If multiplication is repeated addition, then shouldn't division

it is faggot

>>8542605
It is. a/b is how many times you can subtract b from a.

It is

>>8542608
>>8542611
>>8542621

Fucking retards, being outsmarted by me, a monkey poster.

a * b = a + (a + (a + ...)) (with b a's)

By my definition,

a / b = a - (a - (a - ...)) (with b a's).

You fucking plebs can't even understand what repeated subtraction is.

>>8542634
You are proving our point.

>>8542605
of b is divisible a then division is repeated substraction.

For example, let a be any arbitrary number and let b=ka, for some arbitrary number k.

then b/a = ka/a =(k + (k+.... a times...+k)) - (k+.... a times...+k) = k

So if you can divide two integers then this holds.

If you want to divide non-divisible numbers this obviously doesn't hold, just like how if you raise a number to an integer power, that will just be repeated multiplication, but if you do it to a rational power then you have to do more shit, and if you do it to an irrational power then you better get out your fucking wolframalpha because only numerical methods will help you.

>>8542634
4/2=2-2=0

>>8542634
So you made this whole thread just to feel good about yourself?

>>8542605
Defining division as
[math]
f: \mathbb{R^2} \rightarrow \mathbb{R}
[/math]
such that
[math]
f(a, b) = a - \sum_{n = 1}^{b-1} a
[/math]
where
[math]
b > 1
[/math]
results in 0 or negative numbers when dividing by integers larger than 1. This is not useful.

>>8542825
Yes?

>>8542605
Let's see.
sub(x,y) = x - y
Order matters so there are two kinds of repeated subtraction.
rSub1(x,y) = sub( rSub1(x,y-1) , x )
rSub2(x,y) = sub( x , rSub2(x,y-1) )

rSub1(x,1) = x
rSub1(x,2) = x - x = 0
rSub1(x,3) = 0 - x = -x
rSub1(x,4) = -x - x = -2x
rSub1(x,5) = -2x - x = -3x

so rSub1(x,y) = 2x-xy

rSub2(x,1) = x
rSub2(x,2) = x - x = 0
rSub2(x,3) = x - 0 = x
rSub2(x,4) = x - x = 0
rSub2(x,5) = x - 0 = x

rSub2(x,y) is a boring periodic function

>>8542990
>rSub1(x,y) = 2x-xy
Oops, it should be 2y-xy

>>8542990
And here's repeated division.

div(x,y) = x / y

rDiv1(x,y) = div(rDiv1(x,y-1),x)
rDiv2(x,y) = div(x,rDiv1(x,y-1))

rDiv1(x,1) = x
rDiv1(x,2) = x/x = 1
rDiv1(x,3) = 1/x^1 = x^-1
rDiv1(x,4) = 1/x^2 = x^-2
rDiv1(x,5) = 1/x^3 = x^-3

rDiv1(x,y) = x^(2-y)

rDiv2(x,1) = x
rDiv2(x,2) = x/x = 1
rDiv2(x,3) = x/1 = x
rDiv2(x,4) = x/x = 1
rDiv2(x,5) = x/1 = x

rDiv2(x,y) is a periodic function just like rSub2, alternating between the multiplicative identity (1) and x

substraction is the inverse of addition, division is the inverse of multiplication

>>8542605

>If multiplication is repeated addition, then shouldn't division be repeated subtraction?

The reason it shouldn't be is that subtraction is the inverse operation for addition and division can't be the inverse operation of subtraction if you define it as iterated subtraction. Also you can argue subtraction is more like just a shadow of addition than it is some distinct operation in need of an iterated version of itself. The important thing is having inverse operations for the operations of substance which are addition and multiplication.

Every time I come to this board I feel hopefully stupid.

>>8542605
>If multiplication is repeated addition, then shouldn't division be repeated subtraction?
but division is multiplication.

a/b === a * (b^-1)

>>8545076
>division is multiplication
operation-A yielding the same result
as operation-B does not imply A is B

>>8545046
Amen brother, you deserve an upvote :^)

>>8542605
It is repeated subtraction.

div(a,b)
{
q=0;
r=a;
while a>b
{
q = q + 1;
r = a - b;
}
return (q,r) // quotient, remainder pair
}

More interestingly compute a^b (for positive integers, the rest can be added in as special cases) as

a^b
{
x = 1;

while (b>0)
{
x = x + x + x + x ... + x; // (a times)
b = b - 1;
}

return x
}

disclaimer: these algorithms might be fucked up because I'm typing them up in the middle of the night and I don't give a fuck