Define multiplication of equivalence classes as follows: via (zx² - {(0,0)}) × (Zײ – {(0,0)}) →→→ Zײ – {(0,0)} ([xy], [z: w]) → [x:y] · [z: w] =(([x: y], [z: w])) = [xz: yw]. . Prove is a function. 13
Define multiplication of equivalence classes as follows: via (zx² - {(0,0)}) × (Zײ – {(0,0)}) →→→ Zײ – {(0,0)} ([xy], [z: w]) → [x:y] · [z: w] =(([x: y], [z: w])) = [xz: yw]. . Prove is a function. 13
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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![Relations toward functions.
Recall, a relation RC X X Y is simply a subset of X x Y.
Recall also our construction of a relation
R := {((x1, Y₁), (x2, Y2)) ≤ (Zײ – {(0, 0)})ײ |x1Y2 = ¥1x2} C (Zײ – {(0,0)})*
(15)
toward a (successful) construction of the famous rational numbers, denoted by
Q to indicate "quotients” of integers, and that this relation happens to be an
equivalence relation. Thus, by a result you prove during this midterm exam,
the equivalence classes of RC (Zx² - {(0,0)}) *² partition Zײ - {(0,0)}.
X2
Refer to the equivalence class of an element (x, y) € Zײ – {(0,0)} via R by
[ay]. For example, the equivalence class of (1, 2) is
[1:2] = {..., (-3, −6), (-2,-4), (-1, -2), (1, 2), (2, 4), (3,6), ...}.
Define multiplication of equivalence classes as follows:
via
(Zײ – {(0,0)}) × (Zײ – {(0,0)}) — Zx² - {(0,0)}
→
([x : y], [z : w]) → [x : y] · [z : w] := ·(([x : y], [z : w])) := [xz : yw].
(16)
Prove is a function. 13
So far, there is no reason to excise the equivalence class [10] from our tentative
construction of the rational numbers. Since [10] is likened to,
we may think
of (and, indeed, any quotient of a nonzero quantity and zero) as ∞. However,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1797707a-0d96-41d6-8f45-19c20e886f5b%2Fb6b756e3-0f3e-4a9a-b85c-7a91d48f099f%2Flqrhmq_processed.png&w=3840&q=75)
Transcribed Image Text:Relations toward functions.
Recall, a relation RC X X Y is simply a subset of X x Y.
Recall also our construction of a relation
R := {((x1, Y₁), (x2, Y2)) ≤ (Zײ – {(0, 0)})ײ |x1Y2 = ¥1x2} C (Zײ – {(0,0)})*
(15)
toward a (successful) construction of the famous rational numbers, denoted by
Q to indicate "quotients” of integers, and that this relation happens to be an
equivalence relation. Thus, by a result you prove during this midterm exam,
the equivalence classes of RC (Zx² - {(0,0)}) *² partition Zײ - {(0,0)}.
X2
Refer to the equivalence class of an element (x, y) € Zײ – {(0,0)} via R by
[ay]. For example, the equivalence class of (1, 2) is
[1:2] = {..., (-3, −6), (-2,-4), (-1, -2), (1, 2), (2, 4), (3,6), ...}.
Define multiplication of equivalence classes as follows:
via
(Zײ – {(0,0)}) × (Zײ – {(0,0)}) — Zx² - {(0,0)}
→
([x : y], [z : w]) → [x : y] · [z : w] := ·(([x : y], [z : w])) := [xz : yw].
(16)
Prove is a function. 13
So far, there is no reason to excise the equivalence class [10] from our tentative
construction of the rational numbers. Since [10] is likened to,
we may think
of (and, indeed, any quotient of a nonzero quantity and zero) as ∞. However,
![we might like for our construction of the rational numbers to enjoy the arithmetic
https://en.wikipedia.org/wiki/Field(mathematics)Classicae finition.
attributes of a field, see
Is our current construction of the rational numbers (including ∞ :=
[1 : 0]), and with binary operations addition and multiplication (as
we've defined them) a field (as per the “Classic definition" written
on Wikipedia)? (Provably) verify as many axioms as possible, and
provide explicit counterexamples to any axioms which are violated. If
any of the axioms defining a field are violated due to the inclusion of
∞ := [10], is the problem remedied by simply removing ∞ := = [1 : 0]?
13 To prove is a function entails a demonstration that, for all ([x : y], [z : w]) and ([a : b], [c :
d]), if ([x : y], [z : w]) ([a b], [c d]), then ([a b], [c: d]) = ·([x : y], [z: w]). Recall our
classroom proof that the analogous
:
definition of addition of equivalence classes is a function.
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1797707a-0d96-41d6-8f45-19c20e886f5b%2Fb6b756e3-0f3e-4a9a-b85c-7a91d48f099f%2Fimp84db_processed.png&w=3840&q=75)
Transcribed Image Text:we might like for our construction of the rational numbers to enjoy the arithmetic
https://en.wikipedia.org/wiki/Field(mathematics)Classicae finition.
attributes of a field, see
Is our current construction of the rational numbers (including ∞ :=
[1 : 0]), and with binary operations addition and multiplication (as
we've defined them) a field (as per the “Classic definition" written
on Wikipedia)? (Provably) verify as many axioms as possible, and
provide explicit counterexamples to any axioms which are violated. If
any of the axioms defining a field are violated due to the inclusion of
∞ := [10], is the problem remedied by simply removing ∞ := = [1 : 0]?
13 To prove is a function entails a demonstration that, for all ([x : y], [z : w]) and ([a : b], [c :
d]), if ([x : y], [z : w]) ([a b], [c d]), then ([a b], [c: d]) = ·([x : y], [z: w]). Recall our
classroom proof that the analogous
:
definition of addition of equivalence classes is a function.
=
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