Partial Partial Homework I Key
• My Problem 1, 3 points: It’s tempting to pretend you’re in the reals and
claim that (x + y)/2 is between x and y, but the axioms for a field don’t
have any “2” in them!
However, this can provide inspiration. For any arbitrary field F let us
DEFINE 2 = 1 + 1. Now suppose that we have two distinct elements x
and y, and WLOG x < y. Note that since 0 < 1 we have 0 < 2 and so
2−1 exists and 0 < 2−1 (look up the relevant axioms in the book).
Let z = 2−1 x + 2−1 y. It’s tempting to sling everything around like fractions, but you have to be careful. Each step below is justified by the
axioms of a field:
2z
= 2(2−1 x + 2−1 y)
= x+y
> x+x
=
2x.
Multiply both sides of 2z > 2x by 2−1 to obtain z > x. A similar argument
shows z < y.
• My Problem 2, 2 points: Given x < y, let x1 = (x+y)/2; we know from the
previous part that x < x1 < y. Let x2 = (x + x1 )/2, so x < x2 < x1 < y.
In general let xk = (x+xk−1 )/2, so x < xk < xk−1 < xk−2 < · · · < x1 < y.
We generate an infinite sequence x1 , x2 , . . . between x and y.
If you want to be official and use induction, set x0 = y and define xk+1 =
(x + xk )/2. From Problem 1 we have x < xk+1 < xk for all k, and so a
infinite sequence y > x1 > x2 > x3 > · · · > x. Note the xk are all distinct.
An alternate proof is by contradiction: suppose there are only n elements
a1 < a2 < · · · an with x < a1 and an < y. But between any of the a1 and
a2 (or any of the adjacent ak ) there is another element of F (citing the
first problem), contradiction!
• (My Problem 3, 2 points) Well, this can be a pain, can’t it? A brute force
approach is tedious.
But note that our field is just Z arithmetic modulo 2. Given that the
associative, distributive, commutative properties hold for Z, it follows they
hold for F. For example, since x(y + z) = xy + xz for the integers means
it holds modulo 2. No need to check all cases in F.
The only things that really need checking are the additive and multiplicative identities—obvious—and existence of inverses, also clear from the
data given.
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