Consider the set ℕ× ℕ. We will informally think of the pair (x, y) ∈ ℕ× ℕ as representing the integer x − y. Of course, there are many different ways of representing the same integer. To fix this, we define a relation ∼ on ℕ × ℕ by (x, y) ∼ (n, m) if and only if x + m = n + y. a) Prove that ∼ is an equivalence relation. b) We shall abbreviate ℕ× ℕ / ∼ by the symbol ℤ*. Prove that the function f : ℤ* → ℤ given by f([x, y]) = x − y is well-defined and bijective.
Consider the set ℕ× ℕ. We will informally think of the pair (x, y) ∈ ℕ× ℕ as representing the integer x − y. Of course, there are many different ways of representing the same integer. To fix this, we define a relation ∼ on ℕ × ℕ by (x, y) ∼ (n, m) if and only if x + m = n + y. a) Prove that ∼ is an equivalence relation. b) We shall abbreviate ℕ× ℕ / ∼ by the symbol ℤ*. Prove that the function f : ℤ* → ℤ given by f([x, y]) = x − y is well-defined and bijective.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the set ℕ× ℕ. We will informally think of the pair (x, y) ∈ ℕ× ℕ as representing the integer x − y. Of course, there are many different ways of representing the same integer. To fix this, we define a relation ∼ on ℕ × ℕ by (x, y) ∼ (n, m) if and only if x + m = n + y.
a) Prove that ∼ is an equivalence relation.
b) We shall abbreviate ℕ× ℕ / ∼ by the symbol ℤ*. Prove that the function f : ℤ* → ℤ given by f([x, y]) = x − y is well-defined and bijective.
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