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Changed my commentary on failure to be cursive

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Philippe Zwietering
2023-05-16 15:48:46 +02:00
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@@ -204,13 +204,14 @@ See the assignment PDF for the full assignment specification and theorem.
\frac{x_1*x_1}{p_1***p_N*q_1***q_M} &= \frac{x_2*x_2}{v_1***v_n*w_1***w_m} \frac{x_1*x_1}{p_1***p_N*q_1***q_M} &= \frac{x_2*x_2}{v_1***v_n*w_1***w_m}
\end{align} \end{align}
I'm kinda stuck at this point. I see that this is definitely injective, since the way the exponents \textit{I'm kinda stuck at this point.
I see that this is definitely injective, since the way the exponents
are defined, you will always know which prime factors belong to the numerator or to the denominator. are defined, you will always know which prime factors belong to the numerator or to the denominator.
But I fail to prove this using the direct definition of $f$ like we could do for the natural numbers. But I fail to prove this using the direct definition of $f$ like we could do for the natural numbers.
This is because the products of the denominators in the last equation are not unique. This is because the products of the denominators in the last equation are not unique.
So maybe I simplified them too much and shouldn't try and write them in terms of $x_1$ and $x_2$ like So maybe I simplified them too much and shouldn't try and write them in terms of $x_1$ and $x_2$ like
we did earlier, and try and focus more on just the exponents, but I feel it becomes really hard we did earlier, and try and focus more on just the exponents, but I feel it becomes really hard
to show that $x_1 = x_2$ that way. to show that $x_1 = x_2$ that way.}
\item[Surjectivity:] We want to show that $f$ is onto, i.e. $f(\{ q > 0 : q \in \Q \}) = \N$. \item[Surjectivity:] We want to show that $f$ is onto, i.e. $f(\{ q > 0 : q \in \Q \}) = \N$.