What is the negation of this?

[jumbogala]

1. The problem statement, all variables and given/known data
For all integers y, there is an integer x such that x^2 + x = y.


2. Relevant equations



3. The attempt at a solution
Is it there exists an integer y such that for all integers x, x^2 + x = y

OR

There exists an integer y such that for all integers x, x^2 + x DOES NOT EQUAL y?

I believe it is the second one but I'm not sure. I'm not trying to actually prove this.

[NascentOxygen]

EDIT: I mistook your answer as a restatement of the question.

[SammyS]

I smart a** answer is "It is not the case that, for all integers y, there is an integer x such that x^2 + x = y."

However, your preferred choice, the second one, is correct.

If your first one is true, it could still be the case that the original statement is also true.

[HallsofIvy]

Generally, the negation of "if p then q" is "q and not p" (q is true and p is not true).

Your original statement, "For all integers y, there is an integer x such that x^2 + x = y" is the same as "if y is an integer, then there is an integer, x, such that x^2+ x= y" so its negative would be "there exist an integer, y, such that for no integer, x, is it true that x^2+ x= y", which is the same as your second statement.

[ehild]

[QUOTE=HallsofIvy;3544835]Generally, the negation of "if p then q" is "q and not p" (q is true and p is not true).
/QUOTE]

Is not it "p and not q instead"?

ehild

[DivisionByZro]

An easy way to tackle these types of problems is to put it in quantifier notation:

"For all integers y, there is an integer x such that x^2 + x = y."

Becomes:


(\forall y \in \mathbb{Z})( \exists x\in\mathbb{Z})\backepsilon (x^2+x=y)


Of which the negation is:


(\exists y \in \mathbb{Z})( \forall x\in\mathbb{Z})\backepsilon (x^2+x\neq y)


Note: I use /backepsilon for my "such that"; if there is a more common notation for it, I would love to know. :D

[NascentOxygen]

Note: I use /backepsilon for my "such that"; if there is a more common notation for it, I would love to know. :D
Have you seen anyone else using that? :smile:
I always say the colon ":" as "such that". How else could it be pronounced? Okay, "where", also.


(\exists y \in \mathbb{Z})( \forall x\in\mathbb{Z}) : (x^2+x\neq y)

[DivisionByZro]

Have you seen anyone else using that? :smile:
I always say the colon ":" as "such that". How else could it be pronounced? Okay, "where", also.


(\exists y \in \mathbb{Z})( \forall x\in\mathbb{Z}) : (x^2+x\neq y)


Ah yes, I should probably use either " | " or " : "; from set-builder notation. This makes it less confusing since epsilon already means something different.
And for your question, I've actually seen some people using a backwards epsilon for their "such that". It is odd to see.

[HallsofIvy]

Generally, the negation of "if p then q" is "q and not p" (q is true and p is not true).


Is not it "p and not q instead"?

ehild
Yes, you are right. How silly of me.