Prove that 
.
Proof. By Theorem I.3, we have
, so
![\[ (a+b) + (-a-b) = (a+b) + (-a) + (-b) = (a+(-a))+(b+(-b)) = 0+0 = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-42535171fe4c38349fedffd06fb15c0b_l3.png "Rendered by QuickLaTeX.com")
Thus, indeed 
is the additive inverse of 
; hence, by definition
![\[ -(a+b) = -a-b. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-283183474182f5a0852d539351a4fdff_l3.png "Rendered by QuickLaTeX.com")
Prove -(a+b) = -a-b
Proof. By Theorem I.3, we have
, so
Proof. By Theorem I.3, we have
, so
Stumbling Robot
A Fraction of a DotProof. By Theorem I.3, we have
, so
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