September 12, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework02
First, if a < 1, then we can simply let b = 1 since a

With this definition of t, setting b = a + t we see b^{2} = (a + t)^{2} = a^{2} + 2at + t^{2} < a^{2} + 2at + t where the last inequality
comes from the fact that 0 < t < 1. Continuing, we have

Thus since b = a + t and both a and t are positive, we have a^{2} < b^{2} < 2. Finally, t is rational since it is constructed by a
combination of multiplication and addition of rational numbers; this implies that rationality of b since it is the sum of two
rationals, a and t.

Define t = . Since 0 < c < 2 < c

Now let d = c - t. The above equations imply that 0 < d < c and also allow for the use of the inequality in the following equation.

Given this equation, we now have 2 < d^{2} < c^{2}, and furthermore, since d is equal to some combination of rational numbers
which are added/multiplied together, then it too is rational.

Let F be an ordered field containing elements x and y.

Let x < y. Then the definition of an ordered field and Rudin’s Proposition 1.18 yield x + x < x + y, implying 2x < x + y and thus x < . Similarly we have obtain x + y < y + y, x + y < 2y, and < y. Combining these results we have

as desired.

First suppose x > 0. Then Rudin’s Proposition 1.18 (b) implies x(x) > x(0) which is equivalent to x

Now suppose x < 0. Rudin’s Proposition 1.18 (c) implies x(x) > x(0), i.e. x^{2} > 0.

Proof. Let F be an ordered field with a,b ∈ F such that a > 0 and b > 0. Adding b to the both sides of a > 0 yields a + b > b, but b > 0, so a + b > 0. __

Let x,y not be both identically zero. Without loss of generality, assume x≠0. The previous problem then implies x^{2} > 0.
So if y = 0, then x^{2} + y^{2} = x^{2} > 0. If y≠0, then the previous problem gives y^{2} > 0 which the above lemma then yields
x^{2} + y^{2} > 0. Hence, in any case, x^{2} + y^{2}≠0.

Part (b) of this problem implies that for any x,y we have (x - y)

When does equality occur?
Stepping backwards through this proof, we see that x^{2} + y^{2} = 2xy when (x - y)^{2} = 0. Then part (b) of this problem
implies that x - y = 0. Thus we have equality when x = y.

If x < y, then 0 < y -x. Hence the archimedean property of the reals yields an integer n > 0 such that n(y - x) > . Since > 0 we have

0 | < | < n(y - x) | |||||

0 | < | ∕n | < y - x | ||||

x | < | x + ∕n | < y |

If x < y, then 0 < y -x. So again, the archimedean property gives us an integer n > 0 such that ny -nx > 1. This implies, since consecutive integers have a difference of one, that there must some integer m with nx < m < ny. Thus x < m∕n < y, and m∕n is rational.

Let A ⊂ ℝ with a = inf A and bsupA. If x,y ∈ A and x < y then we would have a ≤ x < y ≤ b, and so it’s not possible for a set with two or more elements to have a supremum that’s less than or equal to the infimum. Hence only singleton sets have the desired property.

Let A ⊂ ℝ be bound above by α and B ⊂ ℝ be bounded below by β. Then a ≤ α for all a ∈ A and β ≤ b for all b ∈ B, however, since A∩B ⊂ A and A∩B ⊂ B, then we must have that β ≤ x ≤ α for all x ∈ A ∩ B. In other words, A ∩ B is bounded.

Let z,w,v ∈ ℂ be complex numbers.

We know for a,b ∈ ℂ that |a + b|≤|a| + |b|, which implies |a + b|-|b|≤|a|. Since a and b are arbitrary, then we can find x,y ∈ ℂ when a = x - b and b = y, then |(x - y) + y|-|y|≤|x - y| which implies

With this, we then have

taking advantage of the fact that |x| = |- x| for all x ∈ ℂ in the rightmost equality.

The following region is the set of points, note that the edges of the region are not included.

We have the following sequence of equations due to the fact that |a| = |a|, |ab| = |a||b|, |w| = 1, and a - b = a -b for any a,b ∈ ℂ.

Let x,y,z ∈ ℝ and define

If |x - z|≤|y - z| or |x - z|≤|x - y|, then the fact that f(t) = t∕(1 + t) is an increasing function implies the triangle inequality for d.

So assume that |x - z| is greater than both |x - y| and |y - z|. Since |x - z|≤|x - y| + |y - z| we have

with the rightmost inequality comming from our initial assumption. Thus the triangle inequality holds for this d(⋅,⋅).

The proof for this is identical to the proof of in the previous part of this problem after substituting in the function |⋅-⋅| for g(⋅,⋅). This is because the only property of |⋅-⋅| that was used in the previous proof was that it upholds the triangle inequality, which g(⋅,⋅) also does.

Let f

By the following, the zero constant is the additive identity:

By the following, the one constant is the multiplicative identity:

Then

so is closed under addition. Since polynomial addition is commutative and associative, then addition is commutative and associative in . Finally, since f(x) + (-f(x)) = 0, additive inverses exist in .

Since

then is closed under multiplication. Since polynomial multiplication is commutative and associative, then multiplication
is commutative and associative in . Finally, if f_{1}(x)∕f_{2}(x) is not zero, then

and so multiplicative inverses exist in .

Finally, since polynomial addition and multiplication obides by the distributive law, then so does addition and multiplication in .

So is indeed a field.