Multiplication via squaring and addition












10












$begingroup$


Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.




Peano Arithmetic has the following two axioms:




  1. $x cdot 0 = 0$

  2. $x cdot y = x cdot (y-1) + x$




So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.



I tried a few things and noticed that one has:



$$2xy = (x+y)^2-x^2-y^2 text{ and } 4xy = (x+y)^2-(x-y)^2$$



Close to $xy$ but still not what I am looking for. And actually this uses subtraction...



Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbb{N}, +, cdot^2)$.






logic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
    $endgroup$
    – coffeemath
    22 hours ago






  • 4




    $begingroup$
    @MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
    $endgroup$
    – Noah Schweber
    21 hours ago












  • $begingroup$
    @NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
    $endgroup$
    – Sqyuli
    21 hours ago
















10












$begingroup$


Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.




Peano Arithmetic has the following two axioms:




  1. $x cdot 0 = 0$

  2. $x cdot y = x cdot (y-1) + x$




So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.



I tried a few things and noticed that one has:



$$2xy = (x+y)^2-x^2-y^2 text{ and } 4xy = (x+y)^2-(x-y)^2$$



Close to $xy$ but still not what I am looking for. And actually this uses subtraction...



Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbb{N}, +, cdot^2)$.






logic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
    $endgroup$
    – coffeemath
    22 hours ago






  • 4




    $begingroup$
    @MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
    $endgroup$
    – Noah Schweber
    21 hours ago












  • $begingroup$
    @NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
    $endgroup$
    – Sqyuli
    21 hours ago














10












10








10


2



$begingroup$


Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.




Peano Arithmetic has the following two axioms:




  1. $x cdot 0 = 0$

  2. $x cdot y = x cdot (y-1) + x$




So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.



I tried a few things and noticed that one has:



$$2xy = (x+y)^2-x^2-y^2 text{ and } 4xy = (x+y)^2-(x-y)^2$$



Close to $xy$ but still not what I am looking for. And actually this uses subtraction...



Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbb{N}, +, cdot^2)$.






logic







share|cite|improve this question











$endgroup$




Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.




Peano Arithmetic has the following two axioms:




  1. $x cdot 0 = 0$

  2. $x cdot y = x cdot (y-1) + x$




So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.



I tried a few things and noticed that one has:



$$2xy = (x+y)^2-x^2-y^2 text{ and } 4xy = (x+y)^2-(x-y)^2$$



Close to $xy$ but still not what I am looking for. And actually this uses subtraction...



Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbb{N}, +, cdot^2)$.






logic




logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 21 hours ago







Sqyuli

















asked 23 hours ago









SqyuliSqyuli

327111




327111








  • 2




    $begingroup$
    2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
    $endgroup$
    – coffeemath
    22 hours ago






  • 4




    $begingroup$
    @MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
    $endgroup$
    – Noah Schweber
    21 hours ago












  • $begingroup$
    @NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
    $endgroup$
    – Sqyuli
    21 hours ago














  • 2




    $begingroup$
    2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
    $endgroup$
    – Mauro ALLEGRANZA
    22 hours ago






  • 2




    $begingroup$
    Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
    $endgroup$
    – coffeemath
    22 hours ago






  • 4




    $begingroup$
    @MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
    $endgroup$
    – Noah Schweber
    21 hours ago












  • $begingroup$
    @NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
    $endgroup$
    – Sqyuli
    21 hours ago








2




2




$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
22 hours ago




$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
22 hours ago




2




2




$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
22 hours ago




$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
22 hours ago




2




2




$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
22 hours ago




$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
22 hours ago




4




4




$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
21 hours ago






$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbb{N}; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
21 hours ago














$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
21 hours ago




$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbb{N}, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
21 hours ago










1 Answer
1






active

oldest

votes


















23












$begingroup$

Per your comment, the precise question you're asking is:




Is multiplication definable in the structure $(mathbb{N}; +,cdot^2)$?




The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.




This is a bit unsatisfying; can we do better?



Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:




Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?




The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write ${partialoverpartial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$




$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.



$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.






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    active

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    23












    $begingroup$

    Per your comment, the precise question you're asking is:




    Is multiplication definable in the structure $(mathbb{N}; +,cdot^2)$?




    The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.




    This is a bit unsatisfying; can we do better?



    Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:




    Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?




    The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write ${partialoverpartial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$




    $^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.



    $^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.






    share|cite|improve this answer











    $endgroup$


















      23












      $begingroup$

      Per your comment, the precise question you're asking is:




      Is multiplication definable in the structure $(mathbb{N}; +,cdot^2)$?




      The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.




      This is a bit unsatisfying; can we do better?



      Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:




      Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?




      The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write ${partialoverpartial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$




      $^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.



      $^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.






      share|cite|improve this answer











      $endgroup$
















        23












        23








        23





        $begingroup$

        Per your comment, the precise question you're asking is:




        Is multiplication definable in the structure $(mathbb{N}; +,cdot^2)$?




        The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.




        This is a bit unsatisfying; can we do better?



        Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:




        Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?




        The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write ${partialoverpartial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$




        $^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.



        $^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.






        share|cite|improve this answer











        $endgroup$



        Per your comment, the precise question you're asking is:




        Is multiplication definable in the structure $(mathbb{N}; +,cdot^2)$?




        The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.




        This is a bit unsatisfying; can we do better?



        Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:




        Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?




        The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write ${partialoverpartial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$




        $^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.



        $^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 21 hours ago

























        answered 21 hours ago









        Noah SchweberNoah Schweber

        126k10151290




        126k10151290






























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