Multiplication via squaring and additionCan multiplication be defined in terms of divisibility?Is Hilbert's...
Canadian citizen, on US no-fly list. What can I do in order to be allowed on flights which go through US airspace?
What type of investment is best suited for a 1-year investment on a down payment?
Sometimes a banana is just a banana
Plagiarism of code by other PhD student
What are the issues with an additional (limited) concentration slot instead of Bladesong?
If nine coins are tossed, what is the probability that the number of heads is even?
lead or lag function to get several values, not just the nth
How do you say "powers of ten"?
What is this waxed root vegetable?
How can I handle a player who pre-plans arguments about my rulings on RAW?
How to substitute values from a list into a function?
What does @RC mean in SSDT SQL Server Unit Testing?
Which sins are beyond punishment?
Source for Cremation Specifically Not Jewish
At what level can a party fight a mimic?
What should the omniscient narrator call a character?
When was drinking water recognized as crucial in marathon running?
Second-rate spelling
A right or the right?
Get length of the longest sequence of numbers with the same sign
Where is the fallacy here?
How can I create a Table like this in Latex?
How to make a *empty* field behaves like a *null* field when it comes to standard values?
Can throughput exceed the bandwidth of a network
Multiplication via squaring and addition
Can multiplication be defined in terms of divisibility?Is Hilbert's second problem about the real numbers or the natural numbers?How is exponentiation defined in Peano arithmetic?Where do I go wrong with Presburger “multiplication”?Why are addition and multiplication included in the signature of first-order Peano arithmetic?The (un)decidability of Robinson-Arithmetic-without-Multiplication?How can addition be non-recursive?Meaning of the word “axiom”How much of first order statements can we derive purely from the definitions in arithmetic?Is the definability axiom schema consistent with ZF?
$begingroup$
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:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
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
asked yesterday
SqyuliSqyuli
327111
$endgroup$
|
show 4 more comments
$begingroup$
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:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
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
asked yesterday
SqyuliSqyuli
327111
$endgroup$
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
yesterday
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
yesterday
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
yesterday
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
yesterday
$begingroup$
@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
yesterday
|
show 4 more comments
$begingroup$
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:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
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
asked yesterday
SqyuliSqyuli
327111
$endgroup$
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:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
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
asked yesterday
SqyuliSqyuli
327111
asked yesterday
SqyuliSqyuli
327111
edited yesterday
Sqyuli
asked yesterday
SqyuliSqyuli
327111
asked yesterday
SqyuliSqyuli
327111
asked yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
$begingroup$
@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
yesterday
|
show 4 more comments
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
yesterday
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
yesterday
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
yesterday
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
yesterday
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
yesterday
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
yesterday
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered yesterday
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3134965%2fmultiplication-via-squaring-and-addition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered yesterday
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
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.
answered yesterday
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
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.
answered yesterday
Noah SchweberNoah Schweber
126k10151290
$endgroup$
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.
answered yesterday
Noah SchweberNoah Schweber
126k10151290
edited yesterday
answered yesterday
Noah SchweberNoah Schweber
126k10151290
answered yesterday
Noah SchweberNoah Schweber
126k10151290
answered yesterday
Noah SchweberNoah Schweber
126k10151290
126k10151290
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3134965%2fmultiplication-via-squaring-and-addition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
yesterday
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
yesterday
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
yesterday
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
yesterday
$begingroup$
@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
yesterday