Prove that for every integer n>=8, there exists nonnegative integers a and b, such that n =3a+5b
The Attempt at a Solution
I'm trying to understand the proof of this. It goes as follows:
I am having a hard time figuring out what is going on in any of the three cases.
Firstly, why did they pick 3q, 3q+1, and 3q+2?
They could do this proof using n = 5q, 5q+1, 5q+2, 5q+3, 5q+4 also correct? It would just have more cases, and therefor, be more work?
Then, about the cases.
Case I: n = 3q
I see what they did here. Since they picked to represent their integer as 3q, they supposed that b = 0, and then it's easy to show there exists an a, since a = q at that point. And a >=3 because n = 3q needs to be greater than 8.
Case II: n = 3q+1
Here is where I am troubled.
How do they make the jump to claim that n = 3(q − 3) + 10 ??
What is the thought process behind this?
I would appreciate any explanation that you could provide. Thanks!