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I'm sure its a very basic question but I can't understand why the first of these statements is true and the second is false as the professor says.

1: $$\forall x \in \mathbb{R}, \exists y \in \mathbb{R},\, x + y > 0$$

2: $$\exists x \in \mathbb{R}, \forall y ∈ \mathbb{R}, x + y > 0$$

I know the first is true as you write:

$$y = 1 − x \in \mathbb{R}$$

I don't however understand why I cant do the reverse for the second and write:

$$x = 1 − y \in \mathbb{R}$$

He says that the second is false because its negation is true, the negation being:

$$\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, x + y ≤ 0$$

I understand that this is true as you can write:

$$y = −x \in \mathbb{R}$$

I also understand that is a negation is true then the statement should be false but I cant understand how the two original statements differ?

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    1: Take any $x$. Surely, you will find some $y$ such that the sum is positive (anything larger than $-x$). 2: The claim is there there is one special $x$ such that the sum of $x$ with all possible $y$ is positive. That's not right. We'll always find a $y$ such that $x+y$ is not positive, no matter what $x$ we picked. – 2017-01-03
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    In both statements, first you choose $x$, then you choose $y$. So you can't have $x$ depend on $y$ as you try to do with $x=1-y$. – 2017-01-03

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The second statement says that there is some single real number $x$ with the following property: no matter what real number you add to $x$, the sum is positive. This is clearly false, because whatever $x$ is, $-x$ is a real number, and adding $-x$ to $x$ does not yield a positive sum.

The first is a much weaker statement. It does not say that there is a single real number $x$ whose sum with every real number is positive. It says that for each real number $x$ there is at least one real number $y$ whose sum with $x$ is positive, and this $y$ can be different for different values of $x$.

To put it a bit differently, the first statement says that no matter what $x\in\Bbb R$ you choose, I can find a $y\in\Bbb R$ such that $x+y>0$, but my choice of $y$ can depend on what $x$ you give me. The second says that I can find an $x\in\Bbb R$ such that no matter what $y\in\Bbb R$ you choose, the sum $x+y$ will be positive. I have to choose one $x$ first that will ‘work’ no matter what $y$ you then decide to give me, and that’s simply not possible.

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    It can be very useful in situations like this to think about the graph showing which pairs $(x,y)$ satisfy the condition (in this case, the inequality $x + y > 0$). The ones which do are those lying above and to the right of the diagonal line $y = -x$. So it's clear that, like Brian says, *for each* individual $x$, *there is some* value of $y$ which satisfies the condition (just choose $y > -x$); but *there is **not** some* value of $y$ which *for every* individual $x$ satisfies the condition (since such $y$ would have to be larger than $-x$ for all possible values of $x$). – 2017-01-03
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    The bits in italics are just the quantifiers in your question, and can be directly replaced by them to give the formal versions of the statements. What's more, this sort of reasoning applies equally well to situations where $x$ and $y$ are not real numbers but some other type(s) of mathematical object. – 2017-01-03
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You don't need the negative for the second statement. Just note that 2) means that exists some $ x $ that works with every $y$. This is false for $ y =-x $ because in this case we have $ x+(-x)=0$.

Also these examples show that you can't "commute" $\forall $ and $\exists $.

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In plain English, the first statement says, "If you have a real number $x$, then you can always find a real number $y$ such that their sum $x+y$ is positive." This is obviously true: you just take $y$ to be any number bigger than $-x$.

On the other hand, the second statement says, "There is a real number $x$ such that adding any real number $y$ to it will make the sum $x+y$ positive." Is this true? Well the negation is: "If you have a real number $x$, then you can always find a real number $y$ such that their sum $x+y$ is negative or zero." This is true: just take $y$ to be any number less than or equal to $-x$. Since the negation is true, the original statement is false.

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In the first statement you say that for all elements in $\mathbb{R}$, you can find another element such that their sum is positive.

In the second statement you say that there is a specific element in $\mathbb{R}$ such that the sum with any other element of $\mathbb{R}$ is positive.

The second statement doesn't make any sense, because you don't get to choose the $y$, as you're trying to do. In the first statement, given any $x$, you can find $y$.