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If $(m_1, n_1)$ and $(m_2, n_2)$ are two equal rational numbers, i.e. $m_1*n_1^{-1} = m_2 * n_2^{-1}$, so $m_1*n_2 = m_2*n_1$.

And if $m_1$ and $n_1$ are coprime, prove that $n_2 * n_1^{-1} = m_2 * m_1^{-1} = k \in \mathbb Z$.

The main problem is to show that k is an integer.

Nikita Tkachuk
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  • What is $n_1^{-1}$ and $m_1*n_1^{-1}$? Isn't your definition of equality of $(m_1,n_1)=(m_2,n_2)$ that $m_1n_2=m_2n_1$? – plop Aug 16 '22 at 17:09
  • Numbers of the form $m_1n_1^{-1}$, where $m, n \in \mathbb Z$ are called rational, $m_1n_1^{-1} = m_2 * n_2^{-1}$ is equivalent to $m_1n_2=m_2n_1$ – Nikita Tkachuk Aug 16 '22 at 17:21
  • Multiply both sides by $n_1n_2$ and get $m_1n_2=m_2n_1$ since $aa^{-1}=1$. Similarly, for the second part, multiply each side by $n_1m_2^{-1}$ and get $m_2 m_1^{-1}=n_2*n_1^{-1}$ – mode_er Aug 16 '22 at 17:23
  • @mode_er, I have proven that but how to prove that k is an integer? – Nikita Tkachuk Aug 16 '22 at 17:27

1 Answers1

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We consider the equation $m_1*n_2=m_2*n_1$

Since $m_1$ and $n_1$ are coprime, and $$m_2*n_1 \text{ mod } m_1 \equiv m_1*n_2 \text{ mod } m_1\equiv 0$$ Then, $$m_2 \equiv 0 \text{ mod } m_1$$

This implies $m_2 = m_1*a$ for some $a \in Z$. Thus,

$m_2 * m_1^{-1} = m_1*a * m_1^{-1}=a=k\in Z$, completing the proof.

mode_er
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