Final review question...ugh.
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Suppose an isomorphism existed, f:Z->Q. What would map to 1/2?
Say f(n) = 1/2, and f(m) = 2. Since it's an isomorphism, these must exist. Then f(n)f(m) = f(nm) = 1/2*2 = 1, so nm maps to 1. But also, f(nm) = f(m+m+...+m [n times]) = f(m) + f(m) + ... + f(m) [n times] = nf(m) = 2n = 1, so n is not an integer, a contradiction. There are many similar methods.
Say f(n) = 1/2, and f(m) = 2. Since it's an isomorphism, these must exist. Then f(n)f(m) = f(nm) = 1/2*2 = 1, so nm maps to 1. But also, f(nm) = f(m+m+...+m [n times]) = f(m) + f(m) + ... + f(m) [n times] = nf(m) = 2n = 1, so n is not an integer, a contradiction. There are many similar methods.