My Math Forum (Q, +) not isomorphic with (Q*, x)

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 March 4th, 2012, 05:34 AM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 (Q, +) not isomorphic with (Q*, x) Prove: Group Q with operation addition is not isomorphic with Q* (Q without 0) with operation multiplication. I have this function: f: Q ---> Q* : x ---> 10 ^ x and I showed that f(xy) = f(x) f(y) f-1: Q ---> Q* : x ---> log(x) is the inverse relation. Yet it is not a isomorfism... why? I have two vague ideas, namely that in Q, na/nb = a/b, so the relation would not be a bijective one. (But if you call 10^(a/b) and 10^(na/nb) two different things, would the bijection still hold?) Second idea: 10 ^ x is not a rational number for some x, so it would not be in Q*, hence no bijection. But I have no idea how to prove that a number is not rational. I need some hints on how to complete this proof! Many thanks! Kappie
 March 4th, 2012, 05:42 AM #2 Member   Joined: Jul 2011 From: Trieste but ever Naples in my heart! Italy, UE. Posts: 62 Thanks: 0 Are you sure? Who is domain of $\log_{10}$?
 March 4th, 2012, 07:33 AM #3 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: (Q, +) not isomorphic with (Q*, x) How many elements of order two does (Q,+) have? How about (Q*,x)?
 March 4th, 2012, 07:41 AM #4 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Re: (Q, +) not isomorphic with (Q*, x) I thought of the function myself, btw. It seemed a logical step to take in going from addition to multiplication, but I was afraid to use e as base number because e is irrational (and therefore isn't in Q -- right?). I don't think Q or Q* have elements of order two... am I right?
 March 4th, 2012, 07:43 AM #5 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Re: (Q, +) not isomorphic with (Q*, x) Oh and of course I forgot about the domain of log(x)... Then it should be: x ---> log|x| if x is positive x ---> -log|x| if x is negative
 March 4th, 2012, 08:45 PM #6 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: (Q, +) not isomorphic with (Q*, x) Does the map taking rational numbers $x \mapsto 10^x$ hit all of the nonzero rational numbers? Are all of the outputted elements rational?
March 4th, 2012, 09:23 PM   #7
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Re: (Q, +) not isomorphic with (Q*, x)

Quote:
 Originally Posted by Kappie in Q, na/nb = a/b, so the relation would not be a bijective one.
Exactly -- you had it from the start.

March 4th, 2012, 10:11 PM   #8
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Re: (Q, +) not isomorphic with (Q*, x)

Quote:
 Originally Posted by CRGreathouse Exactly -- you had it from the start.
Ok, thanks, I will work that out.

So I conclude it's inherent to Q that there are infinite elements that have the same value, and therefore there can be no bijective images from Q? Or is there some catch?
For example I've heard that R ---> (0,1) is a bijective image, because there are as many numbers between 0 and 1 as are between minus infinity and infinity.

And one more thing. If Q is a subgroup of R, does R have all those same elements with the same value in it? Or do the doubles come from the definition of Q? (a/b : a,b in Z, b != 0)

March 5th, 2012, 05:34 AM   #9
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Re: (Q, +) not isomorphic with (Q*, x)

Quote:
 Originally Posted by Kappie So I conclude it's inherent to Q that there are infinite elements that have the same value
I don't know what you mean here... all elements of Q are finite.

Quote:
 Originally Posted by Kappie For example I've heard that R ---> (0,1) is a bijective image, because there are as many numbers between 0 and 1 as are between minus infinity and infinity.
There is a bijection between those two, yes. Both have cardinality beth-1. You can even make the bijection explicit.

Quote:
 Originally Posted by Kappie And one more thing. If Q is a subgroup of R, does R have all those same elements with the same value in it? Or do the doubles come from the definition of Q? (a/b : a,b in Z, b != 0)
Since Q is a subgroup of R, you can do arithmetic with the elements of Q in the group R and you'll get the same results as if you did the arithmetic in Q.

 March 5th, 2012, 12:53 PM #10 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Re: (Q, +) not isomorphic with (Q*, x) I ment that e.g. 3/4, 6/8, 9/12 are all in Q and have the same value. I also thought of a quick way to prove this (and the first poster already mentioned it): it is to show that -1 has order 2 in (Q*, x), and no such element exists in (Q, +).

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