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Ed Quanta
Jan28-04, 09:48 AM
I have to prove that x*0= 0 where x is any integer. I can do this pretty easily using the proposition that m(-1)=-m but I am not allowed to use this. In addition, I am unable to prove m(-1)=-m unless I accept the fact that anything times zero equals zero. Can anyone give me a hint or push on how to show this? I am just going around in circles.

matt grime
Jan28-04, 10:12 AM
0.m=(0+0).m =0.m+0.m

subtract m.0 from both sides.

Ed Quanta
Jan28-04, 10:18 AM
Thanks, Im an idiot. I kept trying to represent 0 as (m + -m)

epkid08
Jul20-09, 12:24 AM
When x is an integer, multiplication can be defined as:

x*a = \sum^x_{n=1} a

When a = 0:

x*0 = \sum^x_{n=1} 0 = \underbrace{0 + 0 + ... + 0}_x = 0