# 1 Six proofs of the infinity of primes

### p.3 Second Proof

First equation: Use, for example:

$2^{2^{n}}= 2^{2(2^{n-1})}= 2^{(2^{n-1}+2^{n-1})}= 2^{2^{n-1}}\cdot2^{2^{n-1}}= {\left(2^{2^{n-1}}\right)}^{2}$

### p.4 Lagrange's Theorem

A group (G, $\cdot$) must satisfy four axioms:

• closure $\forall a, b \in G: a \cdot b \in G$
• associativity $(a \cdot b) \cdot c = a \cdot (b \cdot c)$
• identity: $\exists e \in G \ni a \cdot e = e \cdot a = a \forall a \in G$
• inverse: \forall $a \in G \exists b \in G \ni a \cdot b = e$

Groups do not necessarily have commutativity; those that do ($a \cdot b = b \cdot a$) are Abelian groups. (And if something is purple and commutes it is of course an Abelian grape.)

A multiplicative troup muses multiplicative notation (as above) for its binary operation. I get the sense that Abelian groups are more commonly associated with an addition-like binary operation but never mind. The point is that a multiplicative example is not trivial; for example try arithmetic multiplication modulo n on the integers ${0, 1, 2, ..., n - 1}$. This fails the axioms by having no workable inverse. Going a step further to n roots of unity, however, works fine as a multiplicative group as the last remark in the Lagrange's Theorem sidebar suggests. To understand this short proof I need some additional terminology however, specifically the terms:

• coset
• equivalence relation
• equivalence class

A more verbose version of the proof of Lagrange's Theorem now reads:

---left off here hey!---

### p.5, 6 Sixth Proof

While this proof is very From The Book it is presented in such an economical matter that the aesthetic element is lost on me for want of understanding the economy. This is my own limit, but as an exercise in storytelling I will present here the essential pieces first, and then recreate the proof without less economy.

Suppose a set of numbers C is to be generated from two 'constructor' sets of numbers G and H according to the following rules:

• G is a set of the first k primes, i.e. all distinct numbers; and from G we can draw 0, 1, 2, ... up to k elements to build a product a. If we draw 0 elements the result is a = 1, not a = 0. The total number of a values we can produce is (summing a row of Pascal's triangle) $2^k$. For example if k = 4 we have G = { 2, 3, 5, 7 } and the values for a are { 1, 2, 3, 5, 6, 7, 10, 14, 15, 18, 21, 30, 35, 42, 70, 210 }, 16 possible values.
• H is a set of squares of integers { 1, 4, 9, 16, 25, ..., $q^2$ } for some q. Such a number in H is denoted here as $b^2$ for the corresponding integer b.

How many numbers can we construct from these two sets? There are $2^k$ values for a and q values for $b^2$ and the two sets are distinct except that both contain 1. Hence the total number of unique numbers that can be constructed of the form of the form $a \; b^2$ is $2^k \; q$.

Whew... that was a long way to go for such a simple point. Notice that using this method of construction we can cover all the integers (provided we let G and H get big enough). That is, no numbers are skipped over in this construction scheme:

<pos> Giving the a and then the b values:

1 = 1 x 1 2 = 2 x 1 3 = 3 x 1 4 = 1 x 4 5 = 5 x 1 6 = 6 x 1 7 = 7 x 1 8 = 2 x 4 9 = 1 x 9 (b squared = 3 x 3) 10 = 10 x 1 (a = 2 x 5) 11 = 11 x 1 12 = 3 x 4 13 = 13 x 1 14 = 14 x 1 (a = 2 x 7)

...etcetera... </pos>

That is, all integers may be written as a product of mutually unique primes (the a factor) times a square integer (which may be 1). If the primes in a occurred twice they would be subsumed into b squared. If the primes in a occur an even number of times: The same, just so.

And so now we have a constructivists' view of building the integers in this peculiar format, so that it is complete and so that the number of constructions can be connected to the size of the constructor sets G and H.