Please prove: (-a)b = -(ab); (-1)b = -b. Please use axioms only?

Question

Please, also prove this: (-a)(-b) = ab. Again, please use axioms... Thanx a lot...

Answer 1

using b•0=0 (already proved this for you)
b•0=b(1-1)=b+(-1)•b=0. (distributive property a(b+c)=ab+ac)
therefore -b+b+(-1)b=(-1)b=0+-b=-b (adding inverse, equivalent to cancellation property)

Now using (-1)x=-x for all x:
(-a)b=((-1)a)b=(-1)(ab)=-ab (associativity (ab)c=a(bc))

Answer 2

-a = -1 X a
therefore, (-1 X a) X b = a X b X -1 = -1ab, then take -1 out to get -(ab)

(-1)b = -1 X b = -b

(-a)(-b) = (-1 X a)(-1 X b) [expand] = a X b X -1 X -1 [-1 X -1 = 1] = a X b X 1 = a X b = ab

Answer 3

a < b ?a < ? b Multiply both facet of the inequality through ?b ?a * ?b < ?b * ? b ?(ab) < ?(b^2) ?(ab) < b Now bypass decrease back to the second one line. ?a < ?b Multiply both facet of the inequality through ?a ?a * ?a < ?b * ?a ?(a^2) < ?(ab) a < ?(ab) So we've putting jointly both inequalities we in simple terms proved, we get a < ?(ab) < b. wish that helps! extra be conscious: yet otherwise you need to prepare that is this: a < b a * a < b * a a^2 < ab a < b a * b < b * b ab < b^2 So, a^2 < ab < b^2 Take the sq. root. a < ?(ab) < b :)

Answer 4

Will I also get your diploma here too?

Answer 5

you could always do your own homework :P