Confusions about finding equivalence classes for the set of natural numbers corresponding to equivalence relation a+b is even.

[shivajikobardan]

 

[pka]

I find the post very hard to read.
I think it means an equivalence [imath]\mathcal{R}[/imath] on the positive integers [imath]\mathbb{Z}^+=\{1,2,3,\cdots\}[/imath]
where [imath](j,k)\in\mathcal{R}\iff~j+k\text{ is even. }[/imath]
Thus there are only two equivalent classes: [imath]\bf [1]~\&~[2][/imath], the odds & the evens.

 

[Dr.Peterson]

It would be far clearer if you would type your questions in full sentences, rather than just circle parts of what you wrote.

As I understand it, you are asking whether k should be specified as an integer or a natural number; I don't see that it matters, since all numbers involved are positive anyway. Why do you think it might matter?

Then you ask whether a, in [a], must be a natural number. Since the notation means the equivalence class containing a, a must be in the set on which the relation is defined, which is N. (At one place you called it A, which I think was an error.) Similarly, the set [a] must be a subset of the entire set, which is N, not Z. There is no need ever to have written a negative number in your work (which a reason not to use Z, though I don't think it's invalid to have done so).

 

[shivajikobardan]

It would be far clearer if you would type your questions in full sentences, rather than just circle parts of what you wrote.

As I understand it, you are asking whether k should be specified as an integer or a natural number; I don't see that it matters, since all numbers involved are positive anyway. Why do you think it might matter?

If k contains negative integers, I need to use that in calculation ie in [1]={2k-1} for example.

Then you ask whether a, in [a], must be a natural number. Since the notation means the equivalence class containing a, a must be in the set on which the relation is defined, which is N. (At one place you called it A, which I think was an error.) Similarly, the set [a] must be a subset of the entire set, which is N, not Z. There is no need ever to have written a negative number in your work (which a reason not to use Z, though I don't think it's invalid to have done so).

that A was a part of definition, here in this example A=N. I write definition and solve the problem thought that it would make clear (but it did the opposite haha).

 

[pka]

If k contains negative integers, I need to use that in calculation ie in [1]={2k-1} for example.

The point is that there are NO negative integers in what you posted.
Your post says that [imath]\mathcal{R}[/imath] is a relation on the positive integers [imath]\mathbb{N}^+[/imath].
If that is not correct then you should correct the post.

 

[shivajikobardan]

The point is that there are NO negative integers in what you posted.
Your post says that [imath]\mathcal{R}[/imath] is a relation on the positive integers [imath]\mathbb{N}^+[/imath].
If that is not correct then you should correct the post.

ok

 

[Deleted member 4993]

It will be very helpful for our tutors, if you would "skip-a-line" between the lines, in your hand-written response.

 

[Shiloh]

Or, better yet, type your response rather than hand-writing it!

Every even number can be written 2n. The sum of any two even numbers, 2n+ 2m= 2(n+ m), is always even.
Ever odd number can be written 2n+ 1. The sum of any two odd numbers, 2n+ 1+ 2m+ 1= 2n+ 2m+ 2= 2(n+ m+ 1), is always even.

The sum of an odd number and an even number, 2n+ 1+ 2m= 2(n+ m)+ 1, or 2n+ 2m+ 1= 2(n+ m)+ 1, is always odd.

The equivalence "a, b positive integers such that a+ b is even" has two equivalence classes- the set of all even positive integers and the set of all odd positive integers.