0.3 Mappings
1.
a. Not a mapping: α(1) = − 1 is not in 
.
.
c. Not a mapping: 
is not in 
.
is not in .
e. Not a mapping: α(6) = α(2 · 3) = (2, 3) and α(6) = α(1 · 6) = (1, 6).
g. Not a mapping: α(2) is not defined.
2.
a. Bijective. α(x) = α(x1) implies 3 − 4x = 3 − 4x1, so x = x1, and α is one-to-one. Given 
, 
, so α is onto.
, , so α is onto.
c. Onto: If m 
N, then m = α(2m − 1) = α(2m). Not one-to-one: In fact we have α(1) = 1 = α(2).
N, then m = α(2m − 1) = α(2m). Not one-to-one: In fact we have α(1) = 1 = α(2).
e. One-to-one: α(x) = α(x1) implies (x + 1, x − 1) = (x1 + 1, x1 − 1), whence x = x1. Not onto: (0, 0) ≠ α(x) for any x because (0, 0) = (x + 1, x − 1) would give x = 1 and x = − 1.
g. One-to-one: α(a) = α(a1) implies (a, b0) = (a1, b0) implies a = a1. Not onto if |B| ≥ 2 since no element (a, b) is in α(A) for b ≠ b0.
3.
a. Given c C, let c = βα(a) with a A (because βα is onto). ...
C, let c = βα(a) with a A (because βα is onto). ...Get Introduction to Abstract Algebra, Solutions Manual, 4th Edition now with the O’Reilly learning platform.
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