# What is the negation of this?

1. Oct 6, 2011

### 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.

2. Oct 6, 2011

### Staff: Mentor

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

Last edited: Oct 6, 2011
3. Oct 6, 2011

### SammyS

Staff Emeritus
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.

4. Oct 7, 2011

### 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.

5. Oct 7, 2011

6. Oct 7, 2011

### 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

Last edited: Oct 7, 2011
7. Oct 7, 2011

### Staff: Mentor

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

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

Last edited: Oct 7, 2011
8. Oct 7, 2011

### DivisionByZro

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.

9. Oct 7, 2011

### HallsofIvy

Yes, you are right. How silly of me.