# Proof: Show that R is not isomorphic to R*

1. Mar 10, 2009

### hsong9

1. The problem statement, all variables and given/known data
Show that R is not isomorphic to R*.

2. Relevant equations

3. The attempt at a solution
Let f:R -> R* such that f(x) = e^x for x in R
f(a+b) = e^(a+b), but f(a)f(b) = e^(ab).
Therefore, f(a+b) != f(a)f(b).
not isomorphism.

2. Mar 10, 2009

### foxjwill

Re: Isomorphism

I'm assuming your R is the group of reals under addition and R* is the group of nonzero reals under multiplication.

So, to your question. All you've shown is that the particular function f:R->R* defined by f(x)=e^x isn't an isomorphism. However, this does not imply that R isn't isomorphic to R^*. For example, consider the groups $$G=\{e,a^2,a^3,\ldots\}$$ and $$(\mathbb{Z},+)$$. The bijection g:G->Z defined by
$$f(a^n)= \begin{cases} 2&\text{if } n=1\\ 1&\text{if } n=2\\ n&\text{else} \end{cases}$$
is clearly not an isomorphism, yet $$\mathbb{Z}\cong G$$.

The usual process to determine that two groups are not isomorphic is to find some algebraic property (i.e. one that would be conserved under isomorphism) that is not common to each group. For R and R*, try examining the properties of the element $$-1\in \mathbb{R}^*$$. Is there an element in R with the same properties (under addition)?

3. Mar 10, 2009

### Staff: Mentor

Re: Isomorphism

What is R*? From the context, it has to be {x in R : x > 0}. I believe this is usually written as R+. What you're trying to show is that f (where f(x) = e^x) is an isomorphism between the group {R, +} and the group {R+, *}, where + and * are the usual operators for real addition and real multiplication.

There are errors in your work.

f(a+b) = e^(a + b). How else can you write this using the laws of exponents?
f(a)f(b) != e^(ab).
f(a)f(b) = (e^a)(e^b). How else can this be written, again using the laws of exponents?

4. Mar 10, 2009

### foxjwill

Re: Isomorphism

R* is usually the nonzero reals under multiplication.

5. Mar 10, 2009

### Staff: Mentor

Re: Isomorphism

OK, I just don't remember seeing that notation. (It's been a looooooooong while!)

6. Mar 11, 2009

### hsong9

Re: Isomorphism

hmm...
f:R -> R* such that f(x) = -x
f(a + b) = -(a+b) however,
f(a)f(b) = ab
so f(a + b) != f(a)f(b)... correct??

7. Mar 11, 2009

### Staff: Mentor

Re: Isomorphism

How can this be a function from R to R*? For example, f(1) = -1 $\notin$ R*. The only way to get output values that are positive is if the domain is {x | x < 0}.

It is true, however that f(a + b) $\neq$ f(a)f(b).

8. Mar 11, 2009

### foxjwill

Re: Isomorphism

again, all you've shown here is that one particular bijection isn't an isomorphsim.

When two groups are isomorphic, what that means is that they satisfy all the same algebraic properties--in other words, the group operation for one works the exact same way as the group operation for the other. So, to show that two groups are not isomorphic, either (a) prove that every possible bijection between the two is not an isomorphism, or (b) show that one group has an algebraic property (i.e. a property that is true of some group irrespective of what you call its elements) that the other does not. Clearly, (a) isn't feasible in general, so we usually try (b).

When I say look for an element in (R,+) that has the same properties as -1 in (R*,), I mean examine the various properties of it. Specifically, what's the order of -1 in R*? Is there an element in (R,+) that has the same order?

9. Mar 11, 2009

### matt grime

Re: Isomorphism

You've already been told that R* is the non-zero real numbers: -1 is definitely a nonzero real number.

10. Mar 11, 2009

### matt grime

Re: Isomorphism

It isn't a bijection from R to R*. It isn't even a function from R to R*: f(0) = -0 =0.

11. Mar 11, 2009

### hsong9

Re: Isomorphism

Ok, so..
if g=-1 then gg = 1 in R*. This means -1 is itself inverse.
however, 0 is idnentity of R.....
I need to use above things for solving problem.
right?

12. Mar 12, 2009

### matt grime

Re: Isomorphism

You've shown the order of -1 is 2. Orders are preserved by isomorphisms (prove this if it is not immediately obvious). Does R under addition have an element of order 2? What would it mean if x in R had order 2?