence Tao: Analysis 1.
As you may be aware, certain objects are introduced bit by bit, so if i am not 'allowed' to use something yet, please understand.



Show that for every real number $x$ there is exactly one integer $N$ such that



$ N leq x < N + 1$



If $x = 0$ then we can take $N = 0$.
If $x > 0$ then $x$ is the formal limit of some Cauchy sequence $(a_n)$, of which is positively bounded away from zero. Since $(a_n)$ is Cauchy it is bounded by some rational $M$, this implies that $x leq M$, if we take $M < N + 1$ for some integer $N$ then $x < N + 1$.



I seem to think this part is fine, but thinking is dangerous.



I now need to show that $ N leq x $.
I know that the sequence $(a_n)$ is bounded away from zero, so every term of the sequence $a_n > c$ for some rational $c$ But i don't think this gets me any further.



Surely i could use this as i know x is the formal limit of the sequence of rationals $(a_n)$ but i cant quite make the connection.



EDIT:



I know i have to prove that this is true when x < 0, but i have not completed from x > 0.



I know i can use Cauchy sequences, division algorithm... its pretty hard for me to say what i can't use because I am an undergraduate. I'm sorry if that makes it impossible to answer for some, it's very awkward i must admit. What i can tell you is the next chapter is on the least upper bound property.










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$endgroup$












  • $begingroup$
    Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
    $endgroup$
    – Tim Raczkowski
    Feb 9 '15 at 14:39










  • $begingroup$
    It's something thatneeds to be proven, @TimRaczkowski.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:39










  • $begingroup$
    There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
    $endgroup$
    – user214138
    Feb 9 '15 at 14:40










  • $begingroup$
    Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:41






  • 2




    $begingroup$
    Would you mind editing into the question which facts you're allowed to use and which facts you're not?
    $endgroup$
    – Neal
    Feb 9 '15 at 14:43
















5












$begingroup$


I am reading Terence Tao: Analysis 1.
As you may be aware, certain objects are introduced bit by bit, so if i am not 'allowed' to use something yet, please understand.



Show that for every real number $x$ there is exactly one integer $N$ such that



$ N leq x < N + 1$



If $x = 0$ then we can take $N = 0$.
If $x > 0$ then $x$ is the formal limit of some Cauchy sequence $(a_n)$, of which is positively bounded away from zero. Since $(a_n)$ is Cauchy it is bounded by some rational $M$, this implies that $x leq M$, if we take $M < N + 1$ for some integer $N$ then $x < N + 1$.



I seem to think this part is fine, but thinking is dangerous.



I now need to show that $ N leq x $.
I know that the sequence $(a_n)$ is bounded away from zero, so every term of the sequence $a_n > c$ for some rational $c$ But i don't think this gets me any further.



Surely i could use this as i know x is the formal limit of the sequence of rationals $(a_n)$ but i cant quite make the connection.



EDIT:



I know i have to prove that this is true when x < 0, but i have not completed from x > 0.



I know i can use Cauchy sequences, division algorithm... its pretty hard for me to say what i can't use because I am an undergraduate. I'm sorry if that makes it impossible to answer for some, it's very awkward i must admit. What i can tell you is the next chapter is on the least upper bound property.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
    $endgroup$
    – Tim Raczkowski
    Feb 9 '15 at 14:39










  • $begingroup$
    It's something thatneeds to be proven, @TimRaczkowski.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:39










  • $begingroup$
    There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
    $endgroup$
    – user214138
    Feb 9 '15 at 14:40










  • $begingroup$
    Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:41






  • 2




    $begingroup$
    Would you mind editing into the question which facts you're allowed to use and which facts you're not?
    $endgroup$
    – Neal
    Feb 9 '15 at 14:43














5












5








5





$begingroup$


I am reading Terence Tao: Analysis 1.
As you may be aware, certain objects are introduced bit by bit, so if i am not 'allowed' to use something yet, please understand.



Show that for every real number $x$ there is exactly one integer $N$ such that



$ N leq x < N + 1$



If $x = 0$ then we can take $N = 0$.
If $x > 0$ then $x$ is the formal limit of some Cauchy sequence $(a_n)$, of which is positively bounded away from zero. Since $(a_n)$ is Cauchy it is bounded by some rational $M$, this implies that $x leq M$, if we take $M < N + 1$ for some integer $N$ then $x < N + 1$.



I seem to think this part is fine, but thinking is dangerous.



I now need to show that $ N leq x $.
I know that the sequence $(a_n)$ is bounded away from zero, so every term of the sequence $a_n > c$ for some rational $c$ But i don't think this gets me any further.



Surely i could use this as i know x is the formal limit of the sequence of rationals $(a_n)$ but i cant quite make the connection.



EDIT:



I know i have to prove that this is true when x < 0, but i have not completed from x > 0.



I know i can use Cauchy sequences, division algorithm... its pretty hard for me to say what i can't use because I am an undergraduate. I'm sorry if that makes it impossible to answer for some, it's very awkward i must admit. What i can tell you is the next chapter is on the least upper bound property.










share|cite|improve this question











$endgroup$




I am reading Terence Tao: Analysis 1.
As you may be aware, certain objects are introduced bit by bit, so if i am not 'allowed' to use something yet, please understand.



Show that for every real number $x$ there is exactly one integer $N$ such that



$ N leq x < N + 1$



If $x = 0$ then we can take $N = 0$.
If $x > 0$ then $x$ is the formal limit of some Cauchy sequence $(a_n)$, of which is positively bounded away from zero. Since $(a_n)$ is Cauchy it is bounded by some rational $M$, this implies that $x leq M$, if we take $M < N + 1$ for some integer $N$ then $x < N + 1$.



I seem to think this part is fine, but thinking is dangerous.



I now need to show that $ N leq x $.
I know that the sequence $(a_n)$ is bounded away from zero, so every term of the sequence $a_n > c$ for some rational $c$ But i don't think this gets me any further.



Surely i could use this as i know x is the formal limit of the sequence of rationals $(a_n)$ but i cant quite make the connection.



EDIT:



I know i have to prove that this is true when x < 0, but i have not completed from x > 0.



I know i can use Cauchy sequences, division algorithm... its pretty hard for me to say what i can't use because I am an undergraduate. I'm sorry if that makes it impossible to answer for some, it's very awkward i must admit. What i can tell you is the next chapter is on the least upper bound property.







analysis






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share|cite|improve this question













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share|cite|improve this question








edited Feb 9 '15 at 14:45

























asked Feb 9 '15 at 14:32







user214138



















  • $begingroup$
    Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
    $endgroup$
    – Tim Raczkowski
    Feb 9 '15 at 14:39










  • $begingroup$
    It's something thatneeds to be proven, @TimRaczkowski.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:39










  • $begingroup$
    There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
    $endgroup$
    – user214138
    Feb 9 '15 at 14:40










  • $begingroup$
    Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:41






  • 2




    $begingroup$
    Would you mind editing into the question which facts you're allowed to use and which facts you're not?
    $endgroup$
    – Neal
    Feb 9 '15 at 14:43


















  • $begingroup$
    Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
    $endgroup$
    – Tim Raczkowski
    Feb 9 '15 at 14:39










  • $begingroup$
    It's something thatneeds to be proven, @TimRaczkowski.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:39










  • $begingroup$
    There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
    $endgroup$
    – user214138
    Feb 9 '15 at 14:40










  • $begingroup$
    Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:41






  • 2




    $begingroup$
    Would you mind editing into the question which facts you're allowed to use and which facts you're not?
    $endgroup$
    – Neal
    Feb 9 '15 at 14:43
















$begingroup$
Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
$endgroup$
– Tim Raczkowski
Feb 9 '15 at 14:39




$begingroup$
Hmm. Is this really how Tao introduces the integer part? Seems unnecessarily complicated.
$endgroup$
– Tim Raczkowski
Feb 9 '15 at 14:39












$begingroup$
It's something thatneeds to be proven, @TimRaczkowski.
$endgroup$
– Thomas Andrews
Feb 9 '15 at 14:39




$begingroup$
It's something thatneeds to be proven, @TimRaczkowski.
$endgroup$
– Thomas Andrews
Feb 9 '15 at 14:39












$begingroup$
There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
$endgroup$
– user214138
Feb 9 '15 at 14:40




$begingroup$
There is one Exercise earlier on in the book, in which x is rational, thats nice and easy, but this one is a little tricky for me.
$endgroup$
– user214138
Feb 9 '15 at 14:40












$begingroup$
Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
$endgroup$
– Thomas Andrews
Feb 9 '15 at 14:41




$begingroup$
Using Cauchy sequences is trick for this sort of problem, because the integer part of each of $0.9,0.99,cdots,0.99dots9,cdots$ is $0$, but the integer part of the limit is $1$.
$endgroup$
– Thomas Andrews
Feb 9 '15 at 14:41




2




2




$begingroup$
Would you mind editing into the question which facts you're allowed to use and which facts you're not?
$endgroup$
– Neal
Feb 9 '15 at 14:43




$begingroup$
Would you mind editing into the question which facts you're allowed to use and which facts you're not?
$endgroup$
– Neal
Feb 9 '15 at 14:43










6 Answers
6






active

oldest

votes


















2












$begingroup$

Let $x=(x_1,x_2,dots,x_n,dots)$ be a Cauchy sequence of rational numbers, and let $epsilon=frac{1}{2}$. Then there exists an $M$ such that if $n,mgeq M$, $|x_n-x_m|<epsilon$. That means, in particular, $xin [x_M-frac{1}{2},x_M+frac{1}{2}]$.



Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-frac{1}{2}$, and that $N_0leq x<N_0+2$. Then either $N_0leq x<N_0+1$ or $N_0+1leq x<N_0+2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
    $endgroup$
    – user214138
    Feb 9 '15 at 14:57












  • $begingroup$
    Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 14:59












  • $begingroup$
    Is that last part of the last comment a typo?
    $endgroup$
    – user214138
    Feb 9 '15 at 15:01






  • 1




    $begingroup$
    No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
    $endgroup$
    – Thomas Andrews
    Feb 9 '15 at 15:02



















3












$begingroup$

If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $Nleq x$ because otherwise $N+1$ wouldn't be the first element.



Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.






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$endgroup$





















    0












    $begingroup$

    The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.




    Let $d = inf{|N-x| : Nin mathbb{Z}}$. This is a nonempty set of
    reals so it has an infimum. If $d = 0$, then $xinmathbb{Z}$ and
    you're done.




    On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.



    Now write $x = lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_minmathbb{Z}$ and $r_min [0,1)$.



    Now can you prove that $A_m = A_n$ for all $m,n>M$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
      $endgroup$
      – user214138
      Feb 9 '15 at 15:04



















    0












    $begingroup$

    Given the sequence $(x_i)=x$ consider the sequences $(lfloor x_irfloor)=lfloor x rfloor $and $(lceil x_irceil)=lceil xrceil$. Where $lfloor x_irfloor$ is the gratest integer $<x_i$ and $lceil x_irceil$ is the lowest integer $>x_i$.



    You can proof that if $(x_i)$ is Cauchy, than $lfloor x rfloor$ and $lceil xrceil$ are Cauchy sequences that define successive integers.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.



      Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.



      Thus we have that N is less than or equal to x and x is less than N + 1, as desired.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        First show that such an N exists, then show that it is unique.



        $x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).




        1. For the easiest case, if $x=0$, we have $N = 0$.


        2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k leq x$ for any integer $kgeq 0$. Use $k=0$ as the base case. $0 leq x$ is true, as $x$ is positive. Assume that $kleq x$ for some $kgeq 0$, then consider $k+1 leq x$. By our assumption that $N$ does not exist, $k+1leq x$ must be true. This closes the induction.



        But we know the conclusion is false. There must exist an integer $M$ such that $Mgt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can therefore conclude that such an $N$ exists.




        1. A similar induction can be constructed when x is negative.*


        Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 leq x lt n_1 + 1$ and $n_2 leq x lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).



        $$
        begin{align}
        &n_1 le n_2 \
        &implies n_1 + 1 leq n_2 \
        &implies n_1 + 1 leq x && text{as $n_2 leq x$} \
        &implies text{false} &&text{as we assumed $n_1 + 1 > x$}
        end{align}
        $$



        Thus, there exists a unique integer $N$ such that $N leq x lt N + 1$.



        * Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.






        share|cite|improve this answer









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          6 Answers
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          6 Answers
          6






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let $x=(x_1,x_2,dots,x_n,dots)$ be a Cauchy sequence of rational numbers, and let $epsilon=frac{1}{2}$. Then there exists an $M$ such that if $n,mgeq M$, $|x_n-x_m|<epsilon$. That means, in particular, $xin [x_M-frac{1}{2},x_M+frac{1}{2}]$.



          Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-frac{1}{2}$, and that $N_0leq x<N_0+2$. Then either $N_0leq x<N_0+1$ or $N_0+1leq x<N_0+2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
            $endgroup$
            – user214138
            Feb 9 '15 at 14:57












          • $begingroup$
            Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 14:59












          • $begingroup$
            Is that last part of the last comment a typo?
            $endgroup$
            – user214138
            Feb 9 '15 at 15:01






          • 1




            $begingroup$
            No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 15:02
















          2












          $begingroup$

          Let $x=(x_1,x_2,dots,x_n,dots)$ be a Cauchy sequence of rational numbers, and let $epsilon=frac{1}{2}$. Then there exists an $M$ such that if $n,mgeq M$, $|x_n-x_m|<epsilon$. That means, in particular, $xin [x_M-frac{1}{2},x_M+frac{1}{2}]$.



          Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-frac{1}{2}$, and that $N_0leq x<N_0+2$. Then either $N_0leq x<N_0+1$ or $N_0+1leq x<N_0+2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
            $endgroup$
            – user214138
            Feb 9 '15 at 14:57












          • $begingroup$
            Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 14:59












          • $begingroup$
            Is that last part of the last comment a typo?
            $endgroup$
            – user214138
            Feb 9 '15 at 15:01






          • 1




            $begingroup$
            No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 15:02














          2












          2








          2





          $begingroup$

          Let $x=(x_1,x_2,dots,x_n,dots)$ be a Cauchy sequence of rational numbers, and let $epsilon=frac{1}{2}$. Then there exists an $M$ such that if $n,mgeq M$, $|x_n-x_m|<epsilon$. That means, in particular, $xin [x_M-frac{1}{2},x_M+frac{1}{2}]$.



          Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-frac{1}{2}$, and that $N_0leq x<N_0+2$. Then either $N_0leq x<N_0+1$ or $N_0+1leq x<N_0+2$.






          share|cite|improve this answer









          $endgroup$



          Let $x=(x_1,x_2,dots,x_n,dots)$ be a Cauchy sequence of rational numbers, and let $epsilon=frac{1}{2}$. Then there exists an $M$ such that if $n,mgeq M$, $|x_n-x_m|<epsilon$. That means, in particular, $xin [x_M-frac{1}{2},x_M+frac{1}{2}]$.



          Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-frac{1}{2}$, and that $N_0leq x<N_0+2$. Then either $N_0leq x<N_0+1$ or $N_0+1leq x<N_0+2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 9 '15 at 14:51









          Thomas AndrewsThomas Andrews

          130k12147298




          130k12147298












          • $begingroup$
            By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
            $endgroup$
            – user214138
            Feb 9 '15 at 14:57












          • $begingroup$
            Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 14:59












          • $begingroup$
            Is that last part of the last comment a typo?
            $endgroup$
            – user214138
            Feb 9 '15 at 15:01






          • 1




            $begingroup$
            No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 15:02


















          • $begingroup$
            By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
            $endgroup$
            – user214138
            Feb 9 '15 at 14:57












          • $begingroup$
            Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 14:59












          • $begingroup$
            Is that last part of the last comment a typo?
            $endgroup$
            – user214138
            Feb 9 '15 at 15:01






          • 1




            $begingroup$
            No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
            $endgroup$
            – Thomas Andrews
            Feb 9 '15 at 15:02
















          $begingroup$
          By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
          $endgroup$
          – user214138
          Feb 9 '15 at 14:57






          $begingroup$
          By 'theorem for rational numbers' your referring to the exercise that proved the theorem in the book that i mentioned in the comments? i.e $ n leq x < n + 1$ where x is ratonal
          $endgroup$
          – user214138
          Feb 9 '15 at 14:57














          $begingroup$
          Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
          $endgroup$
          – Thomas Andrews
          Feb 9 '15 at 14:59






          $begingroup$
          Yes, so use that theorem to find $N_0$ so that $N_0leq x_M-frac{1}{2}<N_0+1$. Then show that $N_0leq xleq N_0+2$.
          $endgroup$
          – Thomas Andrews
          Feb 9 '15 at 14:59














          $begingroup$
          Is that last part of the last comment a typo?
          $endgroup$
          – user214138
          Feb 9 '15 at 15:01




          $begingroup$
          Is that last part of the last comment a typo?
          $endgroup$
          – user214138
          Feb 9 '15 at 15:01




          1




          1




          $begingroup$
          No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
          $endgroup$
          – Thomas Andrews
          Feb 9 '15 at 15:02




          $begingroup$
          No, it is $2$. You only know that $N_0leq x_M-frac{1}{2}<N_0+1$. But we only know that $x_M-frac{1}{2}leq xleq x_M+frac{1}{2}$.
          $endgroup$
          – Thomas Andrews
          Feb 9 '15 at 15:02











          3












          $begingroup$

          If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $Nleq x$ because otherwise $N+1$ wouldn't be the first element.



          Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $Nleq x$ because otherwise $N+1$ wouldn't be the first element.



            Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $Nleq x$ because otherwise $N+1$ wouldn't be the first element.



              Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.






              share|cite|improve this answer









              $endgroup$



              If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $Nleq x$ because otherwise $N+1$ wouldn't be the first element.



              Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 9 '15 at 14:37









              CarolCarol

              41827




              41827























                  0












                  $begingroup$

                  The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.




                  Let $d = inf{|N-x| : Nin mathbb{Z}}$. This is a nonempty set of
                  reals so it has an infimum. If $d = 0$, then $xinmathbb{Z}$ and
                  you're done.




                  On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.



                  Now write $x = lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_minmathbb{Z}$ and $r_min [0,1)$.



                  Now can you prove that $A_m = A_n$ for all $m,n>M$?






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                    $endgroup$
                    – user214138
                    Feb 9 '15 at 15:04
















                  0












                  $begingroup$

                  The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.




                  Let $d = inf{|N-x| : Nin mathbb{Z}}$. This is a nonempty set of
                  reals so it has an infimum. If $d = 0$, then $xinmathbb{Z}$ and
                  you're done.




                  On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.



                  Now write $x = lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_minmathbb{Z}$ and $r_min [0,1)$.



                  Now can you prove that $A_m = A_n$ for all $m,n>M$?






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                    $endgroup$
                    – user214138
                    Feb 9 '15 at 15:04














                  0












                  0








                  0





                  $begingroup$

                  The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.




                  Let $d = inf{|N-x| : Nin mathbb{Z}}$. This is a nonempty set of
                  reals so it has an infimum. If $d = 0$, then $xinmathbb{Z}$ and
                  you're done.




                  On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.



                  Now write $x = lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_minmathbb{Z}$ and $r_min [0,1)$.



                  Now can you prove that $A_m = A_n$ for all $m,n>M$?






                  share|cite|improve this answer









                  $endgroup$



                  The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.




                  Let $d = inf{|N-x| : Nin mathbb{Z}}$. This is a nonempty set of
                  reals so it has an infimum. If $d = 0$, then $xinmathbb{Z}$ and
                  you're done.




                  On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.



                  Now write $x = lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_minmathbb{Z}$ and $r_min [0,1)$.



                  Now can you prove that $A_m = A_n$ for all $m,n>M$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 9 '15 at 14:51









                  NealNeal

                  24k24087




                  24k24087












                  • $begingroup$
                    I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                    $endgroup$
                    – user214138
                    Feb 9 '15 at 15:04


















                  • $begingroup$
                    I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                    $endgroup$
                    – user214138
                    Feb 9 '15 at 15:04
















                  $begingroup$
                  I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                  $endgroup$
                  – user214138
                  Feb 9 '15 at 15:04




                  $begingroup$
                  I am really sorry for wasting your time Neal, the next chapter is on supremum(least upper bound) and infimum.
                  $endgroup$
                  – user214138
                  Feb 9 '15 at 15:04











                  0












                  $begingroup$

                  Given the sequence $(x_i)=x$ consider the sequences $(lfloor x_irfloor)=lfloor x rfloor $and $(lceil x_irceil)=lceil xrceil$. Where $lfloor x_irfloor$ is the gratest integer $<x_i$ and $lceil x_irceil$ is the lowest integer $>x_i$.



                  You can proof that if $(x_i)$ is Cauchy, than $lfloor x rfloor$ and $lceil xrceil$ are Cauchy sequences that define successive integers.






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    Given the sequence $(x_i)=x$ consider the sequences $(lfloor x_irfloor)=lfloor x rfloor $and $(lceil x_irceil)=lceil xrceil$. Where $lfloor x_irfloor$ is the gratest integer $<x_i$ and $lceil x_irceil$ is the lowest integer $>x_i$.



                    You can proof that if $(x_i)$ is Cauchy, than $lfloor x rfloor$ and $lceil xrceil$ are Cauchy sequences that define successive integers.






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      Given the sequence $(x_i)=x$ consider the sequences $(lfloor x_irfloor)=lfloor x rfloor $and $(lceil x_irceil)=lceil xrceil$. Where $lfloor x_irfloor$ is the gratest integer $<x_i$ and $lceil x_irceil$ is the lowest integer $>x_i$.



                      You can proof that if $(x_i)$ is Cauchy, than $lfloor x rfloor$ and $lceil xrceil$ are Cauchy sequences that define successive integers.






                      share|cite|improve this answer









                      $endgroup$



                      Given the sequence $(x_i)=x$ consider the sequences $(lfloor x_irfloor)=lfloor x rfloor $and $(lceil x_irceil)=lceil xrceil$. Where $lfloor x_irfloor$ is the gratest integer $<x_i$ and $lceil x_irceil$ is the lowest integer $>x_i$.



                      You can proof that if $(x_i)$ is Cauchy, than $lfloor x rfloor$ and $lceil xrceil$ are Cauchy sequences that define successive integers.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Feb 9 '15 at 15:18









                      Emilio NovatiEmilio Novati

                      52.2k43474




                      52.2k43474























                          0












                          $begingroup$

                          Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.



                          Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.



                          Thus we have that N is less than or equal to x and x is less than N + 1, as desired.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.



                            Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.



                            Thus we have that N is less than or equal to x and x is less than N + 1, as desired.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.



                              Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.



                              Thus we have that N is less than or equal to x and x is less than N + 1, as desired.






                              share|cite|improve this answer









                              $endgroup$



                              Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.



                              Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.



                              Thus we have that N is less than or equal to x and x is less than N + 1, as desired.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Apr 13 '16 at 18:44









                              KoleKole

                              1




                              1























                                  0












                                  $begingroup$

                                  First show that such an N exists, then show that it is unique.



                                  $x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).




                                  1. For the easiest case, if $x=0$, we have $N = 0$.


                                  2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k leq x$ for any integer $kgeq 0$. Use $k=0$ as the base case. $0 leq x$ is true, as $x$ is positive. Assume that $kleq x$ for some $kgeq 0$, then consider $k+1 leq x$. By our assumption that $N$ does not exist, $k+1leq x$ must be true. This closes the induction.



                                  But we know the conclusion is false. There must exist an integer $M$ such that $Mgt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can therefore conclude that such an $N$ exists.




                                  1. A similar induction can be constructed when x is negative.*


                                  Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 leq x lt n_1 + 1$ and $n_2 leq x lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).



                                  $$
                                  begin{align}
                                  &n_1 le n_2 \
                                  &implies n_1 + 1 leq n_2 \
                                  &implies n_1 + 1 leq x && text{as $n_2 leq x$} \
                                  &implies text{false} &&text{as we assumed $n_1 + 1 > x$}
                                  end{align}
                                  $$



                                  Thus, there exists a unique integer $N$ such that $N leq x lt N + 1$.



                                  * Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    First show that such an N exists, then show that it is unique.



                                    $x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).




                                    1. For the easiest case, if $x=0$, we have $N = 0$.


                                    2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k leq x$ for any integer $kgeq 0$. Use $k=0$ as the base case. $0 leq x$ is true, as $x$ is positive. Assume that $kleq x$ for some $kgeq 0$, then consider $k+1 leq x$. By our assumption that $N$ does not exist, $k+1leq x$ must be true. This closes the induction.



                                    But we know the conclusion is false. There must exist an integer $M$ such that $Mgt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can therefore conclude that such an $N$ exists.




                                    1. A similar induction can be constructed when x is negative.*


                                    Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 leq x lt n_1 + 1$ and $n_2 leq x lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).



                                    $$
                                    begin{align}
                                    &n_1 le n_2 \
                                    &implies n_1 + 1 leq n_2 \
                                    &implies n_1 + 1 leq x && text{as $n_2 leq x$} \
                                    &implies text{false} &&text{as we assumed $n_1 + 1 > x$}
                                    end{align}
                                    $$



                                    Thus, there exists a unique integer $N$ such that $N leq x lt N + 1$.



                                    * Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      First show that such an N exists, then show that it is unique.



                                      $x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).




                                      1. For the easiest case, if $x=0$, we have $N = 0$.


                                      2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k leq x$ for any integer $kgeq 0$. Use $k=0$ as the base case. $0 leq x$ is true, as $x$ is positive. Assume that $kleq x$ for some $kgeq 0$, then consider $k+1 leq x$. By our assumption that $N$ does not exist, $k+1leq x$ must be true. This closes the induction.



                                      But we know the conclusion is false. There must exist an integer $M$ such that $Mgt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can therefore conclude that such an $N$ exists.




                                      1. A similar induction can be constructed when x is negative.*


                                      Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 leq x lt n_1 + 1$ and $n_2 leq x lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).



                                      $$
                                      begin{align}
                                      &n_1 le n_2 \
                                      &implies n_1 + 1 leq n_2 \
                                      &implies n_1 + 1 leq x && text{as $n_2 leq x$} \
                                      &implies text{false} &&text{as we assumed $n_1 + 1 > x$}
                                      end{align}
                                      $$



                                      Thus, there exists a unique integer $N$ such that $N leq x lt N + 1$.



                                      * Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.






                                      share|cite|improve this answer









                                      $endgroup$



                                      First show that such an N exists, then show that it is unique.



                                      $x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).




                                      1. For the easiest case, if $x=0$, we have $N = 0$.


                                      2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k leq x$ for any integer $kgeq 0$. Use $k=0$ as the base case. $0 leq x$ is true, as $x$ is positive. Assume that $kleq x$ for some $kgeq 0$, then consider $k+1 leq x$. By our assumption that $N$ does not exist, $k+1leq x$ must be true. This closes the induction.



                                      But we know the conclusion is false. There must exist an integer $M$ such that $Mgt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can therefore conclude that such an $N$ exists.




                                      1. A similar induction can be constructed when x is negative.*


                                      Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 leq x lt n_1 + 1$ and $n_2 leq x lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).



                                      $$
                                      begin{align}
                                      &n_1 le n_2 \
                                      &implies n_1 + 1 leq n_2 \
                                      &implies n_1 + 1 leq x && text{as $n_2 leq x$} \
                                      &implies text{false} &&text{as we assumed $n_1 + 1 > x$}
                                      end{align}
                                      $$



                                      Thus, there exists a unique integer $N$ such that $N leq x lt N + 1$.



                                      * Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 28 '18 at 1:44









                                      KevinKevin

                                      1012




                                      1012






























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