The abc conjecture: there are a finite number of c (= a +
b), for
c > rad (abc)^ (1+ ε)
Or
c < K ε x rad (abc)^ (1+ ε) for ALL c.
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, 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
The +/- can be replaced with + only while push – into the d1
(or f1, e1) if necessary.
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.
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
(meeting scenario 1), S2, S3 and S4.
Let S = S1 + S2 + S3 + S4
Now, {rad (abc) < c} = {for any c (not a prime, f1 = 0),
is S finite?}
A cop-out way for the answer is by tossing a coin (selecting
an arbitrary c): head = true; tail = false.
Then, the P (S) = {tail (50%), head (50%)} after infinite
many tosses.
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}, the P (S) = {tail (< 50%), head (> 50%)} of each toss.
After N tosses, P (S, N) = {tail (~ 0), head (~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).
The above process can be proved in four steps: induction
(operational) progressive process.
First, making the above simple tossing (selecting an
arbitrary c doing the actual search) process into a game, as follow.
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}
Theorem 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
First principle (the indeterminacy): when ΔX = 0, the
average of ΔP = 0 after n games (n x T tosses) when n is a large number.
Theorem 2: 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 theorem 1 and 2: in the above,
we have proved n = 1 and n = n. Now, by proofing that n = n + 1 is true, the
theorem is proved with induction.
Fourth, going beyond the induction: is there a math ghost
rascal which can sabotage the above induction proof?
The answer is no: a cheating game cannot be sabotaged even
by a ghost rascal; see the ghost rascal conjecture.
Ghost-rascal
conjecture --- 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 conjecture and the Ultimate Reality
That is, the induction proof of theorem 2 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.
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.
Corollary 1: 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 ΔP (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
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