Example 1.1. Two R-automorphisms of C are the identity z 7→ z and complex conjugation z 7→ z. We will show they are the only ones. If σ : C→ C is an R-automorphism, then for any real a and b we have σ(a+ bi) = σ(a) +σ(b)σ(i) = a+ bσ(i), so σ is determined by σ(i) and i = −1 =⇒ σ(i) = σ(−1) =⇒ σ(i) = −1 =⇒ σ(i) = ±i. If σ(i) = i, then σ(z) = z for all z ∈ C and if σ(i) = −i, then σ(z) = z for al...

In this research, Quantum-mechanical calculations were performed at the HF method with the 6-31+G*basis set and at the B3LYP method with the 6-31+G* basis set in the gas phase and five solvents such as water, DMSO, methanol, ethanol and dichloromethane at six temperatures. According to these theoretical results of IR, we extracted thermo chemical parameters such as enthalpy (∆H Kcal/mol), Gibbs...

F theory and M theory are formulated as gauge theories of area preserving diffeomorphism algebra. Our M theory is shown to be 1-brane formulation rather than 0-brane formulation of M theory of Banks, Fischler, Shenker and Susskind and the F theory is shown to be 1-brane formulation rather than -1-brane formulation of type IIB matrix theory of Ishibashi, Kawai, Kitazawa and Tsuchiya. Area preser...

A signed graph Σ=(G,σ) is a where the function σ assigns either 1 or −1 to each edge of simple G. The adjacency matrix Σ, denoted by A(Σ), defined canonically. In recent paper, Wang et al. extended spectral bounds Hoffman and Cvetković for chromatic number graphs. They proposed an open problem related balanced clique largest eigenvalue graph. We solve strengthened version this problem. As bypro...

A famous result by McKinsey and Tarski ([5] and [6]) provides an interpretation of intuitionistic logic into the modal logic S4. The algebras for S4 are Boolean algebras with an interior operator. One wonders if it is possible to perform a similar construction starting from a logic which is different from classical logic, say a logic extending Full Lambek Calculus FL. The algebraic counterpart ...

70 100 100 100 100 h Discrete: p (n) 1 = 6 12 = 0.5 p (n) 1|1 = 3 3 = 1, p (n) 1|2 = 3 3 = 1, p (n) 1|3 = 0 3 = 0, p (n) 1|4 = 0 3 = 0 Continuous: p (n) 1 = 1 12 (σ(100) + σ(70) + . . .+ σ(−70) + σ(−100)) ≈ 0.5 p (n) 1|1 = 1 3 (σ(100) + σ(70) + σ(100)) ≈ 1 p (n) 1|2 = 1 3 (σ(100) + σ(70) + σ(100)) ≈ 1 p (n) 1|3 = 1 3 (σ(−100) + σ(−70) + σ(−100)) ≈ 0 p (n) 1|4 = 1 3 (σ(−100) + σ(−70) + σ(−100)) ...