Why are the integers and rationals not isomorphic under addition?

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SUMMARY

The integers under addition (denoted as Z) are not isomorphic to the rationals under addition (denoted as Q). This conclusion is established by demonstrating that Q cannot be generated by a single element, which is a requirement for cyclic groups. The discussion highlights the density of the rationals and the existence of rational numbers that cannot be expressed as integer multiples of any proposed generator, leading to contradictions that affirm the non-isomorphism. The key takeaway is that while Z is cyclic, Q is not, confirming their distinct group structures.

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Homework Statement



Prove that the integers (under addition) are not isomorphic to the rationals (under addition).

Homework Equations



Two groups are isomorphic if there is an isomorphism between them.

If there is an isomorphism from G to H, f : G --> H, then G is cyclic iff H is cyclic.

A group G is cyclic if \{ x^n | n \in \mathbb{Z} \} = G, for some x \in G .


The Attempt at a Solution



The integers are generated by <1>. We can show that Z and Q are not isomorphic if we show that the rationals cannot be generated. Thus assume they are. Then there is an a such that

&lt;a&gt; = Q [\tex].<br /> 0a = 0, 1a = \frac{l}{m} , 2a = \fract{2l}{m}[\tex].&lt;br /&gt; &lt;br /&gt; Because the rationals are dense there is a b \in Q s.t. \frac{l}{m} &amp;amp;lt; b &amp;amp;lt; \frac{2l}{m} [\tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; We must show that b = ka = \frac{kl}{m}, thus \frac{l}{m} &amp;amp;amp;lt; \frac{kl}{m} &amp;amp;amp;lt; \frac{2l}{m} [\tex]. &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Now I don&amp;amp;amp;#039;t know what to do. The above is not a contradiction. Any ideas?
 
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Suppose Q is cyclic.

Let p/q be a generator.

Can you find a rational number which is not an integer multiple of p/q?
Think of a rational involving p and q somehow.
 
samkolb said:
Suppose Q is cyclic.

Let p/q be a generator.

Can you find a rational number which is not an integer multiple of p/q?
Think of a rational involving p and q somehow.

sure, continuing

Observe p/2q

Then,

p/2q = p/q where p,q are not equal to zero. (if p is zero then the set is finite).

From the above conclude that
pq=2pq, therefore 1=2.
done.
 
No, no. <p/q> is the set of number k*p/q where k is an integer. Set kp/q=p/(2q) and derive a different contradiction.
 
Dick said:
No, no. <p/q> is the set of number k*p/q where k is an integer. Set kp/q=p/(2q) and derive a different contradiction.

Thanks! Yeah, I don't know what I was thinking there. I went and got some food, came back and hit myself in the head on that one.
 

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