Search results
Results From The WOW.Com Content Network
In the 10th century, the Iraqi mathematician Al-Hashimi worked with numbers as such, called "lines" but not necessarily considered as measurements of geometric objects, to prove algebraic propositions concerning multiplication, division, etc., including the existence of irrational numbers. [11]
One of the widely used types of impossibility proof is proof by contradiction.In this type of proof, it is shown that if a proposition, such as a solution to a particular class of equations, is assumed to hold, then via deduction two mutually contradictory things can be shown to hold, such as a number being both even and odd or both negative and positive.
Group (mathematics) Halting problem. insolubility of the halting problem; Harmonic series (mathematics) divergence of the (standard) harmonic series; Highly composite number; Area of hyperbolic sector, basis of hyperbolic angle; Infinite series. convergence of the geometric series with first term 1 and ratio 1/2; Integer partition; Irrational ...
Goldbach’s Conjecture. One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes ...
Reverse mathematics is a program in mathematical logic that seeks to determine which axioms are required to prove theorems of mathematics. [5] The field was founded by Harvey Friedman . Its defining method can be described as "going backwards from the theorems to the axioms ", in contrast to the ordinary mathematical practice of deriving ...
One of many examples from algebraic geometry in the first half of the 20th century: Severi (1946) claimed that a degree-n surface in 3-dimensional projective space has at most (n+2 3 )−4 nodes, B. Segre pointed out that this was wrong; for example, for degree 6 the maximum number of nodes is 65, achieved by the Barth sextic , which is more ...
To prove that π is transcendental, we prove that it is not algebraic. If π were algebraic, π i would be algebraic as well, and then by the Lindemann–Weierstrass theorem e π i = −1 (see Euler's identity) would be transcendental, a contradiction. Therefore π is not algebraic, which means that it is transcendental.
Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9. [3] Proof: Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these ...