- #1
Math100
- 780
- 220
- Homework Statement
- Establish the following statement:
Each integer n>11 can be written as the sum of two composite numbers.
[Hint: If n is even, say n=2k, then n-6=2(k-3); for n odd, consider the integer n-9.]
- Relevant Equations
- None.
Proof:
Suppose n is an integer such that ## n>11 ##.
Then n is either even or odd.
Now we consider these two cases separately.
Case #1: Let n be an even integer.
Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##.
Consider the integer ## n-6 ##.
Note that ## n-6=2k-6 ##
=## 2(k-3) ##.
This means ## n=2(k-3)+6 ##
=## 2m+6 ##,
where ## m=k-3 ## is an integer.
Thus, ## 2m ## and ## 6 ## are two composite numbers.
Case #2: Let n be an odd integer.
Then we have ## n=2k+1 ## for some ## k\in\mathbb{z} ##.
Consider the integer ## n-9 ##.
Note that ## n-9=2k+1-9 ##
=## 2k-8 ##
=## 2(k-4) ##.
This means ## n=2(k-4)+9 ##
=## 2n+9 ##,
where ## n=k-4 ## is an integer.
Thus, ## 2n ## and ## 9 ## are two composite numbers.
Therefore, each integer ## n>11 ## can be written as the sum of two composite numbers.
Suppose n is an integer such that ## n>11 ##.
Then n is either even or odd.
Now we consider these two cases separately.
Case #1: Let n be an even integer.
Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##.
Consider the integer ## n-6 ##.
Note that ## n-6=2k-6 ##
=## 2(k-3) ##.
This means ## n=2(k-3)+6 ##
=## 2m+6 ##,
where ## m=k-3 ## is an integer.
Thus, ## 2m ## and ## 6 ## are two composite numbers.
Case #2: Let n be an odd integer.
Then we have ## n=2k+1 ## for some ## k\in\mathbb{z} ##.
Consider the integer ## n-9 ##.
Note that ## n-9=2k+1-9 ##
=## 2k-8 ##
=## 2(k-4) ##.
This means ## n=2(k-4)+9 ##
=## 2n+9 ##,
where ## n=k-4 ## is an integer.
Thus, ## 2n ## and ## 9 ## are two composite numbers.
Therefore, each integer ## n>11 ## can be written as the sum of two composite numbers.