Why can't division be converted to multiplication like how subtraction can be converted to addition? Why do you have to go to an iterative series cope for division algorithms? Why are inverse operations so fucking weird?

>>16707018 (OP)
It's just notation. When dealing only with addition, one can trivially assign the inverse operation to mean subtraction
a – b = a + b^-1
It's merely for convenience that b^-1 is written as (-b)
a/b = a * b^-1
But if we wanted to, we could just as easily denote division by the ÷ symbol and say
a ÷ b = a * b^-1
And denote b^-1 := (÷b) such that
a ÷ b = a * (÷b)
It's just for historical reasons we don't write it like this

>>16707018 (OP)
Because you can't divide by zero, duh.

The problem is one of boundaries.

1/7 = 0.(142857) Parentheses indicate repeating decimal digits.

5*35 = 175

SQRT(5) = 2.23606797749979... (unbounded)

1E+7 =1000000 (1x10^7 digits)

56! = 7.10998587804863*10^74 (74 digits)

There are multiplication and exponential small formulas that become impossibly long digit strings, but a larger bunch of division operators become unbounded faster.

You can leave a division formula bounded by simply leaving it "as is", 1/7 stays as it is; the SQRT(5) can remain as it is. CAS math calculators tend to keep simultaneous digit math plus symbolic math systems running together. It increases speed and accuracy.

For example, let's say you've got:
5*4 + 8*3 +6 then group by multiplication
(5*4) + (8*3) +6 then join
(4*(5*1)+(2*3))+6 which becomes
(4*(5+6))+6 or (4*11)+6 or 44+6 =50

Or you can (5*4) + (8*3) +6
20+24 +6 = 44 +6=50

Not much improvement for this small example, but imagine if you were doing matrix math and grouped 100 paired equations that had to be multiplied with a factor of 4 in them which then added or subtracted later.

>>16707031
Maybe you can't, but I can.

>>16707018 (OP)
Dividing is the same as multiplying by the reciprocal, though, so all you have to do is convert the divisor to its reciprocal and multiply.

>>16707096
That involves dividing