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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!

## #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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