The system is inconsistent, in real math-grad-speak. There is no answer. Stop trying.
I know this because I asked this bear who happens to have a masters in algebra.
hmmm, you got me wrong,
[0] is the class of numbers that divided by 2 yield 0,
[1] is the class of numbers that divided by 2 yield 1.
so, if you have, for example, 3, its class is [3] = [1], or 3 (mod 2) = 1
if you add 2 to 3, 2+3=5, [5] = [1], or 5(mod 2) = 1
any number plus two stays in its own class, so any y + 2 = y
its modular arithmetic.
Hope that clears it up
If you divide any integer A by 2, you can get two possible remainders, that is:
0, wich means that A is even (can be written as 2n, for some integer n),
we will call these numbers [0] (if A/2 yields no remainder, [A] = [0]
or 1, wich means that A is odd and can be written as 2n+1 for some integer n.
we will call these [1]
Now, lets study both possibilities:
if [A] = [0], then A =2n for some n
[A + 2] = [2n + 2] = [2(n+1)], wich is evidently even, so [A + 2] = [0] = [A]
if [B] = [1]
[A + 2] = [(2n + 1) + 2] = [2n + 2 + 1] = [2(n+1) + 1] , an even number plus one is always odd, so [B + 2] = [1] = [B]
Ehh, I guess I could figure it out if I tried for a couple minutes.
However, when it comes to math, while I'm not bad at it, I tend to steer away, especially when you go advanced on my ass ._.
Well, if you are interested, wikipedia has a pretty good article on it:
http: //en . wikipedia . org /wiki/Modular_arithmetic
Math can be discouraging, but it's very rewarding. Can change the way you see the world.
when an equation has no possible solution it basically means that the equation is "bad" or wrong. So, again, what you said and what I said is completely the same.
Actually, when solving limits in Calculus, you can substitute infinity for fractions because anything divided by infinity is effectively zero, because it is infinitesimally small.