September 10, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework01
Since r is rational, both 1∕r and -r are rational. Thus, if r + x were rational then

would imply x is rational by the closure of addition in the field of rationals; but since x is assumed to be irrational, so must be r + x.

Similarly, rx must be irrational since

and the field of rationals is closed under multiplication.

Proof. Assume for later contradiction that 3 is rational. Then we can find integers m,n which are not both even such that m∕n = . Therefore

This equation then demands that if n^{2} is even, then so must be m^{2} (and therefore m), and similarly if n^{2} is odd, then so
must be m^{2} (and therefore m). Because m and n were assumed to not both be even, then m and n must
both be odd. Thus we can set m = 2p + 1 and n = 2q + 1. Furthermore the above equation implies that

Since = 2 the previous problem and the above lemma together imply that is irrational.

Let x≠0 and xy = xz, then

This follows from the previous part for z = 1.

Again, this follows from the first part of this problem for z = 1∕x.

Let x≠0, then

For x,y ∈ ℝ

Geometrically, if |x| and |y| are the two lengths of two adjacent sides of a parallelogram, then |x + y| and |x-y| would be the lengths of the two diagonals. So this equation reveals that the sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its sides.

Letting (λ

| (5.1) |

Now assuming that |x - a| = λ|x - b| for some point x ∈ ℝ^{k} we get the following sequence of equations.

What happens when λ = 1?
When λ = 1, we’d be interested in all points x ∈ ℝ^{k} where |x-a| = |x-b|. However the points that satisfy this are the
points in the hyper-plane that perpendicularly bisects the line segment, which of course has no notion of a radius since
it is a hyper-plane and not a hyper-sphere. Hence it makes no sense to inquire about a sphere of radius
r = λ(λ^{2} - 1)^{-1}|b - a|; this is dually supported by the denominater, (λ^{2} - 1), having a value of zero when
λ = 1. In a silly way of thinking, it’s as if the sphere gets so big (as λ approaches 1) that it “becomes a
plane”.

Lemma 6.1. If S_{1},S_{2},S_{3},… are finite sets that are mutually disjoint, then ∪_{j=1}^{∞}S_{i} is countable.

Proof. First note that if any S_{i} is the empty set it can simply be neglected. So fix an ordering of each of the elements of
each set S_{i} and label them as follows

Let S_{1},S_{2},S_{3},… each be a finite set. The problem here is that these sets are not mutually disjoint, but nevertheless, we
can define the sets T_{1},T_{2},T_{3},… as T_{1} = S_{1} and T_{n} = S_{n} -∩_{j=1}^{n-1}S_{j} and furthermore, ∩_{j=1}^{n-1}T_{j} = ∩_{j=1}^{n-1}S_{j}. So due
to the above lemma, ∩_{j=1}^{n-1}T_{j} is countable and therefore so is A = ∩_{j=1}^{n-1}S_{j}.

Define the sets S

Now with this definition, the set A = ∪_{i=1}^{∞}S_{i} will contain only algebraic numbers, but it will furthermore contain all of
the algebraic numbers since if z is the root of a_{n}x^{n} + + a_{1}x + a_{0} and some a_{i} has |a_{i}| > n, then z will be in the S_{N}
where N = max(|a_{0}|,…,|a_{n}|) since z will also be a root of 0x^{N} + 0x^{N-1} + + 0x^{n+1} + a_{n}x^{n} + + a_{1}x + a_{0}. Also,
because a polynomial of degree n has at most n roots and the set ℤ ∩ [-n,n] is finite for any n, then each of the sets S_{i}
are finite. Therefore, the previous problem gives us that A (the set of algebraic numbers) is a countable
set.

Let b > 1 be real and m,n,p,q ∈ ℤ, r ∈ ℚ such that r = m∕n = p∕q

Since m∕n = p∕q, then mq = np and through use of the laws of integer exponents we obtain

So it makes sense to define b^{r} = ^{1∕n}

Let s = a∕b for a,b ∈ ℤ. Then we have

which in turn allows

Define B(r) = {b

Define B(x,y) = {b

The set of real numbers is uncountable, whereas the set of algebraic numbers is countable, as per problem seven. So there are “too many” real numbers for all of them to be algebraic.

No, the irrational numbers are not countable.

Each rational number r = p∕q with p,q ∈ ℤ, q≠0, is the root of f(z) = qz -p, implying that the rational numbers are a subset of the countable algebraic numbers, i.e. they’re countable. So because the rational numbers are countable, the union of the rational and irrational numbers are the reals, and the reals are uncountable, then the irrational numbers are necessarily uncountable.