We have considered the natural numbers (A=Z-Y) and (B=Z-X) in the equation X n +Y n =Z n and we have found out two each methods about Fermat's Last Theorem proof.Jae Yul LeeYou Jin LeeManagement Decision
A simplified proof of a theorem on the difference of the moore–penrose inverses of two positive semi–difinte matricesgeneralized inverseMoore–Penrose inversePositive semi–definite matrixSimplified proofs are given of a standard result that establishes positive semidefiniteness of the difference of the...
whereu(t,x)andρ(t,x)are time-dependent functions on the unit circleS=R/Z, the real dimensionless constantσ∈Ris a parameter which provides the competition, or balance, in fluid convection between nonlinear steepening and amplification due to stretching.μ(u)=∫Sudxdenotes its mean andγi...
the protocol runs in two phases. In the first phase, each party generates one group element and corresponding NIZK proof and sends them – along with itsID– to the other party. In the second phase, upon receiving the first message, both parties...
According to Earnshaw’s theorem, there is no free-space point of stable equilibrium for a paramagnet or ferromagnet in a magnetic field [31]. Still, stability can be attained if the magnet is constrained to move in one or two dimensions. Here, in order to allow movement in two dimensions...
Theorem 1 The function \(Hd(L_{P})\) is bounded, a strictly monotonically increasing and concave function, where \(Hd(L_{P})\) is about independent constant variable x. Proof Because \(0\le \sharp L_{P}\le 2\tau +1\), then \(\frac{\sharp L_{P}^{2}-1}{(2\tau +1)^...
Thus a plane-wave-like pump beam is needed for a larger violation of local realism theorem. The less precise alignment between A2 and A3 will lead to a less overlapping between the slit states of the down-converted two photons. However, such influence only causes a reduction in the ...
As noted in the proof of Theorem 7, the only jobs that may have a damage ratio above 1 are those remaining assigned to the fast machine. We have DRmax(q)≤a2+b1a2+b2. It is sufficient to prove a2+b1a2+b2≤2s2s−1, or equivalently (2s−1)b1≤a2+2sb2. By s(a1+b1)≤b2...
Proof of Theorem 2 This follows the same argument as Proof of Theorem 1, except that whenever it happens that a=b, or a>b, on the following iterate the vector ||Xi|| is bounded according to Lemma 6. Since by Lemma 5 these conditions occur on average 1 / 3 of the time, the resul...
Proof Refer to Appendix 2. Although the constraints in the subproblem (P2-1) are convex according to Theorem 2, the objective function is still nonconvex. To solve this issue, we transform the objective function into a concave one by using the fractional programming technique developed in [22...