# Predicates and quantifier ..

• Apr 5th 2010, 10:14 PM
Rozana
Predicates and quantifier ..
Can you help me in this question ..

http://www.mathhelpforum.com/math-he...1&d=1270534445
• Apr 5th 2010, 10:46 PM
Rozana
• Apr 5th 2010, 10:47 PM
Rozana
please if can give me the steps to solve this question ..
• Apr 6th 2010, 12:55 AM
PiperAlpha167
Quote:

Originally Posted by Rozana
Can you help me in this question ..

http://www.mathhelpforum.com/math-he...1&d=1270534445

Apparently this is a multiple choice question, judging from your other posts (which make it clearer).
But still, the wording is a bit puzzling.

By domain of Q, do mean the extension of Q?
If so, then what is the domain under discussion? The set of the first three natural numbers, {0,1,2}?

If so, and we write out the expansion of that existentially quantified formula, it should look like:

~Q(0,0,0) V ~Q(0,0,1) V ~Q(0,0,2)

Then clearly, none of the four will fill the bill.

But of course, I may be completely misreading the question; it won't be the first time.
Sorry, I have to go. I can look back later, but maybe before that someone else will help.
• Apr 6th 2010, 01:06 AM
Rozana
Quote:

In the question 4

I think the answer in none and it should be (f,f,t)

in the statement x=-x is true because 0=(-0)=0

do you agree with me ??
• Apr 6th 2010, 01:12 AM
Rozana
I need some hints to solve question 3 ..

and how can I evaluate ~Q(0,0,1)?
• Apr 6th 2010, 03:50 AM
Rozana
Quote:

Originally Posted by Rozana

In the question 4

I think the answer in none and it should be (f,f,t)

in the statement x=-x is true because 0=(-0)=0

do you agree with me ??

sorry the true answer is (t,f,t)

but another question i don't understand how to solve
• Apr 6th 2010, 07:28 AM
MoeBlee
Quote:

Originally Posted by Rozana
but another question i don't understand how to solve

TFT

Ax x+1 > x
is obviously true, where 'x' ranges over real numbers.

Ax 2x = 3x, is false where 'x' ranges over real numbers, since, e.g. 2*1 does not equal 3*1.

Ex x = -x, is true where 'x' ranges over real numbers, since 0 = -0, so there is at least one real number x for which x = -x is true.

/

a.

You're being asked to translate Ay Q(0 y 0) into a conjunction or disjunction. But the only values allowed for 'y' are 0 or 1.

So for Q(0 y 0) to hold for all allowed values of 'y' is to say that it holds for both 0 and for 1, when 0 or 1, respectively replace 'y'.

So, in these circumstances,

Ay Q(0 y 0)

becomes

Q(0 0 0) /\ Q(0 1 0)

That's all there is to it, really.
• Apr 6th 2010, 07:51 AM
Hello Rozana

$Q(x,y,z)$ is an example of a propositional function; that is, a function that returns a truth value (i.e. the value True or False) that will depend upon the value(s) of any parameter(s) that are supplied to it.

Here, the value of $Q$ will be determined by the values of three parameters, $x, y$ and $z$. The values that $x, y, z$ can take are given in the question - defined as the domain of $Q(x,y,z)$.

Since $x$ can take $3$ different values $(0,1,2)$; and $y$ can take $2$ different values $(0,1)$; and $z$ can also take $2$ values $(0,1)$, there are $3\times2\times2= 12$ possible propositions that can be represented by $Q(x,y,z)$, each one returning the value True or False. They are:
$Q(0,0,0),\; Q(0,0,1),\; Q(0,1,0),\;Q(0,1,1)$

$Q(1,0,0),\; Q(1,0,1),\; Q(1,1,0),\;Q(1,1,1)$

$Q(2,0,0),\; Q(2,0,1),\; Q(2,1,0),\;Q(2,1,1)$

Without knowing the actual details of what $Q$ is, we can't say what value each of these propositions has. However, we can say, that if (as in your original question) $\exists z,\; \neg Q(0,0,z)$ is a true statement, then for one or both of the possible values of $z$ ( $0$ and $1$), $\neg Q(0,0,z)$ is true. In other words, $\neg Q(0,0,0)$ is true or $\neg Q(0,0,1)$ is true (or both). So:
$\exists z,\; \neg Q(0,0,z)$ can be written as $\neg Q(0,0,0) \lor \neg Q(0,0,1)$
****

Question 3 can be answered in the same way:
$\forall y,\;Q(0,y,0)$
means
For all $y$ (that is, $y = 0$ and $y = 1$), $Q(0,y,0)$ is true.
So
$Q(0,0,0)$ is true and $Q(0,1,0)$ is true.
Therefore:
$\forall y,\;Q(0,y,0)$ can be written as $Q(0,0,0) \land Q(0,1,0)$
****

Question 4

Examine the truth of each of the propositions in turn:
$p:\forall x, \; x+1>x$
$p$ is True. It simply says that if we add $1$ to any real number, the answer is greater than the number we started with.
$q:\forall x, \;2x = 3x$
Clearly $q$ is False. To show this, we only need find one value of $x$ for which $2x \ne 3x$. (That's not very hard, is it?)
$r:\exists x,\;x=-x$
$r$ is True. There is a value of $x$ for which $x = -x$; that is $x = 0$.

So the ordered triple $(p,q,r)$ has the truth value $(T, F, T)$.

Does that clear up the problems?