Proving $\mathbb{Q}$ not isomorphic to $\mathbb{R}^{+}$

Question

Question: Prove that $\mathbb{Q}$ under addition is not isomorphic to $\mathbb{R^{+}}$ under multiplication.

I wonder what is a good way to approach this? Would constructing a map be more helpful?

Thank in advance. Hints only*

score 5 · Accepted Answer · answered Jun 24 '17 at 05:59

5

Hint: To have isomorphism, you need a bijection. Why can't a bijection exist?

answered Jun 24 '17 at 05:59

manthanomen

1

Simply because 0 is not in the positive reals. I can take it from here. – Mathematicing Jun 24 '17 at 06:00
Just think about the sets (don't worry about the group structure). Further hint: What is the cardinality of $\mathbb Q$? What is the cardinality of $\mathbb R^+$? – manthanomen Jun 24 '17 at 06:04
My knowledge of sets is pretty rusty. But I think this is related to countably finite and countably infinite. – Mathematicing Jun 24 '17 at 06:06

score 3 · Answer 2 · answered Jun 24 '17 at 05:59

3

Hint: An isomorphism must in particular be a bijection.

answered Jun 24 '17 at 05:59

Eric Wofsey

score 2 · Answer 3 · answered Jun 24 '17 at 06:03

2

Hint: $\;\mathbb{Q}\,$ is countable, $\,\mathbb{R}^+\,$ is not.

answered Jun 24 '17 at 06:03

dxiv

I have to admit this is good. – Mathematicing Jun 24 '17 at 06:04
1

@Mathematicing That's essentially what the other answers were hinting at as well. – dxiv Jun 24 '17 at 06:06
I got it. 0/1 maps to 0 in the natural, 1/1 maps to 1 in the natural, 2/1 maps to 2 in the naturals, and so on. There exists a one-to-one correspondence with the natural. This cannot be said to be the same for the positive reals. – Mathematicing Jun 24 '17 at 06:13
@Mathematicing I don't think you've actually described a bijection to the naturals: what does 3/4 go to? But you should be able to find a proof that $\mathbb{Q}$ is countably infinite very easily in your favorite resource on introductory set theory. – diracdeltafunk Jun 24 '17 at 06:52

score 1 · Answer 4 · answered Jun 24 '17 at 06:47

1

The group $\mathbb{R}^+$ with respect to multiplication is isomorphic to $\mathbb{R}$ with respect to addition, via logarithm and exponential.

The group $\mathbb{Q}$ has the property that for all $x$ and $y$ there exist integers $m$ and $n$, not both zero, such that $$ mx+ny=0 $$

Does $\mathbb{R}$ share this property?

answered Jun 24 '17 at 06:47

egreg

You're not wrong, but this is too involved. A cardinality argument is much easier. – diracdeltafunk Jun 24 '17 at 06:53
@diracdeltafunk A different point of view. The isomorphism between $\mathbb{R}^+$ with multiplication and $\mathbb{R}$ with addition is basic knowledge. Pointing to an algebraic property that holds in one group and not in the other one is the prime method for showing they're not isomorphic. – egreg Jun 24 '17 at 06:58

4 Answers4