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Rules: (1) (a,b)+(c,d)=(ad+bc,bd) (2) (a,b)·(c,d)=(ac,bd) Suppose (a, b) and (c, d) are equivalent in the sense defined above for fractions and that (x, y) is some other fraction: (1) Show that (a, b) + (x, y) and (c, d) + (x, y) are equivalent (2) Show that (a, b) · (x, y) and (c, d) · (x, y) are equivalent

Rules: (1) (a,b)+(c,d)=(ad+bc,bd) (2) (a,b)·(c,d)=(ac,bd) Suppose (a, b) and (c, d) are equivalent in the sense defined above for fractions and that (x, y) is some other fraction: (1) Show that (a, b) + (x, y) and (c, d) + (x, y) are equivalent (2) Show that (a, b) · (x, y) and (c, d) · (x, y) are equivalent

Advanced Engineering Mathematics
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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Rules: (1) (a,b)+(c,d)=(ad+bc,bd) (2) (a,b)·(c,d)=(ac,bd) Suppose (a, b) and (c, d) are equivalent in the sense defined above for fractions and that (x, y) is some other fraction: (1) Show that (a, b) + (x, y) and (c, d) + (x, y) are equivalent (2) Show that (a, b) · (x, y) and (c, d) · (x, y) are equivalent
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Step 1

Solution:

Let us consider, (a,b)+(c,d)=(ad+bc,bd)  and   (a,b)·(c,d)=(ac,bd).

Let us also consider, (a, b) and (c, d) are equivalent in the sense defined above for fractions.

Therefore, (a, b) and (c, d) are equivalent if and only if there exists a k,k such that a,b.k,k=c,d holds.

We now have to show that, 

(1) (a, b) + (x, y) and (c, d) + (x, y) are equivalent

(2) (a, b) · (x, y) and (c, d) · (x, y) are equivalent.

Let us now consider, (a, b) and (c, d). So, there exists a k,k such that a,b.k,k=c,d holds.

Therefore, a,b.k,k=c,d ⇒ ak,bk=c,d ⇒ c=ak, d=bk

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