80-310/610 Formal Logic Fall 2015 Homework 3
... ¬(p ↔ q), ((p → q) → p) → p Use these normal forms to determine whether each formula is a tautology. 3. (a) Show that all of the truth functions (on {0, 1}) can be defined in terms of {→, ⊥}, i.e. that this is a functionally complete set of connectives. (b) Show that {→, ∨, ∧} is not a functionally ...
... ¬(p ↔ q), ((p → q) → p) → p Use these normal forms to determine whether each formula is a tautology. 3. (a) Show that all of the truth functions (on {0, 1}) can be defined in terms of {→, ⊥}, i.e. that this is a functionally complete set of connectives. (b) Show that {→, ∨, ∧} is not a functionally ...
An alternative quadratic formula
... Trigonometric and hyperbolic methods for solving quadratic equations are also circulating: To explain this idea we assume, without loss of generality, that a > 0 in (1). Then, we distinguish the cases c > 0 and c < 0 (the case c = 0 is trivial). p 1. Case. If c > 0 we bring equation (1) by a scaling ...
... Trigonometric and hyperbolic methods for solving quadratic equations are also circulating: To explain this idea we assume, without loss of generality, that a > 0 in (1). Then, we distinguish the cases c > 0 and c < 0 (the case c = 0 is trivial). p 1. Case. If c > 0 we bring equation (1) by a scaling ...
Steps for solving an Empirical Formula Problem
... Steps for solving an Empirical Formula Problem 1. If given % data, assume 100g of sample. If given grams, then use grams and proceed to #2. 2. Convert g to moles for each element in the compound 3. Divide each of these mole amounts by smallest mole amount calculated in #2 Then, if necessary, multipl ...
... Steps for solving an Empirical Formula Problem 1. If given % data, assume 100g of sample. If given grams, then use grams and proceed to #2. 2. Convert g to moles for each element in the compound 3. Divide each of these mole amounts by smallest mole amount calculated in #2 Then, if necessary, multipl ...
Stirlings Formula
... However!! We not only produced a simple approximation for x!, but turned a discrete function having values for integers only, into a continuous function, giving numbers for something like 3,141! - which may or may not make sense. This may have dire consequences. Using the Strirling formula you may, ...
... However!! We not only produced a simple approximation for x!, but turned a discrete function having values for integers only, into a continuous function, giving numbers for something like 3,141! - which may or may not make sense. This may have dire consequences. Using the Strirling formula you may, ...