# Isomorphism between groups of real numbers

1. Jan 20, 2012

### blahblah8724

Apparently there is an isomorphism between the additive group $(ℝ,+)$ of real numbers and the multiplicative group $(ℝ_{>0},×)$ of positive real numbers.

But I thought that the reals were uncountably infinite and so don't understand how you could define a bijection between them?!

2. Jan 20, 2012

### micromass

The isomorphism is $f(x)=e^x$.

Could you explain more about what's bothering you??

3. Jan 20, 2012

### HallsofIvy

There exist a bijection between two sets if and only if they have the same cardinality (that is, essentiallt, the definition of "cardinality"). The fact that two sets are both uncountably infinite doesn't mean they do not have the same cardinality.

4. Jan 20, 2012

### SteveL27

Yes, and you are already very familiar with it: It's just the exponential function y = e^x. For all real numbers a and b we have e^(a+b) = e^a * e^b. That fits the definition of a homomorphism. Then note that e*x is a bijection between the additive reals and the multiplicative positive reals.

Does that help in terms of seeing the concreteness and familiarity of this isomorphism?

ps -- That's why e^0 = 1. A homomorphism always maps the identity to the identity. I vividly remember being in my first abstract algebra class and slogging through homomorphisms and normal subgroups ... then they mentioned that exp and log are isomorphisms ... and I got this AHA moment -- this stuff is actually about something!

pps -- I see Micromass already mentioned e^x. Hopefully I was able to add some detail for the OP's benefit.

Last edited: Jan 20, 2012