Definition 2.1. When a and b are integers, we say a divides b if b = ak for some k ∈ Z. We then write a | b (read as “a divides b”). Example 2.2. We have 2 | 6 (because 6 = 2 · 3), 4 | (−12), and 5 | 0. We have ±1 | b for every b ∈ Z. However, 6 does not divide 2 and 0 does not divide 5. Divisibility is a relation, much like inequalities. In particular, the relation 2 | 6 is not the number 3, e...

Coffee (Coffea arabica L.) seeds are sensitive to desiccation and oxidative stress during drying processes. We investigated the effect of drying and moisture levels on germination-related gene expressions associated with enzymatic systems that prevent oxidative stress in coffee seeds. Coffee seeds collected at physiological maturity were subjected to slow and quick drying to 40, 30, 20, and 12%...

the transport of dissolved contaminants in groundwater is usually described by the advectiondispersion equation with reaction. several numerical methods for solving the one-dimensional are availableincluding finite difference methods, finite volume methods, and finite element methods. stringent conditions,such as small peclet (pe) and courant (cr) numbers, must be satisfied to ensure the accura...

A pattern-based approach to the presentation, codi cation and reuse of property speci cations for nite-state veri cation was proposed by Dwyer and his colleagues in [4,3]. The patterns enable nonexperts to read and write formal speci cations for realistic systems and facilitate easy conversion of speci cations between formalisms, such as LTL, CTL, QRE. In this paper we extend the pattern system...

It is known that Dodgson’s rule is computationally very demanding. Tideman (1987) suggested an approximation to it but did not investigate how often his approximation selects the Dodgson winner. We show that under the Impartial Culture assumption the probability that the Tideman winner is the Dodgson winner tend to 1. However we show that the convergence of this probability to 1 is slow. We sug...

Two people have a full B-gallon jug of wine and two empty jugs of capacity M and S, respectively, where B= M +S. Any jug may be poured into any other jug until either the first one is empty or the second is full. What is the quickest way for them to divide the wine equally? We solve this problem and show that a natural related problem may be solved in polynomial time.

Telecommunication services are bundled together into single applications. These applications allow for connections to many servers and computers, for quick and reliable data access and for document exchange. The invoked tasks can be combined and complex, they may take place in several applications such as copying final data in spreadsheets or word processors using interapplication communication...