Only using: ! ; ( ; ) ; / and digits: 4 ; 5 Can you make an

>>8235027
well i didn't use 5 at least
4!/((4)((4!)(4!)/((4)(4)(4))))

>>8235037
so can my 4 year old.

>>8235027
55!5!44!/4!45!

>>8235048
pls. make it pc compatible.

>>8235037
That's interesting, in that it relies on 4 being 2 squared. It suggest that if you only got primes like 5 to work with, you can do almost nothing.

>>8235037
>>8235067
no, actually, after reevaluating your proposal, I see you produce two 3's in
(4!)(4!) = 4^3 * 3^2
and you have no way of getting ride of them

>>8235027
4/4+4/4=2
check mate

>>8235122
No +'s

>>8235123
4/((.5)(4))

>>8235133
how did you get the dot for .5

>>8235136
its on my keyboard

>>8235133
No .'s

>>8235027
>Only using: ! ; ( ; ) ; / and digits: 4 ; 5
>>8235139

5!/5 is just 4! and five being prime, there is no other thing we could with to divide by 5.
Clearly 2=(4!/4)/3 and so it comes down to producing 3 from 4.
We're pretty fucked when producing primes such as 7 by repreaded factorial, but with multiplication of factorials we end up with powers of 3 we can't get rid of.

I'd like to see a formal proof that we can't produce 3 from 4 and *, / and !.

>>8235138
Jesus

>>8235148
(((332/2/2)!/(332/2/2))/((3/((3/3)/3/3/3))!)/2 =3

>>8235150
sorry that was for another website

>>8235027
I'm off by 1

[math]((5)(5))!/((4!)!)[/math]

an equation is not a thing. stop increasing entropy for no reason

>>8235180
((5)(5))!/((4!)!) = 25
so you are off by 23

>>8235027
5!=2

>>8235027
4!/4/(5!!/5)

>>8235232
yeah but 23 is 1 digit away from being only 2.

>>8235746
Or in terms of 5's: (5!!)!/(5!!/5)