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#1 2010-04-21 13:28:42

Amarylli$
Guest

help in Predicate logic! :)

Hi all,

"\forall" x : N • p is false but  "\exists" x : N • p is true.

#2 2010-04-21 13:30:19

Amarylli$
Guest

Re: help in Predicate logic! :)

Amarylli$ wrote:

Hi all,

\forall x : N • p is false but  \exists x : N • p is true.

for this, i need to define a pair of predicates p and q involving free variable x

how can i go abt doin tht?

thanks! big_smile

#3 2010-04-21 13:49:39

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: help in Predicate logic! :)

I'm not familiar with the notation N • p


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2010-04-21 15:42:49

Amarylli$
Guest

Re: help in Predicate logic! :)

well..what i understand is that i have to create expression somewhat like below:

p <-> x+ y > x^2

so x and y here are the free variables.. hence in the question above..i need to create such expression that it satisfies the logic

and N are natural numbers in this context.

• - predicate

#5 2010-04-22 00:34:37

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: help in Predicate logic! :)

I'm going to guess that N • p means p(x).  If this is the case, all you must do is find a statement that is true for some x, but not all.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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