Yeah there's alot of ways that 1/0 is just a "bug" in mathematics.
The fact is, by quantification it's necessarily infinite, but 0 doesn't have a sign.
Suppose f function
f(x)= (x+1)/x
At the point f(0) it'll have:
f(0) = 1/0 = ?
But f(x) also has another representation by separation
f(x) = (x+1)/x = x/x + 1/x = 1 + 1/x
which on point 0 is
f(0) = 1 + 1/0
So
1/0 = 1 + 1/0
And cutting 1/0 on both sides.
0=1
In fact on the other way around you can replace 1/0 with f(0) and :
f(0) = 1 + f(0)
So
f(0) = 1 + ( 1 + f(0)) = 1 + [ 1 + ( 1 + f(0))]= ... = +oo + f(0)
So cuting the f(0) you have 0 = +oo too
The fact is, by quantification it's necessarily infinite, but 0 doesn't have a sign.
Suppose f function
f(x)= (x+1)/x
At the point f(0) it'll have:
f(0) = 1/0 = ?
But f(x) also has another representation by separation
f(x) = (x+1)/x = x/x + 1/x = 1 + 1/x
which on point 0 is
f(0) = 1 + 1/0
So
1/0 = 1 + 1/0
And cutting 1/0 on both sides.
0=1
In fact on the other way around you can replace 1/0 with f(0) and :
f(0) = 1 + f(0)
So
f(0) = 1 + ( 1 + f(0)) = 1 + [ 1 + ( 1 + f(0))]= ... = +oo + f(0)
So cuting the f(0) you have 0 = +oo too
