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March 4th, 2012, 06:34 AM   #1
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(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
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March 4th, 2012, 06:42 AM   #2
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Are you sure? Who is domain of ?
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March 4th, 2012, 08:33 AM   #3
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Re: (Q, +) not isomorphic with (Q*, x)

How many elements of order two does (Q,+) have? How about (Q*,x)?
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March 4th, 2012, 08:41 AM   #4
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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?
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March 4th, 2012, 08:43 AM   #5
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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
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March 4th, 2012, 09:45 PM   #6
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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?
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March 4th, 2012, 10:23 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.
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March 4th, 2012, 11:11 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)
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March 5th, 2012, 06:34 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.
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March 5th, 2012, 01:53 PM   #10
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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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