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For example, we can prove by induction that all positive integers of the form 2n − 1 are odd. Let P ( n ) represent " 2 n − 1 is odd": (i) For n = 1 , 2 n − 1 = 2(1) − 1 = 1 , and 1 is odd, since it leaves a remainder of 1 when divided by 2 .
The difference of two squares can also be illustrated geometrically as the difference of two square areas in a plane.In the diagram, the shaded part represents the difference between the areas of the two squares, i.e. .
For example, by Gödel's incompleteness theorem, we know that any consistent theory whose axioms are true for the natural numbers cannot prove all first-order statements true for the natural numbers, even if the list of axioms is allowed to be infinite enumerable. It follows that an automated theorem prover will fail to terminate while ...
There exists still another way to approach the fundamental theorem of algebra, due to J. M. Almira and A. Romero: by Riemannian geometric arguments. The main idea here is to prove that the existence of a non-constant polynomial p(z) without zeros implies the existence of a flat Riemannian metric over the sphere S 2. This leads to a ...
An archetypal double counting proof is for the well known formula for the number () of k-combinations (i.e., subsets of size k) of an n-element set: = (+) ().Here a direct bijective proof is not possible: because the right-hand side of the identity is a fraction, there is no set obviously counted by it (it even takes some thought to see that the denominator always evenly divides the numerator).
In mathematics, the Hodge index theorem for an algebraic surface V determines the signature of the intersection pairing on the algebraic curves C on V.It says, roughly speaking, that the space spanned by such curves (up to linear equivalence) has a one-dimensional subspace on which it is positive definite (not uniquely determined), and decomposes as a direct sum of some such one-dimensional ...
An algebraic solution of a polynomial equation is an expression involving the four basic arithmetic operations (addition, subtraction, multiplication, and division), and root extractions. Such an expression may be viewed as the description of a computation that starts from the coefficients of the equation to be solved and proceeds by computing ...
So, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers. For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6).