Negating the statement $\exists x \in \Bbb R$ so that $x$ is not an integer, $x > 2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$ 
There exists a real number $x$ so that $x$ is not an integer, $x > 2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$.

I would like clarification on how to negate this. My idea of negation is for all real numbers $x$, so that $x$ is not an integer, $x>2016$ and  $\lfloor x^2\rfloor = \lfloor x\rfloor ^2$.
I'm tempted to say for all $x$ so that $x$ is an integer, $x> 2016$, but $\lfloor x^2 \rfloor \neq \lfloor x\rfloor ^2$.
 A: Sometimes, representing the situation symbolically can help. Let's define


*

*$p(x)$ means "x is not an integer"

*$q(x)$ means "x is greater than 2016"

*$r(x)$ means "$\lfloor x^2\rfloor = \lfloor x\rfloor^2$"


Now your original statement is
$$\exists x \in \mathbb{R} : p(x) \wedge q(x) \wedge r(x).$$
We can get a simplified version of its negation using the rules of boolean algebra as follows:
$$\begin{align*}
\neg \left[ \exists x \in \mathbb{R} : p(x) \wedge q(x) \wedge r(x) \right]&\iff \forall x \in \mathbb{R} : \neg \left[p(x) \wedge q(x) \wedge r(x)\right] \\
&\iff \forall x \in \mathbb{R} : \neg p(x) \vee \neg q(x) \vee \neg r(x)\\
\end{align*}$$
To translate back into more natural language:

Every real number $x$ has at least one of the following three properties, possibly more: it is an integer, it is less than or equal to 2016, and/or $\lfloor x^2\rfloor \neq \lfloor x \rfloor ^2$.

A: Statement:
"There exists a real number $x$ so that $x$ is not an integer, $x>2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$."
Its negation:
"For all real numbers $x$, $x$ is an integer or $x\leq 2016$ or $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$."

I used the following facts:
The negation of the statement "there exists an $s\in S$ such that $C$" is the statement "for all $s\in S$, not $C$".
The negation of the statement "$A$ and $B$" is the statement "(not $A$) or (not $B$)". 

Going through it step by step:


*

*Not[ There exists a real number $x$, so that: ($x$ is not an integer) and ($x>2016$), and ($\lfloor x^2 \rfloor = \lfloor x \rfloor^2$).]

*For all real numbers $x$, Not[ ($x$ is not an integer) and ($x>2016$) and ($\lfloor x^2 \rfloor = \lfloor x \rfloor^2$)].

*For all real numbers $x$, Not($x$ is not an integer) or Not($x> 2016$) or Not($\lfloor x^2 \rfloor = \lfloor x \rfloor^2$).

*For all real numbers $x$, $x$ is an integer or $x\leq 2016$ or $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.

A: I read this question somewhat differently than the other two answerers. The phrasing "so that" in the original statement is sloppy at best. Usually, you would have something like "$\exists x\in\mathbb{R}$, where $x$ is not an integer, ..." In fact, in a reasonably decent text, I think you would more likely come across something like "$\exists x\in\mathbb{R}\setminus\mathbb{Z}$." All this to say, I do not think that "$x$ is an integer" should be treated as its own statement or claim. A better phrasing of your original statement, I believe, would be something along the lines of the following:

There exists a noninteger real $x$ for which $x>2016$ and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$.

The above statement becomes easy to express in a more symbolically formal way:
$$
(\exists x\in\mathbb{R}\setminus\mathbb{Z})(x>2016\land\lfloor x^2 \rfloor = \lfloor x \rfloor^2).
$$
The symbolic negation then becomes easy:
$$
(\forall x\in\mathbb{R}\setminus\mathbb{Z})(x\leq2016\lor\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2).
$$
Finally, the linguistic equivalent, no doubt more meaningful, becomes the following:

For all noninteger real $x$, either $x\leq2016$ or $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.

