# mathematical model

• February 23rd 2009, 12:29 PM
sanv
mathematical model
hi,

i have to find a mathematical model of the following sentence, but i am stuck.

$
(\forall x \exists y x$

thanks for any help and advice.
• February 23rd 2009, 12:59 PM
clic-clac
The two last parts of your formula say that the binary relation $<$ is irreflexive and transitive.
A strict order, for instance, is a binary relation which is irreflexive and transitive.

So a possible model of $(\forall x\neg(x is:
$(\{0,1,2,3\},<)$ where $<$ is the restriction to $\{0,1,2,3\}$ of the usual strict order on $\mathbb{N}.$

Now, does that model verify the whole formula? (i.e. does it verify the first part?)
• February 23rd 2009, 01:14 PM
sanv
It doesn't, as it does not hold for the values 0?
• February 23rd 2009, 01:32 PM
clic-clac
Well since $0<1,$ the problem doesn't come from $0$ !
But indeed, that's not a model of the whole formula: there is no element strictly greater than $3$ in that set. (if $<$ is a strict order, that's what says $\forall x\exists y(x : for any $x$ there is a $y$ such that $x)

So using a set whose elements are integers, a model would be...
• February 23rd 2009, 01:55 PM
sanv
Hmm, I do not get that to be honest...

Why is it:
Quote:

there is no element strictly greater than http://www.mathhelpforum.com/math-he...f2a7baf3-1.gif in that set.
?
• February 23rd 2009, 02:01 PM
clic-clac
No problem :)

Since a strict order $<$ satisfies the two last parts of your formula, if we find a set and a strict order on that set which satisfy the first part of the formula, we have a model.

So assume $<$ is a strict order, the first part: $\forall x\exists y(x means that if you take any element in your set, then you can find another element which is strictly greater. A consequence is that your set can't have a greatest element. So for instance, a simple model is $(\mathbb{N},<)$

Of course, there are a lot of other models. Another one quite simple is the set of all primes with the same order.
• February 23rd 2009, 02:01 PM
Plato
Quote:

Originally Posted by sanv
Hmm, I do not get that to be honest...

Look, the first condition simply says that the model cannot be bounded above.
While the second says the model must be transitive.
Use the positive integers.
• February 23rd 2009, 02:30 PM
sanv
Thanks, I think I understand that now.

That would mean e.g. the set of all positive odd numbers will also be a possible model?
• February 23rd 2009, 07:57 PM
aliceinwonderland
Quote:

Originally Posted by sanv
That would mean e.g. the set of all positive odd numbers will also be a possible model?

The set of all positive odd numbers alone is not a model of your sentence,

$
(\forall x \exists y x$

You need to understand some basic concepts on what a structure (interpretation) and a model of the first-order language are.

Basically, a structure $\mathfrak{A}$ for a first-order language assigns the meaning to the parameter such that,

1. What collection of things the universal quantifier symbol ( $\forall$) refers to, and
2. What the other parameters( the predicate and function symbols) denotes.

Now, if a sentence is true in $\mathfrak{A}$, then $\mathfrak{A}$ is a model of the sentence.

For example, consider the following sentence of the first-order language

$\exists x \forall y \neg y < x$.

(There is a natural number such that no natural number is smaller.)

The above sentence is true in the structure $\mathfrak{A}$ = $(\mathbb{N}; <)$ with
$|\mathfrak{A}|$ = the set of natural numbers,
$<^{\mathfrak{A}}$ = the set of ordered pairs <m,n> such that m < n.

Similary, as clic-clac mentioned, $(\mathbb{N}; < )$ can be the model of your sentence,

$
(\forall x \exists y x$