Discussion Problem_ (9)

[duck]

Let P(n,m) be a property about two integers n and m. If we want to disprove the claim that "For every integer n, there exists an integer m such that P(n,m) is true", then we need to prove that:

(a)If P(n,m) is true, then n and m are not integers.
(b)For every integer n, there exists an integer m such that P(n,m) is false.
(c)There exists integers n,m such that P(n,m) is false.
(d)There exists an integer n such that P(n,m) is false for all integers m.
(e)There exists an integer m such that P(n,m) is false for all integers n.
(f)For every integer n, and every integer m, the property P(n,m) is false.
(g)For every integer m, there exists an integer n such that P(n,m) is false.

[Sumit]

d)

[Mauli]

yes
i am also getting d)

[maahi]

how to do this ?

[duck]

Hi Maahi.. :)

Just find the negation of the given statement.

[maahi]

how do u rule out b ?

[Kogo]

what is the difference between (d) and (g)?
is there any difference between "for every" and "for all" ?

[duck]

Hi Kogo.. :)

Option (d)There exists an integer n such that P(n,m) is false for all integers m.
i.e There is some "n" for which "no m" could make P(n,m) true.


Option (g)says for every integer m, there is some integer n such that P(n,m) is false.
i.e Choose any "m" and you would be able to find an interger "n" for which P(n,m) doesn't hold.

[duck]

Hi Maahi.. :)

Consider a simple example,
If I say "All prime numbers are odd number".

How would you negate it?
Find some "prime number which is an even number". Eg: 2.

When you're asked to find the negation, you're required to negate the whole statement.
Negation of "For every" is "there exist".