p.28-1 Prove ub = b (and b notequal 0) implies u = 1. 1/b exists (u b) 1/b = b 1/b, applying the associative property to the LHS and noting b 1/b =1 gives the desired result. p.28-3ac a)2x +5 +(-5) = 8 + (-5) (existence of additive inverse, definition of addition. 2x = (8= (-3)) assoc property and def add inverse (add of 0) 1/2 2 x = 1/2 (8+(-3)) definition mult inv, def of multiplication x = 1/2(8+(-3)) assoc property and def mult inv, mult by 1. c) x^2 = 2x i) if x = 0, by theorem for mult by 0 both sides are 0, so 0 is a sol. ii) if x notequal 0, x x 1/x = 2 x 1/x by existence mult inv (x notequal 0) and def mult. by assoc property, def mult inv, def mult by 1, x=2. p.28-12ac a) comm since + is comm, .5(.5(a+b)+c)) not equal .5(a+(.5(b+c)) (.25a vs .5a), so not assoc. no identity since a = .5(a+e) requires e=a, but a can vary. c) a-b = -(b-a), not comm.; (1-2)-3 not equal 1-(2-3), not assoc.; a=a-e requires that the identity is 0, but b=0-b shows that candidate for an identity does not work. p.37-1 a+c leq b+c; b+c < b+d; therefore a+c < b+d (you must give reasons) b) same argument with leq everywhere. p.37-5 by theorem 2.2.5, if a noteq 0, a^2 is positive, similarly b neq 0 entails b^2 pos. positive is closed under addition (This shows If a,b neq 0 then a^2 + b^2 >0). If a neq 0 and b =0 (or vice versa), a^2 + b^2 is positive + 0, which is positive. (So we now have if a or b neq 0, a^2 + b^2 is positive; the contrapositive of which is if a^2 + b^2 is not positive (i.e, 0, since we can easily show it cannot be negative), a=b = 0. It is trivial that a=b=0 => a^2 +b^2=0). p.37-11 (.5(a+b))^2 = .25a^2 + .5ab + .25b^2 = .5(a^2 + b^2) +.5ab - .25(a^2 + b^2) = .5(a^2 + b^2) - .25(a-b)^2 leq .5(a^2 + b^2) since the subtrahend is positive unless a=b in which case it is zero. this gives the desired result.

Sat Sep 6 1997