- #1
mliuzzolino
- 58
- 0
Homework Statement
Let S = {(a, b): a, b [itex]\in \mathbb{Z}[/itex] and b ≠ 0}. An equivalence relation "~" on S is defined by (a, b) ~ (c, d) iff ad = bc.
1) For any b [itex]\in \mathbb{Z} \setminus[/itex] {0}, show that [0/b] = [0/1] and [b/b] = [1/1].
2) For any a, b [itex]\in \mathbb{Z}[/itex] with b ≠ 0, show that [a/b] + [0/1] = [a/b] and [a/b][1/1] = [a/b].
Homework Equations
The Attempt at a Solution
1) First, show that [0/b] = [0/1].
For any value of b not zero, [0/1] is contained in the equivalence class when b = 1: Eb = {[0/b]: b ≠ 0}. Therefore, [0/b] = [0/1].
Next, show that [b/b] = [1/1].
For any value of b not zero, [1/1] is contained in the equivalence class when b = 1: ESUB]b[/SUB] = {[b/b]: b ≠ 0}. Therefore, [b/b] = [1/1].
2) First show that [a/b] + [0/1] = [a/b].
[a/b] + [0/1] = [itex]\frac{a*1 + 0*b}{b * 1}[/itex] = [itex]\frac{a + 0}{b}[/itex] = [a/b].
Next, show that [a/b][1/1] = [a/b].
[a/b][1/1] = [itex]\frac{a(1)}{b(1)}[/itex] = [a/b].
For 1), I'm not exactly sure that I'm doing this correctly. It seems like I should be using this equivalence class as the argument, but I'm not sure I set it up correctly.
For 2) this seems too simple. Am I missing something? Should I be adding the axioms used at each step?