F5:  a·0=0

"For any  a  [in]  R,  we have  a·0=0."  (see ita-2-1-12-pg18)


Consequently "a·0  and  0  are both solutions of the equation  x + a·0 = a·0,  hence equal, by  F3."  (see ita-2-1-12-pg18)

I think this result is deceptively profound.  It's deceptive because it's so much common sense that zero times anything is zero, and it's profound because zero was defined as the neutral element for addition - without any mention at all of multiplication.

It should be noted also from this result that division by zero is not a legitimate operation because there is no real number  x  such that  0·x=1.  This means that zero does not have a multiplicative inverse - which would be needed according to  F4  to define division by zero.

Rosenlicht also notes that  "if a product of several elements in  R  is zero then one of the [elements] must be zero:  for if  ab=0  and [a is not equal to 0] we can multiply both sides by [the multiplicative inverse of a] to get  b=0."  (see ita-2-1-12-pg18)  Note that it is okay to 'multiply both sides by a number' because doing so is a single multiplication - which has a unique result by the definition of multiplication.


Converted [ in part ] by Mathematica      December 4, 2007