The right side being $r$.

PS. I also noticed that at a certain spot I should have written $p_s \ne q_t$ for all $s$ and $t$ (and not $p_i \ne q_j$). ]]>

$p_1\ldots p_i – p_1Qq_2\ldots q_j = r\cdot q_2\ldots q_j$

The right side being $r$. ]]>

In the proof of lemma 1, the image of is indeed a subgroup of as you’ve shown in the second paragraph right after slogan 12. But I don’t think it is because “ is a homomorphism”.

]]>The sets aE, bE, cE, dE, etc. either are the same set or are distinct. This is equivalent to saying that the property of being in the same set is an equivalence relation. Then the transitivity property is proven by means of the associative property of multiplication. ]]>

I’m afraid I don’t know what mathematical point you are referring to when you talk about Lagrange’s theorem and associativity — can you elaborate slightly?

]]>Thanks for this long and detailed post. It is not customary that mathematicians publish an analysis (in the greek sense of the word) of a particular result, even an ancient one, only a synthesis in the form of a proof.

The way you present this proof has a few drawbacks but has among its advantages a preparation for the related results in algebraic number theory and ideal theory of rings.

I feel you should point out the relation of Lagrange’s theorem with associativity of multiplication.

NB: I think you forgot to use the congruence sign (and mod p) in the statement of lemma 1 at the start of your synthesis of the proof (at the end of your post).

]]>In this case, there is the natural algorithm A (which you mentioned in a previous post) that given a number n produces its factorization, by finding the smallest prime p that divides n, outputting it, and continuing with n/p . So, another way to state the theorem is:

Claim 1 (existence): For all n, A(n) is indeed a factorization of n.

Claim 2 (uniqueness): If n = p_1…p_k with p_1<= … <= p_k then A(n) = (p_1,….,p_k)

The inductive proof of Claim 1 is easy. To prove Claim 2 by induction, you need to prove that for n as above, if p< p_1 then p does not divide n. To do this what you need to prove is your Corollary 2 that if p|ab then p|a or p|b (here a=p_1 and b=p_2…p_k, and p does not divide p_1 since they're distinct primes, and you can prove that p does not divide p_2….p_k by induction).

This gives motivation to the statement of Corollary 2 but then one of course needs to prove it, either by the g.c.d algorithm or as you did it.

]]>The proof of the fundamental theorem of arithmetic is easy because you don’t tackle the whole formal ball game at once. Rather you start with the claim you want to prove and gradually reduce it to ‘obviously’ true lemmas like the p | ab thing. Then you search for proofs to those.

]]>I’d prefer something like this: suppose \alpha \neq 0 is a zero divisor in Z_p. Consider the mapping f:Z_p -> G = {\alpha x for x in Z_p}. Clearly the image of this map is a group under + and it must be of strictly smaller order than Z_p since f isn’t injective as f(0)=f(\alpha*x). Now consider the equivalence relation on Z_p given by x ~ y iff x- y = \alpha *z for some z in 0…p-1. Now given an equivalence classes E_x the map g: E_x -> E_y defined by g(z)=z-x+y is surjective as given w \in E_y w -y =\alpha * r so g(\alpha*r +x) = w and injective as if g(z_1) = g(z_2) then z_1 -x +y = z_2 -x +y in Z_p so z_1=z_2 in Z_p. Let the size of an equivalence class be k as they all have the same size and note that k is the size of G as E_0 = G. Now the sum of the sizes of the equivalence classes must be p so r*k = p but k \neq 1 and k \neq p contradicting the primality of p. Hence \alpha \neq 0 is not a zero divisor. Hence if ab = 0 mod p either a = 0 mod p or b = 0 mod p.

But this isn’t real math this is formal crap. Real math just handwaves to get results. If you want a formally valid proof go look at the computer generated stuff since the goal of mathematics is to generate results we are confident in not formal proofs.

As an aside I once did a talk back in school just presenting sketches of something like 10 different proofs there were an infinite number of primes. Amazing the kind of elegant proofs you can get from weird topologies or even analysis.

]]>It is perhaps a minor notational point, but I find it easier to think of your chain of congruences in corollary 2 as a modification of the congruence by left multiplication by . That is, in my mental image of this part of the proof, we start off with this congruence, and it is modified to . We have a sequence of mental images with rules (or mental machinery) for progressing from one to the next. Alternatively, I have to look at the each congruence in the chain in order and try to think of a reason why it’s true and I feel my picture of what’s going on isn’t as good.

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