Reason for the integer case and the rational case to be solved differently












4












$begingroup$



Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.




This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;




first prove this for integer $x$, then for rational $x$, then finally for real $x.$




The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?



Any help would be much appreciated.






real-analysis functions continuity functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
    $endgroup$
    – the_fox
    Dec 17 '18 at 16:19










  • $begingroup$
    the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
    $endgroup$
    – Conrad
    Dec 17 '18 at 16:24












  • $begingroup$
    Maybe exists a direct method. But surely the solution using the hint is easier.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 17:36










  • $begingroup$
    But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
    $endgroup$
    – 299792458
    Dec 17 '18 at 17:40










  • $begingroup$
    Because the integer case is used in the proof of rational case?
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 20:17
















4












$begingroup$



Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.




This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;




first prove this for integer $x$, then for rational $x$, then finally for real $x.$




The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?



Any help would be much appreciated.






real-analysis functions continuity functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
    $endgroup$
    – the_fox
    Dec 17 '18 at 16:19










  • $begingroup$
    the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
    $endgroup$
    – Conrad
    Dec 17 '18 at 16:24












  • $begingroup$
    Maybe exists a direct method. But surely the solution using the hint is easier.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 17:36










  • $begingroup$
    But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
    $endgroup$
    – 299792458
    Dec 17 '18 at 17:40










  • $begingroup$
    Because the integer case is used in the proof of rational case?
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 20:17














4












4








4


1



$begingroup$



Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.




This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;




first prove this for integer $x$, then for rational $x$, then finally for real $x.$




The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?



Any help would be much appreciated.






real-analysis functions continuity functional-equations







share|cite|improve this question











$endgroup$





Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.




This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;




first prove this for integer $x$, then for rational $x$, then finally for real $x.$




The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?



Any help would be much appreciated.






real-analysis functions continuity functional-equations




real-analysis functions continuity functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 16:13









user10354138

7,4322925




7,4322925










asked Dec 17 '18 at 15:38









299792458299792458

527




527












  • $begingroup$
    A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
    $endgroup$
    – the_fox
    Dec 17 '18 at 16:19










  • $begingroup$
    the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
    $endgroup$
    – Conrad
    Dec 17 '18 at 16:24












  • $begingroup$
    Maybe exists a direct method. But surely the solution using the hint is easier.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 17:36










  • $begingroup$
    But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
    $endgroup$
    – 299792458
    Dec 17 '18 at 17:40










  • $begingroup$
    Because the integer case is used in the proof of rational case?
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 20:17


















  • $begingroup$
    A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
    $endgroup$
    – the_fox
    Dec 17 '18 at 16:19










  • $begingroup$
    the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
    $endgroup$
    – Conrad
    Dec 17 '18 at 16:24












  • $begingroup$
    Maybe exists a direct method. But surely the solution using the hint is easier.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 17:36










  • $begingroup$
    But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
    $endgroup$
    – 299792458
    Dec 17 '18 at 17:40










  • $begingroup$
    Because the integer case is used in the proof of rational case?
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 17 '18 at 20:17
















$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19




$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19












$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24






$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24














$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36




$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36












$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40




$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40












$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17




$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17










2 Answers
2






active

oldest

votes


















2












$begingroup$

Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.



Continous Additive functions are linear



Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.



By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$



Hence $f(y)=1+f(y-1)$



Further consider $f(x+y)$



$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$



Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$



Hence $g(x+y)=g(x)+g(y)$



Hence $g(x)=g(1)x$



Which implies $f(x)-1=g(1)x$



Hence $f(x)=1+(f(x)-1)x$



Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$



We conclude $f(x)=1+x$



You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But why $2f(0)=2.1$?
    $endgroup$
    – 299792458
    Dec 19 '18 at 10:22










  • $begingroup$
    @299792458 because $f(0)=1$ I'd given in the question.
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 3:38










  • $begingroup$
    In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
    $endgroup$
    – 299792458
    Dec 23 '18 at 4:18






  • 1




    $begingroup$
    You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
    $endgroup$
    – Milo Brandt
    Dec 23 '18 at 5:29










  • $begingroup$
    @MiloBrandt May Be. I will inform him about this
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 5:48



















3












$begingroup$

You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.



Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach



Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044092%2freason-for-the-integer-case-and-the-rational-case-to-be-solved-differently%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.



    Continous Additive functions are linear



    Coming back to your Orignal Problem
    Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
    Keep it aside for a while.



    By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$



    Hence $f(y)=1+f(y-1)$



    Further consider $f(x+y)$



    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$



    Hence we have $f(x+y)=f(x)+f(y)-1$
    This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$



    Hence $g(x+y)=g(x)+g(y)$



    Hence $g(x)=g(1)x$



    Which implies $f(x)-1=g(1)x$



    Hence $f(x)=1+(f(x)-1)x$



    Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$



    We conclude $f(x)=1+x$



    You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But why $2f(0)=2.1$?
      $endgroup$
      – 299792458
      Dec 19 '18 at 10:22










    • $begingroup$
      @299792458 because $f(0)=1$ I'd given in the question.
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 3:38










    • $begingroup$
      In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
      $endgroup$
      – 299792458
      Dec 23 '18 at 4:18






    • 1




      $begingroup$
      You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
      $endgroup$
      – Milo Brandt
      Dec 23 '18 at 5:29










    • $begingroup$
      @MiloBrandt May Be. I will inform him about this
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 5:48
















    2












    $begingroup$

    Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.



    Continous Additive functions are linear



    Coming back to your Orignal Problem
    Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
    Keep it aside for a while.



    By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$



    Hence $f(y)=1+f(y-1)$



    Further consider $f(x+y)$



    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$



    Hence we have $f(x+y)=f(x)+f(y)-1$
    This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$



    Hence $g(x+y)=g(x)+g(y)$



    Hence $g(x)=g(1)x$



    Which implies $f(x)-1=g(1)x$



    Hence $f(x)=1+(f(x)-1)x$



    Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$



    We conclude $f(x)=1+x$



    You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But why $2f(0)=2.1$?
      $endgroup$
      – 299792458
      Dec 19 '18 at 10:22










    • $begingroup$
      @299792458 because $f(0)=1$ I'd given in the question.
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 3:38










    • $begingroup$
      In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
      $endgroup$
      – 299792458
      Dec 23 '18 at 4:18






    • 1




      $begingroup$
      You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
      $endgroup$
      – Milo Brandt
      Dec 23 '18 at 5:29










    • $begingroup$
      @MiloBrandt May Be. I will inform him about this
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 5:48














    2












    2








    2





    $begingroup$

    Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.



    Continous Additive functions are linear



    Coming back to your Orignal Problem
    Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
    Keep it aside for a while.



    By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$



    Hence $f(y)=1+f(y-1)$



    Further consider $f(x+y)$



    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$



    Hence we have $f(x+y)=f(x)+f(y)-1$
    This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$



    Hence $g(x+y)=g(x)+g(y)$



    Hence $g(x)=g(1)x$



    Which implies $f(x)-1=g(1)x$



    Hence $f(x)=1+(f(x)-1)x$



    Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$



    We conclude $f(x)=1+x$



    You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above






    share|cite|improve this answer











    $endgroup$



    Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.



    Continous Additive functions are linear



    Coming back to your Orignal Problem
    Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
    Keep it aside for a while.



    By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$



    Hence $f(y)=1+f(y-1)$



    Further consider $f(x+y)$



    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
    $$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$



    Hence we have $f(x+y)=f(x)+f(y)-1$
    This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$



    Hence $g(x+y)=g(x)+g(y)$



    Hence $g(x)=g(1)x$



    Which implies $f(x)-1=g(1)x$



    Hence $f(x)=1+(f(x)-1)x$



    Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$



    We conclude $f(x)=1+x$



    You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 5:47

























    answered Dec 18 '18 at 4:26









    Rakesh BhattRakesh Bhatt

    967214




    967214












    • $begingroup$
      But why $2f(0)=2.1$?
      $endgroup$
      – 299792458
      Dec 19 '18 at 10:22










    • $begingroup$
      @299792458 because $f(0)=1$ I'd given in the question.
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 3:38










    • $begingroup$
      In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
      $endgroup$
      – 299792458
      Dec 23 '18 at 4:18






    • 1




      $begingroup$
      You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
      $endgroup$
      – Milo Brandt
      Dec 23 '18 at 5:29










    • $begingroup$
      @MiloBrandt May Be. I will inform him about this
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 5:48


















    • $begingroup$
      But why $2f(0)=2.1$?
      $endgroup$
      – 299792458
      Dec 19 '18 at 10:22










    • $begingroup$
      @299792458 because $f(0)=1$ I'd given in the question.
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 3:38










    • $begingroup$
      In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
      $endgroup$
      – 299792458
      Dec 23 '18 at 4:18






    • 1




      $begingroup$
      You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
      $endgroup$
      – Milo Brandt
      Dec 23 '18 at 5:29










    • $begingroup$
      @MiloBrandt May Be. I will inform him about this
      $endgroup$
      – Rakesh Bhatt
      Dec 23 '18 at 5:48
















    $begingroup$
    But why $2f(0)=2.1$?
    $endgroup$
    – 299792458
    Dec 19 '18 at 10:22




    $begingroup$
    But why $2f(0)=2.1$?
    $endgroup$
    – 299792458
    Dec 19 '18 at 10:22












    $begingroup$
    @299792458 because $f(0)=1$ I'd given in the question.
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 3:38




    $begingroup$
    @299792458 because $f(0)=1$ I'd given in the question.
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 3:38












    $begingroup$
    In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
    $endgroup$
    – 299792458
    Dec 23 '18 at 4:18




    $begingroup$
    In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
    $endgroup$
    – 299792458
    Dec 23 '18 at 4:18




    1




    1




    $begingroup$
    You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
    $endgroup$
    – Milo Brandt
    Dec 23 '18 at 5:29




    $begingroup$
    You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
    $endgroup$
    – Milo Brandt
    Dec 23 '18 at 5:29












    $begingroup$
    @MiloBrandt May Be. I will inform him about this
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 5:48




    $begingroup$
    @MiloBrandt May Be. I will inform him about this
    $endgroup$
    – Rakesh Bhatt
    Dec 23 '18 at 5:48











    3












    $begingroup$

    You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.



    Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach



    Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.



      Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach



      Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.



        Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach



        Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.






        share|cite|improve this answer









        $endgroup$



        You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.



        Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach



        Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 4:42









        Ross MillikanRoss Millikan

        300k24200374




        300k24200374






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044092%2freason-for-the-integer-case-and-the-rational-case-to-be-solved-differently%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more