龚学ABC 猜想的证明

龚学ABC 猜想的证明—供大家参考。

结论:这个ABC 猜想的成立的证明,正是这个宇宙粒子产生的标准模型及其宇宙万物产生的机制。这个宇宙的存在就是这个ABC 猜想的证明!

The proof of the abc conjecture | The Great Vindications

The proof of the abc conjecture

THE PROOF OF THE ABC CONJECTURE

The abc conjecture: there are a finite number of c (= a + b), for

c > rad (abc)^ (1+ ε); ε > 0, a real number

Or

c < K (ε) x rad (abc)^ (1+ ε) for ALL c; K (ε) > 0, a real number.

Let a = d + d1 = (p^n x dp) + d1; dp is the largest prime for d; d1 is the smallest integer for a to have a d.

Example: a = 17 = 18 – 1 = (3^2x 2) -1; 18 = d, d1 = -1, p = 3, dp = 2, n = 2

Doing the same for b and c;

b = e + e1 = (q^m x ep) + e1

c = f + f1 = (w^k x fp) + f1

So, the rad (abc) = rad (a) rad (b) rad (c)

For rad (c) >= c, f1 ≠ 0 is the necessary condition.

For c >= rad (abc), f1 = 0 is the necessary condition; that is, c cannot be a prime.

The sufficient condition (SC): rad (abc) = pqw (dp x ep x fp) < C

Some scenarios can be evaluated for this sufficient condition (SC).

Scenario 1: if d1 = e1 = 0 and there is a h1 (a natural number) while 1 < h1 < min {n, m, k}, then SC = true

Scenario 2: if d1 = 0 and there is a h2 (a natural number) while 1 < h2 < min {n, m, k}, then SC = true

Scenario 3: if e1 = 0 and there is a h3 (a natural number) while 1 < h3 < min {n, m, k}, then SC = true

Scenario 4: all other cases (the uncertainty).

All four cases, the SC = true.

For any give c (with f1 = 0, not a prime), there are S1 (number of cases meeting scenario 1), S2, S3 and S4.

Let S = S1 + S2 + S3 + S4; S can be finite or infinite.

Now, {rad (abc) < c} = {for any c (not a prime, f1 = 0), is S finite?}

For any given c, we can do some actual search for S1, S2, S3.

From our experience, S1 + S2 + S3 is more often as finite than not.

However, there is no way of guaranteeing that S4 is finite.

By not knowing the answer, we can try with tossing a coin on S4 of an arbitrary selected c: head = true; tail = false.

Then, the P (S) = {tail (50%), head (50%)} after infinite many tosses.

Law 1: There are infinite many c (= a + b) for c > rad (abc).

Obviously, law 1 is a physics law, verifiable via experiments.

Now, we can make a cheating weight (ε > 0, a real number) and add it to rad (abc) side as {rad (abc) ^ (1+ ε)}.

With this cheating weight on the rad side as {rad (abc)^ (1+ ε) < c}, then P (S) = {tail (< 50%), head (> 50%)} of each toss. Again, this can be verified physically.

After N tosses, P (S, N) = {P(S, tail)/N = (~ 0), P (S, head)/N = (~100)}; that is, for any c (a real number while a + b = c) there is always a N (ε) for each ε (a real number) to ensure that

{N (ε) x rad (abc)^ (1+ ε) > c}, N (ε) is the number of toss needed for that (ε).

Again, this can be verified physically.

The above process can be proved in four steps: induction (operational) progressive process, a verification TRAIN.

First, making the above simple tossing [selecting an arbitrary c (= a + b) and doing the actual search] process into a game with the following rules.

Every game consists of T (=10) tosses, which produces (i tails, j heads), T = i + j = 10 in this case.

So, P (SC = true) = j/T, P (SC = false) = i/T, P the possibility of SC

ΔP = P (SC = false) – P (SC = true),

If ΔP > 0, abc conjecture is false.

If ΔP < 0, abc conjecture is true

Let G = 1 when ΔP < 0; G = 0 when ΔP >= 0

This game will be repeated N times.

When N = 1, G1 = (0 or 1)

N = 2, G2 = (0 or 1)

N = n, Gn = (0 or 1)

Let G’n = (number of 1) – (number of 0); {(number of 1) + (number of 0) = n}

Definition 1: If G’n > 0 for all n > N (ε), [N (ε) a large number > 0], then abc conjecture is true.

Second, the cheating: a cheating weight ε is added on one side of the tossing coin.

That is:    ΔX = {rad (abc)^ (1+ ε) – rad (abc)} = rad (abc)^ε

Law 2 (the indeterminacy): when ΔX = 0, the average of ΔP = 0 after n games (n x T tosses) when n is a large number.

Law 2 can be verified physically.

Law 3: when ΔX > 0, the average of ΔP < 0 after n games (n x T tosses) when n is large. (This can be proved by actual calculation and search with a finite n).

Third, the induction proof of law 1, 2, 3: these physically proofs establish a verification TRAIN.

Fourth, going beyond the induction: is there a math ghost rascal which can sabotage the above induction TRAIN?

The answer is no: a cheating game cannot be sabotaged even by a ghost rascal; see the ghost rascal law.

Ghost-rascal law — For a coin flipping (tossing) game (head vs tail), T is the number times flip as one ‘game’, N is the number times that that ‘game’ is played. If T >= 10 and N >= 10^500, then no amount of sabotage from a Ghost can change the outcome of this game.

See, Ghost-rascal law and the Ultimate Reality

http://prebabel.blogspot.com/2014/02/ghost-rascal-conjecture-and-ultimate.html

This is a physics law which can be verified by actual verification to any large number N. This Ghost-rascal law is, in fact, the way that nature GENERATES the standard model particles, see Chapter four of this book. For nature, T = 3 and a finite number N is enough to guarantee the generating and confining the particle zoo.

Law 4:the induction TRAIN of Law 1, 2, 3 with a large n cannot be sabotaged by any math ghost rascal.

With this ghost-rascal guarantee, there is always an N (ε) for each ε (a real number) to ensure that {N (ε) x rad (abc) ^ (1+ ε) > c} for ALL c (= a + b), N (ε) is the number of toss needed for that (ε).

The abc conjecture is now proved.

However, this Ghost-rascal law is a physics law which can be verified via physics means. Without this Ghost-rascal law, the abc conjecture cannot be proved.

 

While the human math can only be proved via the lego rules, the nature math can and sometimes only be verified via physical means. In fact, the nature math is the nature laws (same as physics laws).

But what does this abc conjecture mean in the number (or physics) system?

Equation of Wonder: bigger the ΔX, smaller the ΔP < 0.

For every c (= a + b)

Let a = d + d1 = (p^n x dp) + d1; dp is the largest prime for d; d1 is the smallest integer for a to have a d.

b = e + e1 = (q^m x ep) + e1

c = f + f1 = (w^k x fp) + f1

Then {p, q, w, dp, ep, fp, d1, e1, f1, n, m, k} are the players for the dynamics of rad (abc).

Let Q be the dynamics of rad (abc) on those players.

With ΔX (on rad (abc)), there will be a ΔQ.

Definition 2: ΔQ = | h/ ΔP|; the larger |ΔP < 0| is, the stronger the possibility that abc conjecture is true. That is, the larger |ΔP < 0| is, the smaller ΔQ is.

Now, the equation of wonder can be rewritten as:

ΔQ = h/ ΔX or (ΔQ x ΔX = h), h is a real number and should be a constant.

|ΔP < 0| = h/ ΔQ is the possibility of whether there is infinite SC {sufficient condition (SC): rad (abc) = pqw (dp x ep x fp) < c} for an arbitrary c (= a + b).

That is, |ΔP < 0| = h/ ΔQ really defines the internal radical/prime dynamics for SC?

The equation {ΔQ x ΔX = h} shows that ΔQ (internal radical/prime dynamics) is confined by ΔX (the cheating weight).

More info about this Equation of Wonder, see the derivation of physics uncertainty equation via the number system at {Multiverse bubbles are now all burst by the math of Nature, http://prebabel.blogspot.com/2013/10/multiverse-bubbles-are-now-all-burst-by.html }.

评论:

章锋:
怎么比望月新一的简单那么多啊

坚石:
宇宙本身的证明,说简单也简单,说复杂也复杂。

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