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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) f1: 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 ?

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 > logx if x is positive x > logx 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 hit all of the nonzero rational numbers? Are all of the outputted elements rational? 
March 4th, 2012, 09:23 PM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: (Q, +) not isomorphic with (Q*, x) Quote:
 
March 4th, 2012, 10:11 PM  #8  
Newbie Joined: Feb 2012 Posts: 10 Thanks: 0  Re: (Q, +) not isomorphic with (Q*, x) Quote:
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  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: (Q, +) not isomorphic with (Q*, x) Quote:
Quote:
Quote:
 
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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