Thread: induction?

Show that every integer greater than 11 is the sum of two composite integers.

Do I need to use induction for this question? I've tried, but it's not working.
Here's my work so far:

Let P(n): n=a+b for n>11, where a and b are composite integers.

Initial step: P(12) says that 12 is the sum of two composite integers because 12=4+8=6+6 where 4, 6 and 8 are all composite integers.

Induction step: Let k>11 and suppose P(k) is true and k=a+b.
We want to show that P(k+1) is true.
k+1=a+b+1

this is what i've tried, but actually I'm not sure if I'm on the right track. Just give me some suggestions. Thanks very much !!!

Originally Posted by suedenation

Show that every integer greater than 11 is the sum of two composite integers.

Do I need to use induction for this question? I've tried, but it's not working.
Here's my work so far:

Let P(n): n=a+b for n>11, where a and b are composite integers.

Initial step: P(12) says that 12 is the sum of two composite integers because 12=4+8=6+6 where 4, 6 and 8 are all composite integers.

Induction step: Let k>11 and suppose P(k) is true and k=a+b.
We want to show that P(k+1) is true.
k+1=a+b+1

this is what i've tried, but actually I'm not sure if I'm on the right track. Just give me some suggestions. Thanks very much !!!

Cool problem almost like the Goldbach conjecture.
---
Consider 2 cases for n>11.
~~~
n is even:
Then, for n=12 we can write,
n=4+8 both of which are composites and even.
Now, we prove that n+2 can be expressed as even composites.
Thus, for some n>11 even we have,
n=a+b where a and b are even composites then,
n+2=a+(b+2) where a is an even composite and (b+2)>b>2 is an even composite. Thus, this shows by induction a stronger case that any even number can be expressed as a sum of two even composites (not just composites) for n>11.
~~~
n is odd:
Then, for n>11 we can automaticall write,
n=9+(n-9) where 9 is a composite and,
n-9>2 because n>11 thus it is an even number greater then two and thus a composite.


Note: This was not completely a math induction problem only the part of it was.