Disprove why 0 ∉ Z

Question

Disprove this

A value x is said to be an integer when

floor(x) = x, where x ∈ ℝ

floor(x)/x = 1

Therefore

floor(x)/x ∈ Z, where x ∈ ℝ

And since 0 ∈ ℝ

From the definition of an integer,

floor(x)/x ∈ Z, where x ∈ ℝ

if 0 ∈ Z,

floor(0) = 0

Then

floor(0)/0 = 1 ∉ Z

Why do you care about $\lfloor x\rfloor/x\in\mathbb Z$ when $\lfloor x\rfloor/x=1$ is the required property ? — , Dec 11 '15 at 08:20
because you're making the mistake of assuming floor(0) to be 0, but that assumes 0 is an integer. This is really to disprove that floor(0) is actually 0. — Andrew Kor, Dec 11 '15 at 08:23
Where are you getting your definition of "integer"? I've never seen anyone require integers to satisfy $\lfloor x\rfloor/x=1$. — Noah Schweber, Dec 11 '15 at 08:26

score 3 · Answer 1 · answered Dec 11 '15 at 08:24

3

Your definition of "integer" is wrong - we do not require $\lfloor x\rfloor/x$ to equal $1$.

answered Dec 11 '15 at 08:24

Noah Schweber

245,398

1

Why the downvote? This is correct. – Noah Schweber Dec 11 '15 at 08:24
If you add $x=0\lor$, this is not such a wrong definition of an integer. – Dec 11 '15 at 08:26
@YvesDaoust Yes, but then it's a very different definition (and in fact just a consequence of "$\lfloor x\rfloor=x$," so it's redundant). – Noah Schweber Dec 11 '15 at 08:26
How is this right? it's simple algebra. floor(x)/x = x/x which is floor(x)/x = 1 – Andrew Kor Dec 11 '15 at 08:27
@AjeetKljh $x\over x$ is not always $1$ - if $x$ is zero, $x\over x$ is undefined. – Noah Schweber Dec 11 '15 at 08:27
This definition (or property) is useful in the context of this proof, as the proof is all about the ratio $\lfloor x\rfloor/x$. So there is nothing extraterrestrial about it. [Of course well all know that the definition excludes $0$, not counting the fact that the entire proof is wrong.] – Dec 11 '15 at 08:29
@MorganRodgers The OP does not strike me as a troll, necessarily - they have asked other questions which are definitely not nonsense. I think this falls into the "basic error" category, instead. – Noah Schweber Dec 11 '15 at 08:31
@AjeetKljh What does that have to do with anything? Undefined does not equal $1$, and we never assume that it does - how on earth are you concluding that integers must satisfy $\lfloor x\rfloor/x=1$ (unless you're starting with this as part of the definition of integer, in which case where are you getting this definition)? – Noah Schweber Dec 11 '15 at 08:33
@NoahSchweber what exactly is the definition of "undefined" anyway? – Andrew Kor Dec 11 '15 at 08:43
@AjeetKljh I think if you do not understand the definition of "undefined," you should ask a separate question about that (and your specific confusions) - I don't think the comments are a good place to hash this out. Briefly, though, not every mathematical expression is automatically assumed to be meaningful - think of "$x\over y$" as shorthand for "the unique number which, when multiplied by $y$, gives $x$." Such a number may not exist, or may not be unique, in which case that expression is undefined. – Noah Schweber Dec 11 '15 at 08:47

score 0 · Answer 2 · answered Dec 11 '15 at 08:23

0

There are two big flaws:

$\frac00$ is not defined and there's nothing you can conclude about it.
there is no justification of your final $\notin$. On the opposite, $1\in\mathbb Z$.

answered Dec 11 '15 at 08:23

I think the OP was actually using the fact $0\over 0$ is undefined to derive ${0\over 0}\not\in \mathbb{Z}$ and hence $1\not\in\mathbb{Z}$ and hence a contradiction. The only real problem in this proof, is, as stated in other comments, the definition of integer is wrong. If the floor definition were the right definition of integer, then it would actually be a valid proof by contradiction. – cr001 Dec 11 '15 at 08:41
$\lfloor x\rfloor=x$ is strictly equivalent to $x=0\lor\lfloor x\rfloor/x=1$, so there's nothing wrong with the alternative definition (in this fixed version; such a definition being "unfamiliar" isn't a valid argument to refute it). The real issue in the proof is the meaning of $0/0\notin\mathbb Z \land0/0=1\implies 1\notin\mathbb Z$ (?!) – Dec 11 '15 at 08:58
Indeed, if you fix the definition then there is no problem at all. The problem in OP's proof is he uses the definition "If $x$ is an integer then $\lfloor x \rfloor \over x$$=1$" – cr001 Dec 11 '15 at 09:01

Disprove why 0 ∉ Z

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