>>19170529Let us use prefix notation:
>(x * y) = mult(x,y)>neutral = neutral element of mult(,) such that: mult(x, neutral) = xInverse element:
>inv(x) = y such that mult(x,y) = neutralDefinition of div:
>div(x,y) = mult(x,inv(y))div(1,0)
= mult(1,inv(0))
= mult(1,y) such that mult(0,y) = neutral
= mult(1,y) such that mult(z, mult(0,y)) = z
Let mult(x,y) = mult'(x,y) if y != 0 and x otherwise with mult' = ordinary multiplication
Let y = 0:
= mult(1,y) such that mult(z, mult(0,y)) = z
= mult(1,y) such that mult(z, 0) = z
= mult(1,0) such that mult(z, 0) = z
Therefore inv(0) = 0 and therefore
>div(1,0) = mult(1,inv(0)) = mult(1,0) = 1Per definition of mult(,) above:
>mult(1,0) = 1Or, in conventional notation:
>1 * 0 = 1 