There exists a real number $x$ such that for all real numbers $y,\; 2x+y=7.$
When I take any real number $y,$ I can easily compute $x$ from it.
But my discrete math textbook by Susanna Epp says that the given statement is false.
I'm also stumped on how to negate a multiply quantified statement like this.
The statement is saying that there exists a single number $x$ such that the following equations all hold simultaneously:
And so on and so forth, for every real number $y$. But these equations obviously all induce different values of $x$, so no single $x$ can make them all hold true simultaneously.
The order of quantifiers matters. That is the reason why complex assumptions must always be written in mathematical language: because it allows no ambiguity. English language (or whatever your first language is) should only be used to explain things in a simpler to read way, but when in doubt, always refer to the mathematical formula.
Here in math writing, we would have: $\exists x\in \Bbb{R}, \forall y \in \Bbb R, 2x+y=7$
This is trivially false because that $x$ should have to work with any $y$, for example $y=-1$ and $y=1$, and for any $x$, $2x+1\ne 2x-1$...
What is true is: $\forall y\in \Bbb R, \exists x\in \Bbb R, 2x+y=7$: for any $y$ I can solve $2x+y=7$ to get $x=\dfrac{7-y} 2$
There exists a real number $x$ such that for all real numbers $y,\; 2x+y=7.$
my textbook says that this statement is false.
Let's rephrase the given statement:
There is some real number $b$ for which the following property holds:
whichever real number $y$ is picked, $b=\dfrac12(7-y).$
Proving this statement is just a matter of exhibiting at least one such fixed value of $\boldsymbol b.$ Is this possible?
I'm also stumped on how to negate a multiply quantified statement like this.
Your given statement has structure $$\exists x\, \Big(\forall y\; Q(x,y)\Big).$$
Now, to negate a sentence is to find one with the opposite truth value regardless of the context. For example, the negation of "something is papepipopu" (∃x Px) is "everything is not papepipopu" (∀x ¬Px); conversely, ∃x ¬Px negates ∀x Px.
Thus, the required negation is $$\forall x\, \lnot\Big(\forall y\; Q(x,y)\Big),$$ which is $$\forall x\; \exists y\; \lnot Q(x,y),$$ that is,
Regardless of which real number $x$ is picked,
it is possible to find at least one real number $y$ such that $2x+y\ne7.$
There are already a few good answers under your post, showing how to formally express the negation of that quantified statement.
To help you see how it is false, let's phrase it this way: it says that there is a number $(2x)$ to which you can add any number to get $7$, which is clearly false.
Note that $2x$ is uniquely determined by $x$.
