p.28-1  Prove u  b = 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.