the expression:
( (i0*(1*(1%(((i0 + 1) / (i1 + 1))+1))*(1%(((i0 + 1) / (i2 + 1))+1))*(1%(((i0 + 1) / (i3 + 1))+1))*(1%(((i0 + 1) / (i4 + 1))+1)))) + (i1*(1*(1%(((i1 + 1) / (i0 + 1))+1))*(1%(((i1 + 1) / (i2 + 1))+1))*(1%(((i1 + 1) / (i3 + 1))+1))*(1%(((i1 + 1) / (i4 + 1))+1)))) + (i2*(1*(1%(((i2 + 1) / (i0 + 1))+1))*(1%(((i2 + 1) / (i1 + 1))+1))*(1%(((i2 + 1) / (i3 + 1))+1))*(1%(((i2 + 1) / (i4 + 1))+1)))) + (i3*(1*(1%(((i3 + 1) / (i0 + 1))+1))*(1%(((i3 + 1) / (i1 + 1))+1))*(1%(((i3 + 1) / (i2 + 1))+1))*(1%(((i3 + 1) / (i4 + 1))+1)))) + (i4*(1*(1%(((i4 + 1) / (i0 + 1))+1))*(1%(((i4 + 1) / (i1 + 1))+1))*(1%(((i4 + 1) / (i2 + 1))+1))*(1%(((i4 + 1) / (i3 + 1))+1)))))/(0+1*(1%(((i0 + 1) / (i1 + 1))+1))*(1%(((i0 + 1) / (i2 + 1))+1))*(1%(((i0 + 1) / (i3 + 1))+1))*(1%(((i0 + 1) / (i4 + 1))+1))+1*(1%(((i1 + 1) / (i0 + 1))+1))*(1%(((i1 + 1) / (i2 + 1))+1))*(1%(((i1 + 1) / (i3 + 1))+1))*(1%(((i1 + 1) / (i4 + 1))+1))+1*(1%(((i2 + 1) / (i0 + 1))+1))*(1%(((i2 + 1) / (i1 + 1))+1))*(1%(((i2 + 1) / (i3 + 1))+1))*(1%(((i2 + 1) / (i4 + 1))+1))+1*(1%(((i3 + 1) / (i0 + 1))+1))*(1%(((i3 + 1) / (i1 + 1))+1))*(1%(((i3 + 1) / (i2 + 1))+1))*(1%(((i3 + 1) / (i4 + 1))+1))+1*(1%(((i4 + 1) / (i0 + 1))+1))*(1%(((i4 + 1) / (i1 + 1))+1))*(1%(((i4 + 1) / (i2 + 1))+1))*(1%(((i4 + 1) / (i3 + 1))+1)))
is equal to the largest positive integer(and 0) out of i0,i1,i2,i3,i4.
"a/b" denotes floor(a/b) or integer division
"a%b" denotes a mod b
"a+b" and "a*b" is obvious
generated through a python script. I realised that it's not the shortest way to express such a thing, but still. The most efficient expression would still take up acouple lines(i would think). I encourage other people to try to write an expression that is equal to the largest of two numbers using only these operations.