Solution 3

1. Since 0 + 0 = 0 we have (0 + 0).a = 0.a and so by the Distributive law we get 0.a + 0.a = 0.a and adding the additive inverse of 0.a to each side we get 0.a = 0.
   
   Then 0 = (-1).0 = (-1).(1 + (-1)) = -1 + (-1)²and adding 1 to both sides gives 1 = (-1)².
   
   If 0 > 1 then add -1 to both sides to get -1 > 0. Then -1.-1 > 0 1 > 0 and we get a contradiction.
   
   If a > 0 then add -a to both sides to get 0 > -a
   If a > 0 and 0 > b then -b > 0 and so a.(-b) > 0. Then a.(-b) = -(ab) and then -(ab) > 0 0 > ab.
   a > b a - b > 0. If 0 > c from the last result we have 0 > (a - b).c or 0 > ac - bc bc > ac.
   If a > 0 and 0 > a^-1 then from above, a.a^-1< 0 which contradicts 1 > 0. Similarly for the other part.
   
   

2. Verifying that (F, +) and (F-{0}, .) are abelian groups is straightforward algebra. The distributive laws (a(b + c) = ab + ac follows from the same rule in Z.
   
   With the given ordering we have 0 > 1, but then adding 6 to both sides gives a contradiction.
   
   If F is an ordered field under some ordering then (see above) 1 > 0. But then we have 2 > 1, 3 > 2, ... by successively adding 1. We have just seen that this ordering does not satisfy the axioms.
   
   

3. (a) lub = 2^1/3 and is not in the set. The set is not bounded below and so glb does not exist.
   (b) lub = 2. glb = -2. Both lie in the set.
   (c) lub = 1 S and glb = 0 S.
   (d) If x²ⁿ⁺¹= 2 then x is the (2n + 1)th root of 2 which is close to 1 for large n. So lub = ³2 S. The glb = 1 S.
   (e) The set is [0, 1) Q with lub = 1 S and glb = 0 S.
   (f) lub = 1 S and glb = 0 S.
   (g) The biggest such number is 0.888... = ⁸/₉ = lub S. The glb = 0 S.
   (h) lub = 1 S and glb = 0.111... = ¹/₉ S.

4. The reals in (0, 1) whose first decimal digit 9 form the interval (0.00... , 0.900...) which is an interval of length ⁹/₁₀.
   Reals with first and second decimal digits 9 form a union of intervals (0.000... , 0.0900...) (0.100... , 0.1900...) ... (0.800... , 0.8900...) which is a union of 9 intervals each of length ⁹/₁₀₀.
   Reals with first, second and third decimal digits 9 form a union of 9²intervals each of length ⁹/₁₀₀₀, etc.
   Reals with first, second, ... , nth decimal digits 9 form a union of 9^n-1 intervals each of length 9 /10ⁿ with total length (9/10)ⁿ.
   As n this length becomes arbitrarily small and so the set of reals with no 9 in its decimal expansion has measure zero.
   You can visualise this in a different way by thinking how you would choose a "random" real number in the unit interval. You could, for example, roll a 10-sided dice infinitely often to generate its decimal expansion. What would be the chance of never getting a 9 when doing this?