true or False? Prove your answer! a) Suppose h:R→R is not one-to-one, and suppose a, b∈R with a≠b. Then h(a) =h(b). b) Suppose g:R→R is a function such that for every x∈R, there exists y∈R such that g(x)≠y . Then g is not onto.
true or False? Prove your answer! a) Suppose h:R→R is not one-to-one, and suppose a, b∈R with a≠b. Then h(a) =h(b). b) Suppose g:R→R is a function such that for every x∈R, there exists y∈R such that g(x)≠y . Then g is not onto.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter6: More On Rings
Section6.2: Ring Homomorphisms
Problem 6E
Related questions
Question
true or False? Prove your answer!
a) Suppose h:R→R is not one-to-one, and suppose a, b∈R with a≠b. Then
h(a) =h(b).
b) Suppose g:R→R is a function such that for every x∈R, there exists y∈R such that g(x)≠y . Then g is not onto.
Expert Solution
Step 1
(a) The given statement, "Suppose is not one-to-one, and suppose with then " is false.
Counter example:- Let defend by
Since function is not one-one
Take and
At ,
At ,
Here, and
Hence, statement is false.
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