Is there any unique function $c=f(a, b)$ such that if I know only $c$, I can deduce unique $a$ and $b$ from it? [duplicate]

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  • Injective function from $mathbbR^2$ to $mathbbR$?

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I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.



As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
Does there exist such function which can make me guess input $a, b$ as unique?



also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.










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marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35


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


















    up vote
    2
    down vote

    favorite













    This question already has an answer here:



    • Injective function from $mathbbR^2$ to $mathbbR$?

      3 answers



    I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.



    As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
    Does there exist such function which can make me guess input $a, b$ as unique?



    also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.










    share|cite|improve this question















    marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35


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
















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      This question already has an answer here:



      • Injective function from $mathbbR^2$ to $mathbbR$?

        3 answers



      I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.



      As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
      Does there exist such function which can make me guess input $a, b$ as unique?



      also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.










      share|cite|improve this question
















      This question already has an answer here:



      • Injective function from $mathbbR^2$ to $mathbbR$?

        3 answers



      I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.



      As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
      Does there exist such function which can make me guess input $a, b$ as unique?



      also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.





      This question already has an answer here:



      • Injective function from $mathbbR^2$ to $mathbbR$?

        3 answers







      relations function-and-relation-composition






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      edited Sep 9 at 9:44

























      asked Sep 9 at 9:25









      user8428217

      112




      112




      marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35


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






      marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35


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






















          2 Answers
          2






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          up vote
          1
          down vote













          Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
          $$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
          where reconstructing the digits of the arguments is readily possible.






          share|cite|improve this answer



























            up vote
            0
            down vote













            What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
            In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).



            But it can also be shown that there are not continuous ones.



            Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.



            So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$



            This goes against the hypothesis of the injective function $f$.



            If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.






            share|cite|improve this answer




















            • ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
              – user8428217
              Sep 9 at 10:42







            • 1




              Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
              – John Hughes
              Sep 9 at 10:44










            • Notice that I used a lot the continuity, whereas concatenation is not continuous :)
              – Riccardo Ceccon
              Sep 9 at 10:46






            • 1




              ok. now it is clear. thanks
              – user8428217
              Sep 9 at 10:55










            • Please eave an upvote if you want :)
              – Riccardo Ceccon
              Sep 9 at 11:18

















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
            $$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
            where reconstructing the digits of the arguments is readily possible.






            share|cite|improve this answer
























              up vote
              1
              down vote













              Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
              $$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
              where reconstructing the digits of the arguments is readily possible.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
                $$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
                where reconstructing the digits of the arguments is readily possible.






                share|cite|improve this answer












                Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
                $$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
                where reconstructing the digits of the arguments is readily possible.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 9 at 9:38









                Hagen von Eitzen

                267k21260483




                267k21260483




















                    up vote
                    0
                    down vote













                    What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
                    In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).



                    But it can also be shown that there are not continuous ones.



                    Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.



                    So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$



                    This goes against the hypothesis of the injective function $f$.



                    If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.






                    share|cite|improve this answer




















                    • ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                      – user8428217
                      Sep 9 at 10:42







                    • 1




                      Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                      – John Hughes
                      Sep 9 at 10:44










                    • Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                      – Riccardo Ceccon
                      Sep 9 at 10:46






                    • 1




                      ok. now it is clear. thanks
                      – user8428217
                      Sep 9 at 10:55










                    • Please eave an upvote if you want :)
                      – Riccardo Ceccon
                      Sep 9 at 11:18














                    up vote
                    0
                    down vote













                    What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
                    In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).



                    But it can also be shown that there are not continuous ones.



                    Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.



                    So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$



                    This goes against the hypothesis of the injective function $f$.



                    If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.






                    share|cite|improve this answer




















                    • ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                      – user8428217
                      Sep 9 at 10:42







                    • 1




                      Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                      – John Hughes
                      Sep 9 at 10:44










                    • Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                      – Riccardo Ceccon
                      Sep 9 at 10:46






                    • 1




                      ok. now it is clear. thanks
                      – user8428217
                      Sep 9 at 10:55










                    • Please eave an upvote if you want :)
                      – Riccardo Ceccon
                      Sep 9 at 11:18












                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
                    In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).



                    But it can also be shown that there are not continuous ones.



                    Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.



                    So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$



                    This goes against the hypothesis of the injective function $f$.



                    If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.






                    share|cite|improve this answer












                    What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
                    In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).



                    But it can also be shown that there are not continuous ones.



                    Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.



                    So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$



                    This goes against the hypothesis of the injective function $f$.



                    If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 9 at 9:57









                    Riccardo Ceccon

                    875320




                    875320











                    • ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                      – user8428217
                      Sep 9 at 10:42







                    • 1




                      Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                      – John Hughes
                      Sep 9 at 10:44










                    • Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                      – Riccardo Ceccon
                      Sep 9 at 10:46






                    • 1




                      ok. now it is clear. thanks
                      – user8428217
                      Sep 9 at 10:55










                    • Please eave an upvote if you want :)
                      – Riccardo Ceccon
                      Sep 9 at 11:18
















                    • ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                      – user8428217
                      Sep 9 at 10:42







                    • 1




                      Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                      – John Hughes
                      Sep 9 at 10:44










                    • Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                      – Riccardo Ceccon
                      Sep 9 at 10:46






                    • 1




                      ok. now it is clear. thanks
                      – user8428217
                      Sep 9 at 10:55










                    • Please eave an upvote if you want :)
                      – Riccardo Ceccon
                      Sep 9 at 11:18















                    ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                    – user8428217
                    Sep 9 at 10:42





                    ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
                    – user8428217
                    Sep 9 at 10:42





                    1




                    1




                    Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                    – John Hughes
                    Sep 9 at 10:44




                    Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
                    – John Hughes
                    Sep 9 at 10:44












                    Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                    – Riccardo Ceccon
                    Sep 9 at 10:46




                    Notice that I used a lot the continuity, whereas concatenation is not continuous :)
                    – Riccardo Ceccon
                    Sep 9 at 10:46




                    1




                    1




                    ok. now it is clear. thanks
                    – user8428217
                    Sep 9 at 10:55




                    ok. now it is clear. thanks
                    – user8428217
                    Sep 9 at 10:55












                    Please eave an upvote if you want :)
                    – Riccardo Ceccon
                    Sep 9 at 11:18




                    Please eave an upvote if you want :)
                    – Riccardo Ceccon
                    Sep 9 at 11:18


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