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Well, all of us know the real numbers.

$-1000<-122.34<-\pi^e<-\gamma<-0.00000001<-0.0000000000000000001 \in \mathbb{R}$

But if we continue this way, looks like the largest negative number will be

$-0.0000\dots1$, but this number, as defined by Cantor, is $0$.

So can we say the largest negative number is $0$?

If not, why not?


I agree $0$ is not a negative number. How about the question, what's the largest negative number?

Is there no answer to this question? If the answer is that there is no largest negative number, how can we prove this mathematically?

share|cite|improve this question
There is no largest negative number. – Alex Becker Apr 13 '12 at 23:25
This is an example of the fact that with a converging sequence where each element shares a particular property, the limit need not have that property. – Henry Apr 13 '12 at 23:45
It is the negative of the smallest positive number. – André Nicolas Apr 14 '12 at 0:44
Well, I dont know @AsafKaragila opinion. But I think this in an interesting question. If you say that something does not exists in math, you need to prove. Asaf proved and this ended this question. This looks similar to the continum hypothesis to me, for example. But I can survive with the down votes =) – GarouDan Jun 17 '15 at 21:46
@ GaroDan. In the same vein : What is the largest real number smaller than 2016 ? - – JJacquelin Dec 26 '15 at 18:15
up vote 4 down vote accepted

In the real numbers (or the rationals) there is no "largest negative number", much like there is no "smallest positive number".

The order of the real numbers is dense, if we define the negative numbers as $\{x\in\mathbb R\mid x<0\}$ then between every $x$ in that set and $0$ there is another number, so if $x$ is a negative number there is some $y$ which is negative and is slightly larger than $x$. For example, $y=\frac{x}2$.

We have a philosophical conundrum on our hands as well, why should every partial order attain a maximal element? (Indeed, there is no reason!)

share|cite|improve this answer
I agree Asaf, I was thinking exactly about that, your x and y example. And I'm searching for a proof, so can we proof this by contradiction? – GarouDan Apr 13 '12 at 23:38
@GarouDan: No need for contradiction, just take $y=\frac{x}2$. – Asaf Karagila Apr 13 '12 at 23:40
Thx Asaf, I think this ends the enquire.^^ – GarouDan Apr 13 '12 at 23:44

protected by Asaf Karagila Dec 26 '15 at 18:14

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