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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 $\lfloor x^2 \rfloor \neq \lfloor x\rfloor ^2$.

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  • $\begingroup$ No matter how you look at this, the "so that" phrasing is sloppy at best. $\endgroup$ – Daniel W. Farlow Jul 19 '16 at 1:04
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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 $\lfloor x^2\rfloor \neq \lfloor x \rfloor ^2$.

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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 $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.
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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 $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.

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  • $\begingroup$ I agree there is a bit of ambiguity about the original statement. I read it like this: "There exists a real number $x$ so that: $x$ is not an integer, $x>2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$." (All I added was a colon.) $\endgroup$ – smcc Jul 19 '16 at 1:15

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