- #1

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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...