- 北有云溪
-
1: definition of proposition N by integer set of elements in the collection, after four operations are all the even number greater than 2, the set of all even numbers called corresponding solution; after operation is a even number element consisting of the set, called the even-numbered corresponding solutions;
1 Proposition definition N: the set composed of integers, the integers can be convert to all greater than 2even numbers using arithmetic operation, is called the corresponding solution of all even numbers . The set of the composition after computing is a corresponding even is called the corresponding even solution.
Note: four operations, any element in the process of operation can use unlimited. In case of elements:
Note:
Arithmetic operation: any elements in the operation can be used times without limitation . For example, element multiplication:
(1) a collection of elements, if:2x 4= 8;2x 2x 2 =8(2is multiplied by itself. ). So,8 corresponding solutions to {2,4}.
(1) An element of a set, if2x4 = 8;2x2x2 = 8(2 multiplied by itself So, the corresponding ).
(2) each even-numbered corresponding solutions ( set ) and operation. That was all even-numbered corresponding solutions.
(2) All even corresponding solution can be got from each even corresponding solution ( set ) operation.
2 for a meet the following two conditions proposition N solution
2 Determine the solution for meet the following two conditions of proposition N.
(1) any even number greater than 2can be expressed as two mutually quality number elements and form or a prime number with its own form. Algebraic form:2A = B+C;2A = A+A
(1) Any even number greater than 2can be expressed as a form of two relatively prime number elements addition or a form of prime number adding itself . The algebra forms are:2A = B+C;2A = A+A.
(2) all the even number greater than 2can be expressed as the form of elements.
(2) All of the even number greater than 2can be expressed as a form of elements
3reasoning process
3 Reasoning explanation
(1) elements for all integer greater than 1, by removing the portion of the element method for solving. The first of an even number of corresponding solutions, the solution is removed ( sets ) of elements. Then each even number corresponding to the set and operation, obtained by seeking the solution.
(1) We could solve the question using removing part of the element, if all elements of the integer are greater than1. First ask for a solution corresponding an even; and take out part element of the solution ( set ). Then get the solution after operating each corresponded set of the even numbers.
(2) every even number corresponding to the solution of the elements can be removed for: remove the even not coprime numbers.
This is because: meet the required conditions (1), these elements cannot be expressed as the sum of two prime numbers or even two mutual prime number add, remove;
Meet the required conditions (2), the number can be expressed as the form of the quality factor. Don"t need this even not mutually prime elements, remove.
(2) The element, which can be removed from each of the solution of the corresponding even, Delate all the composite numbers not co-prime to the number.
When the required conditions (1) was fit, these elements can " t be expressed as two evens adding or two relatively prime number elements adding and it can be removed;
When the required conditions (2) was fit, the even number can be expressed as a form of its prime factor elements multiplication . This element of the prime factor elements multiplication is not needed, so it can be removed.
1定义命题:由整数组成的集合,集合中元素经过四则运算得到全部大于2的偶数,该集合称为全体偶数对应的解经过运算等于某一偶数的元素所组成的集合,称为该偶数对应的解;
1个命题的定义:集整数,整数可以转换所有大于2的偶数使用算术运算,称为相应的解决方案,所有偶数。本集的组成后,计算相应的甚至被称为相应的均匀溶液。
说明:进行四则运算,任何元素在运算过程中都可以无限次使用以元素相乘为例:
注:
算术操作中的任何元素的操作可以使用次数没有限制。例如,元素的乘法:
(1)一个集合的元素,如果:2×4 =8;2×2×2=8(2自身相乘)。所以,8对应的解为{}4。
(1)一个集合的元素,如果工作=8;2x2x 2=8(2乘以本身)。所以,相应的
(2)每个偶数对应的解(集合进行并运算即得全体偶数对应的解)。
(2)所有相应的解决方案,可以得到从每个相应的解决方案(套)操作。
2求一个满足以下两个条件的命题氮的解
2确定的解决方案,满足以下条件命题n .
(1)任意大于2的偶数可以表示为两个互质数元素和的形式或某一质数加自身的形式。代数形式为:甲=乙+一+=;2
(1)任何大于2的偶数都可以表示为一个二维形式相对素数元素或某种形式的素数也加入。代数形式:甲=二+三;2= + A
(2)全部大于2的偶数都可以表示为元素相乘的形式。
(2)所有的大于2的偶数都可以表示为一种元素
3推理过程
3种解释
(1)元素为全体大于1的整数时,采取去掉部分元素的方法求解。首先求一个偶数对应的解,去掉该解(集合中的部分元素然后每个偶数对应的集合进行并运算,得到所求的解)。
(1)我们可以解决这个问题用拆卸元素的一部分,如果所有元素的整数是大于1。先要解决相应的一个;并采取了部分元素的溶液(集)。然后得到解决后操作每一个对应的偶数。
(2)每一偶数对应的解中可以去掉的元素为:去掉该偶数的不互素合数。
这是因为:满足所要求的条件(1)时,这些元素不能表达为该偶数的两质数相加或两互质数相加,所以去掉;
满足所求的条件(2)时,该偶数可以表达为其质因数相乘的形式不需要这个偶数的不互素的元素,所以去掉。
(2)元素,可以被删除从每个解决了相应的均匀,等所有合成数不互质的数。
当所需的条件(1)符合这些要素,不能表示为两事件增加或相对素数的元素加入,它可以被删除;
当所需的条件(2)适合,甚至都可以表示为一个素数因子元素相乘的形式。这一部分的主要因素元素乘法是不需要的,以便它可以被删除。
- 西柚不是西游
-
ttttttt