If you haven't read the following link, you can read it before reading on:
http://www.jiskha.com/display.cgi?id=1407879480
The idea of solving parts (a) and (b) of the linked question is to establish a methodical way to solve (c) and (d).
The typical approach to solving math questions would be to recall, reread and understand all definitions involved in the question. Relevant definitions for this case would be "equivalence class", "equivalence relation", etc.
Having done this, if you are to demonstrate a general statement and have no clue where to start, you would start with specific cases. Here, you would start with [(1,1)], [(2,2)],etc.
You can then try to use induction to generalize.
Let R be the relation on ℤ+×ℤ+ defined by (a,b)R(c,d) if and only if a−2d=c−2b.
1. Find an equivalence class that has exactly 271 elements.
2. Is it true that for every positive integer n, there is an equivalence class that contains exactly n elements? Explain.
My work for #1:
I found two equivalence classes: [(0, 272)] and [(1,271)]. The way I did this was purely through guessing and checking when a=0 and a=1. I just noticed a pattern where the number of elements is b-1 when a=0 and the number of elements is b when a=1. However, I know there has to be a better way to get this.
I did read something about taking the floor of d when c=1, but I don't think I would have been able to notice that myself, so is there another way to attempt this?
My work for #2:
I feel like my answer for this largely depends on #1 but I didn't do #1 correctly. My gut feeling is that the statement is true. But I'm not entirely sure how to go about proving it.
2 answers
To explain in more detail, we would investigate the relation:
(a-2d)=(c-2b)
Transposing terms, we get the same relation as (a+2b)=(c+2d)
Hence all elements (a,b) in the equivalence class would have the same sum a+2b.
For example, if the sum is 10, we can have 4 elements: (2,4),(4,3),(6,2),(8,1).
Check:
(2,4)R(2,4)..ok
(2,4)R(4,3)..ok
(2,4)R(8,1)..ok
... etc
(8,1)R(8,1)..ok
NOTE: recall that we are in Z+, so zeroes and negatives are not included.
By enumerating elements belonging to each sum, we can figure out the number of elements as a function of the sum, equal to a+2b.
Sum elements
3 (1,1)
4 (2,1)
5 (1,2),(3,1)
6 (2,2),(4,1)
7 (1,3),(3,2),(5,1)
8 (2,3),(4,2),(6,1)
9 (1,4),(3,3),(5,2),(7,1)
10 ...
By extending the table 10 more lines, and examining the pattern, you will be able to deduce the number of elements as a function of the sum=a+2b.
(see also link in previous post)
(a-2d)=(c-2b)
Transposing terms, we get the same relation as (a+2b)=(c+2d)
Hence all elements (a,b) in the equivalence class would have the same sum a+2b.
For example, if the sum is 10, we can have 4 elements: (2,4),(4,3),(6,2),(8,1).
Check:
(2,4)R(2,4)..ok
(2,4)R(4,3)..ok
(2,4)R(8,1)..ok
... etc
(8,1)R(8,1)..ok
NOTE: recall that we are in Z+, so zeroes and negatives are not included.
By enumerating elements belonging to each sum, we can figure out the number of elements as a function of the sum, equal to a+2b.
Sum elements
3 (1,1)
4 (2,1)
5 (1,2),(3,1)
6 (2,2),(4,1)
7 (1,3),(3,2),(5,1)
8 (2,3),(4,2),(6,1)
9 (1,4),(3,3),(5,2),(7,1)
10 ...
By extending the table 10 more lines, and examining the pattern, you will be able to deduce the number of elements as a function of the sum=a+2b.
(see also link in previous post)