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The first case (case I) is to show that there are no primitive solutions (x, y, z) to the equation x p + y p = z p under the condition that p does not divide the product xyz. The second case (case II) corresponds to the condition that p does divide the product xyz. Since x, y, and z are pairwise coprime, p divides only one of the three numbers.
Here a B-valued relation on V B is a function from V B × V B to B. To avoid confusion with the usual equality and membership, these are denoted by ‖ x = y ‖ and ‖ x ∈ y ‖ for x and y in V B. They are defined as follows: ‖ x ∈ y ‖ is defined to be Σ t ∈ Dom(y) ‖ x = t ‖ ∧ y(t) ("x is in y if it is equal to something in y").
Logical equality is a logical operator that compares two truth values, or more generally, two formulas, such that it gives the value True if both arguments have the same truth value, and False if they are different.
unstrict inequality signs (less-than or equals to sign and greater-than or equals to sign) 1670 (with the horizontal bar over the inequality sign, rather than below it) John Wallis: 1734 (with double horizontal bar below the inequality sign) Pierre Bouguer
Define e x as the value of the infinite series = =! = + +! +! +! + (Here n! denotes the factorial of n. One proof that e is irrational uses a special case of this formula.) Inverse of logarithm integral.
X and Y are two whole numbers greater than 1, and Y > X. Their sum is not greater than 100. S and P are two mathematicians (and consequently perfect logicians); S knows the sum X + Y and P knows the product X × Y. Both S and P know all the information in this paragraph. In the following conversation, both participants are always telling the truth:
The predicate calculus goes a step further than the propositional calculus to an "analysis of the inner structure of propositions" [4] It breaks a simple sentence down into two parts (i) its subject (the object (singular or plural) of discourse) and (ii) a predicate (a verb or possibly verb-clause that asserts a quality or attribute of the object(s)).
After Hilbert's tenth problem was solved in 1970, B. F. Caviness observed that the use of e x and ln 2 could be removed. [3] Wang later noted that under the same assumptions under which the question of whether there was x with A(x) < 0 was insolvable, the question of whether there was x with A(x) = 0 was also insolvable.