Proof: Show that R is not isomorphic to R*

[hsong9]

Homework Statement

Show that R is not isomorphic to R*.

Homework Equations

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.

 

[foxjwill]

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)?

 

[Mark44]

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?

 

[foxjwill]

What is R*?

R* is usually the nonzero reals under multiplication.

 

[Mark44]

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

 

[hsong9]

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??

 

[Mark44]

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??

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).

 

[foxjwill]

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??

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?

 

[matt grime]

How can this be a function from R to R*? For example, f(1) = -1 $\notin$ R*.

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

 

[matt grime]

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

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

 

[hsong9]

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?

 

[matt grime]

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?