Can you help me in this question ..

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

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- Apr 5th 2010, 10:14 PMRozanaPredicates and quantifier ..
Can you help me in this question ..

http://www.mathhelpforum.com/math-he...1&d=1270534445 - Apr 5th 2010, 10:46 PMRozana
- Apr 5th 2010, 10:47 PMRozana
please if can give me the steps to solve this question ..

- Apr 6th 2010, 12:55 AMPiperAlpha167
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 AMRozana
- Apr 6th 2010, 01:12 AMRozana
**I need some hints to solve question 3 ..**

and how can I evaluate ~Q(0,0,1)? - Apr 6th 2010, 03:50 AMRozana
- Apr 6th 2010, 07:28 AMMoeBlee
The answer to #4 is:

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.

/

The answer to #3 is:

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 AMGrandad
Hello Rozana

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

Since $\displaystyle x$ can take $\displaystyle 3$ different values $\displaystyle (0,1,2) $; and $\displaystyle y$ can take $\displaystyle 2$ different values $\displaystyle (0,1)$; and $\displaystyle z$ can also take $\displaystyle 2$ values $\displaystyle (0,1)$, there are $\displaystyle 3\times2\times2= 12$ possible propositions that can be represented by $\displaystyle Q(x,y,z)$, each one returning the value True or False. They are:$\displaystyle Q(0,0,0),\; Q(0,0,1),\; Q(0,1,0),\;Q(0,1,1)$Without knowing the actual details of what $\displaystyle Q$ is, we can't say what value each of these propositions has. However, we can say, that if (as in your original question) $\displaystyle \exists z,\; \neg Q(0,0,z)$ is a true statement, then for one or both of the possible values of $\displaystyle z$ ($\displaystyle 0$ and $\displaystyle 1$), $\displaystyle \neg Q(0,0,z)$ is true. In other words, $\displaystyle \neg Q(0,0,0)$ is true or $\displaystyle \neg Q(0,0,1)$ is true (or both). So:

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

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

$\displaystyle \exists z,\; \neg Q(0,0,z)$ can be written as $\displaystyle \neg Q(0,0,0) \lor \neg Q(0,0,1)$****

Question 3 can be answered in the same way:$\displaystyle \forall y,\;Q(0,y,0)$meansFor all $\displaystyle y$ (that is, $\displaystyle y = 0$ and $\displaystyle y = 1$), $\displaystyle Q(0,y,0)$ is true.So

$\displaystyle Q(0,0,0)$ is true and $\displaystyle Q(0,1,0)$ is true.Therefore:

$\displaystyle \forall y,\;Q(0,y,0)$ can be written as $\displaystyle Q(0,0,0) \land Q(0,1,0)$****

Question 4

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

$\displaystyle r:\exists x,\;x=-x$$\displaystyle r$ is True. There is a value of $\displaystyle x$ for which $\displaystyle x = -x$; that is $\displaystyle x = 0$.

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

Does that clear up the problems?

Grandad - Apr 13th 2010, 01:01 AMRozana
Thank you for all especially Grandad..

very clear explanation ..