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, $x &gt; 2016$, 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, $x&gt;2016$ and  $\lfloor x^2\rfloor = \lfloor x\rfloor ^2$.

I'm tempted to say for all $x$ so that $x$ is an integer, $x&gt; 2016$, but $\lfloor x^2 \rfloor \neq \lfloor x\rfloor ^2$.
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:

$$\begin{align*}
\neg \left[ \exists x \in \mathbb{R} : p(x) \wedge q(x) \wedge r(x) \right]&amp;\iff \forall x \in \mathbb{R} : \neg \left[p(x) \wedge q(x) \wedge r(x)\right] \\
&amp;\iff \forall x \in \mathbb{R} : \neg p(x) \vee \neg q(x) \vee \neg r(x)\\
\end{align*}$$

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 $\lfloor x^2\rfloor \neq \lfloor x \rfloor ^2$.
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 $x&gt;2016$ and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$.


The above statement becomes easy to express in a more symbolically formal way:
$$
(\exists x\in\mathbb{R}\setminus\mathbb{Z})(x&gt;2016\land\lfloor x^2 \rfloor = \lfloor x \rfloor^2).
$$
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 $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.