>Employing strictly the properties of the set of rational

>>7806656
It might help if you post the properties that you cut off from the top of the page

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>>7806662

>>7806673
to do 1.(i):
ab+a(-b)
=a(b+(-b)) (by (v))
= a0 (by definition of additive inverses)
=a(0+0) (since 0 is additive inverse)
=a0+a0

in particular this implies a0=0 so a(-b)=-(ab)

>>7806686
how do you prove the transposition part?

>>7806727
What transposition are you referring to?

>>7806730
ab+a(-b)

>>7806744
I'm still not sure what you mean. ab+a(-b) is a number, not a transposition.

>>7806750
I mean how did you get from a(-b)=-(ab) to ab + a(-b) = 0?

>>7806757
-(ab) designates the additive inverse of ab, i.e. the unique number such that ab+(-ab)=0. So I showed ab+a(-b)=0, which implies a(-b)=-(ab).

>>7806760
I get it now

cheers

These are arithmetic axioms.
With these you can prove that

ab = 0 => that either a or b is 0

>>7806656
> Question 3

Shouldn't it be for a != 0?