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ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...

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We consider the unary operation − on R and want to determine those mappings<br />

f of R into it self, which are homomorphisms. the condition is that<br />

f(−x) = −f(x),<br />

which is the condition for f to be an odd function<br />

This result can be generalized to general linear spaces.<br />

Exercise 9: Homomorphisms for unary operations (see nr 8)<br />

Write the condition f : R → R to be a homomorphism wrt the operations<br />

x ↦→ −x and x ↦→ x −1 , and find such a homomorphism. (Hint: f(x) = a x )<br />

Exercise 10: The four fundamental isomorphisms concerning the algebraic operations.<br />

(see nr 8)<br />

Let a be a real constant. Study the four function definitions ax, a x , log a(x), x a<br />

and explain that they define homomorphisms for suitable choices of domain<br />

and codomain. Notice that all combinations of addition and multiplication are<br />

possible.<br />

Exercise 11: Closed wrt to addition. (see nr 14)<br />

Construct an increasing chain of subsets of C which are closed wrt addition in<br />

C<br />

Exercise 12: Show that the set {x+y √ 3 : x, y ∈ Q} is closed both wrt addition<br />

and multiplication .<br />

Show that the same is not true for {x + y 3√ 2 : x, y ∈ Q}. (see nr 14)<br />

Exercise 13: Product of equal factors (see nr 19)<br />

Show that Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}. Show that the product operation<br />

+ × + partially is given in the following table and complete:<br />

(+,+) (0,0) (0,1) (1,0) (1,1)<br />

(0,0) (0,0) (0,1) (1,0) (1,1)<br />

(0,1) (0,1) (0,0) (1,1) (1,0)<br />

(1,0) (1,0) (1,1)<br />

(1,1) (1,1) (0,1)<br />

(·, ·) (0,0) (0,1) (1,0) (1,1)<br />

(0,0) (0,0) (0,0) (0,0) (0,0)<br />

(0,1) (0,0) (0,1) (0,0) (0,1)<br />

(1,0) (0,0) (0,0)<br />

(1,1) (0,0) (1,0)<br />

Exercise 14: Product of different factors (see nr 19)<br />

Make tables for (Z2, +) × (Z3, +) and (Z2, ·) × (Z3, ·), in the same way as in<br />

X13<br />

Example 15: Chinese remainder theorem (see nr 21)<br />

37

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