Recall the following properties of logarithms

Page Produced by Ben Mueller

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Recall the following properties of logarithms:

Ln (x) + Ln (y) = Ln (xy)
Ln (x) – Ln (y) = Ln (x/y)
Ln (x^2) = 2 Ln (x) or more generally, Ln (x^b) = b Ln (x)
Ln (e) = 1
Ln (1) = 0

Note: Ln (x) is only defined when the value of x is greater than zero. Also, the range of the function Ln (x) is negative infinity to infinity.

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These properties will help in understanding the following mapping.

Let’s define a mapping F such that:

F maps from the group of real numbers under multiplication to the group of real numbers under addition.
F(x) = Ln (x)

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Application Problem (Created by Ben Mueller):

Use properties of logarithms to show that mapping F is an isomorphism.

Solution:

Part 1 (Showing that F is a homomorphism):

We must show that F(x*y) = F(x) + F(y). By our mapping, F(x*y) = Ln (x*y). Then by properties of logarithms, Ln (x*y) = Ln (x) + Ln (y). Our mapping then tells us that

Ln (x) + Ln (y) = F(x) + F(y) and we have shown that F (x*y) = F(x) + F(y) completing part 1 showing that F is a homomorphism.

Part 2 (Showing that F is injective or 1-1):

Let’s start by assuming that F(x) = F(y). Then we have that F(x)=Ln(x) and F(y)= Ln(y). Thus if F(x) = F(y), it is true that Ln(x) = Ln(y). By the horizontal line test of the graph of Ln(x), Ln(x) = Ln(y) implies that x = y. This completes part 2 of our proof that our mapping is injective or 1-1.

Part 3 (Showing that F is surjective or onto):

Here we want to show that for every real number r, there is an x such that F(x) = r. Now if F(x) = r, then Ln(x) = r. Since the range of the graph of the function Ln(x) is from negative infinity to infinity, this is true for all values of r. This completes part 3 of our proof that F is surjective or onto.

Application Problem 2:

If it exists, find the kernel of our mapping F.

Solution:

The co-domain of mapping F is the group of real numbers under addition. The identity element in our co-domain is 0. Thus we want to find all real numbers x such that F(x) = 0 in our mapping F. Because F(x) = Ln (x), our problem becomes finding all real numbers such that Ln (x) = 0. Ln (x) = 0 is only true when x = 1 and thus Ker F = {1}.

Interesting note: In our mapping F, 1 was the identity element in our domain (group of all real numbers under multiplication) and 0 in our co-domain (group of all real numbers under addition). Thus we had that F(e) = e’ (e being identity in domain and e’ in co-domain.

Here is a proof to supplement our above note:

Let F be a homomorphism from the group G to the group G’. If e denotes the identity element in G and e’ denotes the identity element in G’, then prove that F(e)=e’.

Proof (Written by Ben Mueller): Since e is the identity element, ee = e and F(e) = F(ee). Now, F is a homomorphism which means that F(ee) = F(e)F(e). From this, F(e) = F(e)F(e) and this implies that F(e) is the identity element in G’ (F(e) is an element in G’). Finally, since the identity element in a group must be unique, e’ = F(e)./(Proof complete)