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I am afraid to even ask this question, but this is something that was asked over a breakfast table by a friend of mine.

How do I represent a number, say 4, by using just zeros as a numeral in tandem with any function or mathematical operator(s)?

Now one way to do this is by using a factorial, since $0!=1$: $$0! + 0! + 0! + 0!$$ Another way uses $\cos 0=1$: $$\cos 0+\cos 0+\cos 0+\cos 0$$ Also, the topmost $1$ on Pascal’s Triangle is the zero$^{th}$ row, zero$^{th}$ entry, i.e., $0C$$_0$ = $1$.

The Partition function , P(n) is , by convention , defined as $1$ for $n=0$, i.e. , $P(0) = 1$.

Is there any other way I could possibly achieve this?I would be more interested to know the functions which when applied on zero produce non-zero results , like the ones mentioned earlier in the question, rather than semantic manipulations or visual representations.

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closed as off-topic by user21820, Matthew Towers, user133281, Jean-Claude Arbaut, Ali Caglayan Oct 16 '16 at 14:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – user21820, Matthew Towers, user133281, Jean-Claude Arbaut, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ Why the hell would you and your friend want this? $\endgroup$ – Billy Rubina Oct 16 '16 at 1:26
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    $\begingroup$ It started off as more of a philosophical debate that zero holds up infinity and then took rather this silly turn. Besides imagine designing a scientific calculator with just a 0 digit-button. $\endgroup$ – naveen dankal Oct 16 '16 at 1:31
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    $\begingroup$ Although not "$0$", I like $|\mathcal{P}(\mathcal{P}(\mathcal{P}(\emptyset)))|$ $\endgroup$ – Bonnaduck Oct 16 '16 at 4:42
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    $\begingroup$ The problem with your question is the meaning of "just zeros". None of your examples consists of ""just zeros". So you would have to define what you mean. $\endgroup$ – Martin Argerami Oct 16 '16 at 14:17
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    $\begingroup$ @naveendankal, you can click on the "reopen" button just below your question in order to have it reopened. $\endgroup$ – Mikhail Katz Oct 19 '16 at 13:40

11 Answers 11

23
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Peano says: with the successor function.

$$4=S(S(S(S(0))))$$

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15
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You could try using unary with $0$. So $4_{10}$ is $0000$ in unary.


Also, this usually works:

       0
     0 0
   0   0
 0     0
0000000000
       0
       0
       0
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  • 1
    $\begingroup$ Your examples separate the symbol $0$ from its standard meaning: the value of zero. In a similar way we could introduce bogo numerals (as opposed to Arabic numerals) with the symbol $0$ meaning four, then just write $0$. $\endgroup$ – Kamil Maciorowski Oct 16 '16 at 0:19
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    $\begingroup$ OP did ask to use only zero, so even if this $0$ isn't the same as the one with the usual arithmetic properties, it wouldn't take anyone more than couple of guesses to tell what $0000$ means. I think. Also, I took the idea of only zero literally. So the first example is only zeros representing 4. $\endgroup$ – Cehhiro Oct 16 '16 at 0:22
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    $\begingroup$ OK, your ways are imaginative and they as well may be what OP wants. I'm just pointing out they are semantic tricks more than math tricks. My feeling is it should be explicitly noted here on Mathematics Stack Exchange. $\endgroup$ – Kamil Maciorowski Oct 16 '16 at 0:37
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    $\begingroup$ @ O. Von Seckendorff : your answer is creative but I am more interested in the Mathematical representation , maybe some function that translates 0 into 1 or something like the Cos and factorial. $\endgroup$ – naveen dankal Oct 16 '16 at 0:48
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    $\begingroup$ If "four" is represented by 0000, then "one" is represented as '0'. Which makes it impossible to distinguish between "zero" and "one" as both are written as '0'. Unless one would like to write zero as '' (empty). However, inventing a symbol for zero usually is considered as a milestone, and this would revert it. TL;DR -> I think that 4 should be written down as '00000' (five zeros). $\endgroup$ – Sjoerd Oct 16 '16 at 3:05
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$$\Large{\vert\{}\normalsize{ 0,\{0\},\{\{0\}\},\large{\{}\normalsize{\{\{0\}\}}}\large{\}}\Large{\} \vert}$$

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  • $\begingroup$ Is that what they call a family tree? $\endgroup$ – Mateen Ulhaq Oct 16 '16 at 7:00
  • $\begingroup$ @MateenUlhaq I've never heard that phrase in math. $\endgroup$ – GFauxPas Oct 16 '16 at 16:16
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Well $ 4 = (0!+0!)^{(0!+0!)} $

And we could take this notation: $$x_n=\sum_{n=0!}^n 0!$$ as we are not explicitly using the number $n$ , but rather this is just short hand.

So now we may express $e$ in terms of $0$ as $$lim_{n\rightarrow \infty}\displaystyle (0!+\frac{0!}{x_n})^{x_n} $$

or if you prefer:

$$ e = (0!+0!)+\cfrac{0!}{0!+\cfrac{0!}{(0!+0!)+\cfrac{0!}{0!+\cfrac{0!}{0!+ \cfrac{0!}{(0!+0!)^{(0!+0!)}+\cdots }}}} } $$

but of course this is all rather ridiculous!

Of course assigning $x_n$ in this way, is essentially the same way as assigning a number such as $4$ or $300$ -- it is arbitrary, which opens all such thought on the proper existence of numbers ... but this is starting to dig far deeper into your silly question!!!

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    $\begingroup$ That's exactly what has been troubling me too with regards to this question. It sounds both stupid and silly and deep and profound at the same time, maybe less mathematically but more philosophically. It leads to a the vulnerability of the existence of numbers and raises a slight doubt whether they are essential or inconsequential. I am glad that you added that last paragraph to your answer because this is exactly what my fear was when I stated this question. An extra up for that para!! $\endgroup$ – naveen dankal Oct 16 '16 at 5:28
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    $\begingroup$ Thus the unreasonable effectiveness of mathematics follows! $\endgroup$ – Hugh Entwistle Oct 16 '16 at 5:32
4
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A solution only using "$0$" once is

$$\large\sqrt{\left\lceil\exp(\exp(\exp(0)))\right\rceil}=4$$

But I admit, that it is not very aesthetic.

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  • $\begingroup$ Use Mathjax with HUGE. It could be nice ! $\endgroup$ – Claude Leibovici Oct 15 '16 at 18:23
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    $\begingroup$ @ClaudeLeibovici Decided to use \large instead, feeling that \huge is overkill. $\endgroup$ – Simply Beautiful Art Oct 16 '16 at 0:54
  • $\begingroup$ It might be a good idea to use $\exp^n$ or $\exp \circ \dots\circ\exp$ to represent successive iterations of $\exp$, and $\lceil\cdot\rceil$ for ceiling. $\endgroup$ – Henricus V. Oct 16 '16 at 6:00
  • $\begingroup$ @Henry W $1)$ I do not know how to produce the ceil braces. (If you know it, you can edit the answer) $2)$ "$exp^n(x)$ would contain other digits. We are only allowed to use "0" as a digit here. $\endgroup$ – Peter Oct 16 '16 at 8:46
3
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Arguably the von-Neumann construction does exactly that: It identifies each natural number with the set of natural numbers below (where $0$ is considered a natural number).

In that construction, of course $1=\{0\}$: The only natural number below $1$ is $0$. Obviously this only uses $0$ (and the set construction, but then, you were obviously willing to use extra constructions as long as they are not themselves $0$).

Furthermore, $2=\{0,1\}$. Now, this obviously uses $1$ besides of $0$. But we've just seen that $1=\{0\}$. So $2=\{0,\{0\}\}$.

Then $3=\{0,1,2\}$. Using the previous results, we get $3=\{0,\{0\},\{0,\{0\}\}\}$.

And $4=\{0,1,2,3\} = \{0,\{0\},\{0,\{0\}\},\{0,\{0\},\{0,\{0\}\}\}\}$

So clearly $4$ (and any other natural number) can be represented using only zero.

But it's even better: I omitted the question: What is $0$? Well, $0$ is the set of natural numbers less than $0$. But there are no natural numebrs less than $0$. Therefore $0$ is the empty set, $0=\emptyset$.

So actually the von-Neumann construction creates all natural numbers literally out of nothing. Given the number $n$, you simply get the next natural number by appending the number $n$ as new element to itself: $n+1=n\cup\{n\}$

So $0$ is the empty set. So you get the next number, $1$, by appending $0$ as element; that is, you've got $1=\{0\}$. To get from $1$ to $2$, you append the element $1$, so $2=\{0,1\}$, and so on.

Moreover, you mention that the question arose from a discussion about infinity. And it turns out that the von-Neumann construction is exactly the way to get to infinity (and beyond!). That's because you can now ask, what is the set of all natural numbers. Well, if we want to consider it a number, it must be a number large enough that all natural numbers. That is, it is infinite. This specific number is commonly called $\omega$.

So it is infinity, right? Well, wait. If it is a number in its own right, then we can make again a set of numbers containing all natural numbers and $\omega$. This set by construction contains all numbers $\le\omega$, so it cannot have any "holes". But then, according to the rule, it should be another number, that is also larger than every natural number, and furthermore larger than $\omega$! Indeed, since we've just appended the element $\omega$ to $\omega$, so what we get should be $\omega+1$. And we can repeat that construction to get $\omega+2$, $\omega+3$, … So we get actually infinitely many additional infinities.

The numbers whose construction I just describes (well, strictly speaking, I only described the construction of the lowest ones) is called the ordinal numbers.

Now you may ask: What does the complete set of ordinal numbers look like? Well, here comes the next surprise: The set of ordinal numbers does not exist! That's seen directly from the construction. Remember how we arrived at $\omega$? Well, we had the set of all natural numbers, and we decided that the very fact that we have that set makes it a new number, that's larger than any of them.

Now imagine we'd have the set of ordinal numbers. Then by the very construction, that set itself should be an ordinal number larger than any ordinal number. But then it cannot be in that set, and therefore that set cannot be the set of ordinal numbers, as we've just found an ordinal number that's not in that set, namely the very set itself!

So you see, we've started with $0$ (actually, with literally nothing), and ended up with so many numbers that they don't even fit in a set!

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My guess is $0^0 =1$ thus $0^0 + 0^0+0^0+0^0 =4$.

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  • 11
    $\begingroup$ $0^0$ is undefined $\endgroup$ – Bernard Masse Oct 15 '16 at 17:49
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    $\begingroup$ I like this because it uses only the zero symbol. While an analyst might feel $0^0$ is undefined in some contexts, a combinatorialist will likely be happy with a value of one assigned. $\endgroup$ – hardmath Oct 15 '16 at 17:56
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    $\begingroup$ Isn't $0^0 = \prod_\text{false}0 = 1$? What else would it mean? $\endgroup$ – GFauxPas Oct 16 '16 at 3:35
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    $\begingroup$ @Peter That limit is definitely not why $0^0$ is conventionally defined to be $1$. The reason is twofold: there is precisely one function $\varnothing\to\varnothing$, and $0^0=1$ allows us to plug $x=0$ into $\sum_n a_n x^n$ reasonably. $\endgroup$ – arctic tern Oct 16 '16 at 5:17
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    $\begingroup$ $0^0=1$ when the exponent is the integer 0. $0^0$ is undefined when the exponent is the real 0. $\endgroup$ – Meni Rosenfeld Oct 16 '16 at 7:47
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Every real number admits a binary expansion. For every occurrence of $1$ in the expansion, write $1=0^0$.

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    $\begingroup$ 1. You've mis-spelled "write". 2. $0^0$ is undefined. Perhaps you meant $1=\lim\limits_{x\to0}x^x$. $\endgroup$ – barak manos Oct 15 '16 at 17:45
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    $\begingroup$ @barakmanos Thanks for pointing out the spelling error. Your other point is wrong. $0^0$ is $1$. This is the standard convention. $\endgroup$ – MathematicsStudent1122 Oct 15 '16 at 17:46
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    $\begingroup$ I've never heard of a unified "standard convention". What I have seen is that in particular settings where such undefined expressions arise, they are defined situationally as the limit of some expression as the parameter in question approaches $0$. $\endgroup$ – AJY Oct 15 '16 at 18:26
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    $\begingroup$ @AJY Do you accept the binomial theorem? Does it hold for $(x+0)^n$? This isn't really up for debate. $\endgroup$ – MathematicsStudent1122 Oct 15 '16 at 18:56
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    $\begingroup$ @AJY Or how about the power series of $\exp$, $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ at $x=0$. Is that also disputed? $\endgroup$ – MathematicsStudent1122 Oct 15 '16 at 18:57
0
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Please take it lightly ,

Add zero to a number let say x so now the new number is x0 , x0-x = yz , now u= sum y+z , now sqrt u + u to the power 0 will be 4. Explanation u will always be 9 so sqrt 9 will be 3 and power 0 will be one so 3+1=4

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-1
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In some languages...

i = 0, i++, i++, i++, i++, i

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-1
$\begingroup$
0 = 0
1 = 0++
2 = (0++)++
3 = ((0++)++)++

and so on. This uses the syntax of C-like languages.

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  • 2
    $\begingroup$ Maybe you could explain your notation? $\endgroup$ – snulty Oct 16 '16 at 11:47

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