Why constant symbols in a language? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why does $phi,(phiRightarrowpsi)$ not semantically entail $psi$ if $phi$ has a free variable and $psi$ doesn't?satisfiability in a structure implies satisfiability in a substructure?Eventually constant variable assignmentsTerm models in group theoryLanguage structure of $mathbbR$ and $L_mathbbR$Structure/Model for first order languageExtending a language by adding a constant symbolDo we really need constant symbols in first-order theories?Logic: one vs many structures for a given languageIntuition behind a structure of a language in mathematical logic [long read but simple]

Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?

Can Pao de Queijo, and similar foods, be kosher for Passover?

Does surprise arrest existing movement?

Is there a concise way to say "all of the X, one of each"?

Do I really need recursive chmod to restrict access to a folder?

What causes the vertical darker bands in my photo?

Right-skewed distribution with mean equals to mode?

The logistics of corpse disposal

When -s is used with third person singular. What's its use in this context?

What makes black pepper strong or mild?

Are my PIs rude or am I just being too sensitive?

I am not a queen, who am I?

How to recreate this effect in Photoshop?

Why one of virtual NICs called bond0?

How do I keep my slimes from escaping their pens?

Is the address of a local variable a constexpr?

What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?

How to draw this diagram using TikZ package?

Disable hyphenation for an entire paragraph

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

ListPlot join points by nearest neighbor rather than order

What happens to sewage if there is no river near by?

Why is black pepper both grey and black?

Is above average number of years spent on PhD considered a red flag in future academia or industry positions?



Why constant symbols in a language?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why does $phi,(phiRightarrowpsi)$ not semantically entail $psi$ if $phi$ has a free variable and $psi$ doesn't?satisfiability in a structure implies satisfiability in a substructure?Eventually constant variable assignmentsTerm models in group theoryLanguage structure of $mathbbR$ and $L_mathbbR$Structure/Model for first order languageExtending a language by adding a constant symbolDo we really need constant symbols in first-order theories?Logic: one vs many structures for a given languageIntuition behind a structure of a language in mathematical logic [long read but simple]










1












$begingroup$


What is the point of constant symbols in a language?



For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?



I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.



Is there something else, which makes constants 'special'?



Other example: Suppose we have the language $L=c$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrakS_n$ over the set $mathbbZ$, where $c$ gets interpreted by $n$.



Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?



For $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ has the property, that $c+c=c$ for example.



I hope you understand what I am asking for.



I think it boils down to:




Is there a difference between the structure $mathfrakS_n$ as L-structure and $mathfrakS_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).




But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)



Thanks in advance.










share|cite|improve this question









$endgroup$











  • $begingroup$
    In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
    $endgroup$
    – Asaf Karagila
    3 hours ago















1












$begingroup$


What is the point of constant symbols in a language?



For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?



I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.



Is there something else, which makes constants 'special'?



Other example: Suppose we have the language $L=c$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrakS_n$ over the set $mathbbZ$, where $c$ gets interpreted by $n$.



Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?



For $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ has the property, that $c+c=c$ for example.



I hope you understand what I am asking for.



I think it boils down to:




Is there a difference between the structure $mathfrakS_n$ as L-structure and $mathfrakS_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).




But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)



Thanks in advance.










share|cite|improve this question









$endgroup$











  • $begingroup$
    In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
    $endgroup$
    – Asaf Karagila
    3 hours ago













1












1








1





$begingroup$


What is the point of constant symbols in a language?



For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?



I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.



Is there something else, which makes constants 'special'?



Other example: Suppose we have the language $L=c$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrakS_n$ over the set $mathbbZ$, where $c$ gets interpreted by $n$.



Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?



For $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ has the property, that $c+c=c$ for example.



I hope you understand what I am asking for.



I think it boils down to:




Is there a difference between the structure $mathfrakS_n$ as L-structure and $mathfrakS_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).




But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)



Thanks in advance.










share|cite|improve this question









$endgroup$




What is the point of constant symbols in a language?



For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?



I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.



Is there something else, which makes constants 'special'?



Other example: Suppose we have the language $L=c$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrakS_n$ over the set $mathbbZ$, where $c$ gets interpreted by $n$.



Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?



For $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ has the property, that $c+c=c$ for example.



I hope you understand what I am asking for.



I think it boils down to:




Is there a difference between the structure $mathfrakS_n$ as L-structure and $mathfrakS_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).




But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)



Thanks in advance.







logic first-order-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









CornmanCornman

3,74821233




3,74821233











  • $begingroup$
    In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
    $endgroup$
    – Asaf Karagila
    3 hours ago
















  • $begingroup$
    In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
    $endgroup$
    – Asaf Karagila
    3 hours ago















$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila
3 hours ago




$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila
3 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.



For example, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_varnothing$-structures would not.



So while "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
    $endgroup$
    – Cornman
    4 hours ago






  • 1




    $begingroup$
    @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    3 hours ago



















1












$begingroup$

Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.



For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.






share|cite|improve this answer









$endgroup$













    Your Answer








    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%2f3189303%2fwhy-constant-symbols-in-a-language%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









    4












    $begingroup$

    An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.



    For example, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_varnothing$-structures would not.



    So while "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
      $endgroup$
      – Cornman
      4 hours ago






    • 1




      $begingroup$
      @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago
















    4












    $begingroup$

    An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.



    For example, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_varnothing$-structures would not.



    So while "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
      $endgroup$
      – Cornman
      4 hours ago






    • 1




      $begingroup$
      @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago














    4












    4








    4





    $begingroup$

    An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.



    For example, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_varnothing$-structures would not.



    So while "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$



    An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.



    For example, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_varnothing$-structures would not.



    So while "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 4 hours ago









    Clive NewsteadClive Newstead

    52.2k474137




    52.2k474137











    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
      $endgroup$
      – Cornman
      4 hours ago






    • 1




      $begingroup$
      @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago

















    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
      $endgroup$
      – Cornman
      4 hours ago






    • 1




      $begingroup$
      @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago
















    $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
    $endgroup$
    – Cornman
    4 hours ago




    $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ can be isomorphic (for $nneq m$) when you take $L=c,<$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
    $endgroup$
    – Cornman
    4 hours ago




    1




    1




    $begingroup$
    @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    3 hours ago





    $begingroup$
    @Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of $ c, < $-structures between $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    3 hours ago












    1












    $begingroup$

    Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.



    For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.



      For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.



        For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.






        share|cite|improve this answer









        $endgroup$



        Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.



        For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Mark KamsmaMark Kamsma

        3818




        3818



























            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%2f3189303%2fwhy-constant-symbols-in-a-language%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

            Isabella Eugénie Boyer Biographie | Références | Menu de navigationmodifiermodifier le codeComparator to Compute the Relative Value of a U.S. Dollar Amount – 1774 to Present.

            Mpande kaSenzangakhona Biographie | Références | Menu de navigationmodifierMpande kaSenzangakhonavoir la liste des auteursm

            Hornos de Moncalvillo Voir aussi | Menu de navigationmodifierm