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MIllenium prize problems.

BrettNortje

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"The P versus NP problem is a major unsolved problem in computer science. Informally speaking, it asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer...

...Consider the subset sum problem, an example of a problem that is easy to verify, but whose answer may be difficult to compute. Given a set of integers, does some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because the subset {−2, −3, −10, 15} adds up to zero" can be quickly verified with three additions. There is no known algorithm to find such a subset in polynomial time (there is one, however, in exponential time, which consists of 2n-n-1 tries), but such an algorithm exists if P = NP; hence this problem is in NP (quickly checkable) but not necessarily in P (quickly solvable).

An answer to the P = NP question would determine whether problems that can be verified in polynomial time, like the subset-sum problem, can also be solved in polynomial time. If it turned out that P ≠ NP, it would mean that there are problems in NP (such as NP-complete problems) that are harder to compute than to verify: they could not be solved in polynomial time, but the answer could be verified in polynomial time." ~ wikipedia

So, we need to be able to solve and check these algorithms before we set the computer up to 'do' them. this means that we need to make an algorithm that solves them quickly, quickly enough for us to verify.

If p = np, then the obvious answers are n = 1, or, that the equation is 'a whole one.' this could not be though, as, n equals a prime number, and one is not a prime number, thus the prime working out to be equal to 'one' would be where p = 0.5, or, [p] = anything to the power of or divided by one.

Seeing as how [p] needs to equal [np] and anything times by [n] or [p] needs to equal [p], then the answer comes with a 'to the power of,' or 'divided by' attached to it.

This means that we need to find the equation that [p] hides in. this would mean that it might be a remainder too! well, i am considering all aspects of it at this point. you could also say [n] = [1], times [1] or even divided by [1]. saying that all together will mean making a new symbol though.
 
Of course, it might mean that n = [1x], which would be an equation inside of an equation. of course, that could mean there are more of these out there, somewhere...
 
"In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and the complex numbers with real part 1/2. It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.

The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann hypothesis, along with Goldbach's conjecture, is part of Hilbert's eighth problem in David Hilbert's list of 23 unsolved problems; it is also one of the Clay Mathematics Institute's Millennium Prize Problems.

The Riemann zeta function ζ(s) is a function whose argument s may be any complex number other than 1, and whose values are also complex. It has zeros at the negative even integers; that is, ζ(s) = 0 when s is one of −2, −4, −6, .... These are called its trivial zeros. However, the negative even integers are not the only values for which the zeta function is zero. The other ones are called non-trivial zeros. The Riemann hypothesis is concerned with the locations of these non-trivial zeros, and states that:

The real part of every non-trivial zero of the Riemann zeta function is 1/2.

Thus, if the hypothesis is correct, all the non-trivial zeros lie on the critical line consisting of the complex numbers 1/2 + it, where t is a real number and i is the imaginary unit." ~ wikipedia

If this is correct, then it why does every odd number have a 0.50 added to it? this would mean that there is a zero at every odd number. this would be where xxx7 halved [1/2] = 3.500... if it is any of the prime numbers, then it would be [3=] 1.50, [5=] 2.50, [7=] 3.50, [9=] 4.50 and [1=] 0.50.

All prime numbers end in either a one, three, five seven or nine. this is evident with all prime numbers that they will end in halves, or, point five if halved.
 
"In mathematical physics, the Yang–Mills existence and mass gap problem is an unsolved problem and one of the seven Millennium Prize Problems defined by the Clay Mathematics Institute, which has offered a prize of US$1,000,000 to the one who solves it.

The problem is phrased as follows:[1]

Yang–Mills Existence and Mass Gap. Prove that for any compact simple gauge group G, a non-trivial quantum Yang–Mills theory exists on 4 R and has a mass gap Δ > 0. Existence includes establishing axiomatic properties at least as strong as those cited in Streater & Wightman (1964), Osterwalder & Schrader (1973) and Osterwalder & Schrader (1975).

In this statement, a Yang–Mills theory is a non-Abelian quantum field theory similar to that underlying the Standard Model of particle physics; 4 R is Euclidean 4-space; the mass gap Δ is the mass of the least massive particle predicted by the theory.

Therefore, the winner must prove that:

Yang–Mills theory exists and satisfies the standard of rigor that characterizes contemporary mathematical physics, in particular constructive quantum field theory,[2][3] and

The mass of the least massive particle of the force field predicted by the theory is strictly positive.

For example, in the case of G=SU(3)—the strong nuclear interaction—the winner must prove that glueballs have a lower mass bound, and thus cannot be arbitrarily light." ~ wikipedia.

This seems to me to be like radius to the power of four, would leave, at the power of 1, would mean that it would be double the radius, or, the diameter, at the power of three would be twice the volume, and, at the power of four would be four times the volume, leaving [4R]. this would mean that the radius is leaving [4] x [360 degrees] radius, but, that would mean that this is also radius to the power [4], coming to [4 degrees].
 
"The Navier–Stokes existence and smoothness problem concerns the mathematical properties of solutions to the Navier–Stokes equations, one of the pillars of fluid mechanics (such as with turbulence). These equations describe the motion of a fluid (that is, a liquid or a gas) in space. Solutions to the Navier–Stokes equations are used in many practical applications. However, theoretical understanding of the solutions to these equations is incomplete. In particular, solutions of the Navier–Stokes equations often include turbulence, which remains one of the greatest unsolved problems in physics, despite its immense importance in science and engineering.

Even much more basic properties of the solutions to Navier–Stokes have never been proven. For the three-dimensional system of equations, and given some initial conditions, mathematicians have not yet proved that smooth solutions always exist, or that if they do exist, they have bounded energy per unit mass.[citation needed] This is called the Navier–Stokes existence and smoothness problem.

Since understanding the Navier–Stokes equations is considered to be the first step to understanding the elusive phenomenon of turbulence, the Clay Mathematics Institute in May 2000 made this problem one of its seven Millennium Prize problems in mathematics. It offered a US$1,000,000 prize to the first person providing a solution for a specific statement of the problem:[1]

Prove or give a counter-example of the following statement:

In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations." ~ wikipedia.

This would be [velocity] - [scalar pressure field] = rough and jagged. this is because the pressure is influenced by the velocity of the gas or liquid.

If it doesn't happen on earth, it won't happen in space. there is 'gravity' out there, coming from orbital bodies. this is evident with jupiter an saturn.

Of course, if there is gravitational pull, or whatever you want to call it, the liquids or gases will not be smooth. this is because they will need velocity to get them into space or leave a container. this is something it cannot do without, and, as we see on earth, the patterns of these things varies from glob to glob.
 
If you can prove that all the zeros of the Riemann zeta function lie on the critical line, using the Hardy-Littlewood conjectures, I'll give you $5.
 
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