Representing any number using only 0

Question

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.

Answer 1

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

Answer 2

Every real number admits a binary expansion. For every occurrence of $1$ in the expansion, write $1=0^0$.

Answer 3

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.

Answer 4

Peano says: with the successor function.

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

Answer 5

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

Answer 6

$$\Large{\vert\{}\normalsize{ 0,\{0\},\{\{0\}\},\large{\{}\normalsize{\{\{0\}\}}}\large{\}}\Large{\} \vert}$$

Answer 7

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!!!

Answer 8

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!

Answer 9

Changing my answer...
Per the question, use only zeroes $$\frac{\color{blue}{zero} + \color{blue}{zero} + \color{blue}{zero} + \color{blue}{zero}}{\color{blue}{zero}}$$ Note: $zero\ne0$ $\color{lightgray}{:-)}$

Answer 10

0 = 0
1 = 0++
2 = (0++)++
3 = ((0++)++)++

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

Answer 11

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