Vtyjkyuk

A lateral thinking answer:

0! 0 0 0, because the binary equivalent of 8 is 1000 :)

$0 + 0 + 0 + 0 ~~!!=~ 8$

because

$ !!= $ is an alternative way of writing $ ne $.

Lateral thinking!

$$0+0+substack0\0=0+0+8=8$$

let me try:

$0! Vert 0 - 0!-0! =8$

$10-1-1=8$


$Vert$ is a concatenation operation

$((0!+0!)^(0!+0!))!!$

Evaluation:

$((0!+0!)^(0!+0!))!!$

$rightarrow ((1+1)^(1+1))!!$

$rightarrow (2^2)!!$

$rightarrow 4!! = 8$

It's different:

$,++$

$0;;;0$

$,++$

$0;;;0$

$,++$


An ASCII art $8$ using only four $0$'s and $+$'s.

0 + 0 + 0 + 0 = !8

because

In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.

It's just a matter of perspective ...

0!/0 + 0!/0 = ∞

My reasoning....

0/0 is undefined so we have to first change the 0's into 1's with 0!

(...and why did you write the infinity symbol sideways in your question?)

$$[+!0]+[0]-!0-!0$$

Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.

In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.

$concat(0!,0) - 0! - 0! = 8$

becomes:

$concat(1,0) - 1 - 1 = 8$

and finally:

$10 - 2 = 8$

cool puzzle!

$0 + 0 + 0 + 0 equiv 8$

Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.

Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.

Here is an answer that doesn't use the semi-factorial or any concatenation.

$$0 + 0 - 0! / 0 = (-8)!$$

The left side is $-1/0$ and the right side is $-infty$.

Plugging the expression into Wolfram.

Question limits the symbols, not the operations. So with the symbol + can make the operator ++.

(++(++(++(++(++(++(++0000))))))) = 8

concatenate(0!, 0) - concatenate(0! + 0!) = 8.

Note that 0! = 1

(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8

Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbbZ/mathbb2Z$. (Note I'm trying to avoid writing $[0] = [8]$...)

This could work too:

(0!+0+0)/0 = ∞

Explanation

(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)

$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.

Further explanation:

The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.

If you turn the problem around

As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

This is doubly evident as otherwise how would we use the $mathbbconcatenation$ function without a comma to separate the arguments?

So the solution is:

$0! - mathbbconcatenation(., 0 + 0! + 0!!) = .8$

$1 - mathbbconcatenation(., 0 + 1 + 1) = .8$

$1 - mathbbconcatenation(., 2) = .8$

$1 - .2 = .8$

$00^00 = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.

If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.

Factorial of Zero equals to 1. And 4!!=8.
Then you can put plus and power to end up the story.
((0!+0!)^(0!+0!))!!

Similar to @Vaelus

$0+0+0+0 leq 8$

Explanation

You can get the $leq$ by adding a $-$ inclined on top of the $=$