- #1
chwala
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- TL;DR Summary
- I am looking at the basics here...Consider the attachment below. My intention is to try and prove ##2.11## from ##2## to ##4## and i would appreciate any constructive feedback.
Ok for ##1##, we also have,
##a⋅0=a⋅(0+0)=a⋅0 + a⋅0 ## We know that ##a⋅0=0 ## by additive cancellation.
For ##2.11##, Number ##2##;
We first show and prove that
##-b=-1⋅b##
adding ##b## on both sides,
##-b+b=0## for the lhs
##-1⋅b +1⋅b=b(-1+1)=b(0)=0## for the rhs
therefore,
##(-a)b=(-1⋅a)b=-1(ab)=-(ab)## using distributive property...
also,
##a⋅(-b)=a⋅-1⋅b=-1⋅(a⋅b)=-1(ab)=-(ab)##
Will look at (3) and (4) later...