Hey, I'm currently trying to understand a proof. And I have trouble understanding a specific part of it which I will highlight. I find this part hard to visualize. Maybe someone can draw me a diagram of what is happening.
Recall that two fractions are equal if they are the same point on the number line. We obvserved that for all nonzero whole numbers \(\displaystyle n\) and \(\displaystyle k\), \(\displaystyle \frac{nk}{n} =\frac{k}{1}\), as both are equal to \(\displaystyle k\). The following generalizes this fact.
Theorem. Given two fractions \(\displaystyle \frac{m}{n}\) and \(\displaystyle \frac{k}{l}\), suppose there is a nonzero whole number \(\displaystyle c\) so that \(\displaystyle k=cm\) and \(\displaystyle l = cn\). Then \(\displaystyle \frac{m}{n} = \frac{k}{l}\)
Proof. Let \(\displaystyle k = cm\) and \(\displaystyle l = cn\) for whole numbers \(\displaystyle c\), \(\displaystyle k\), \(\displaystyle l\), \(\displaystyle m\), and \(\displaystyle n\). We will prove that \(\displaystyle \frac{m}{n} = \frac{k}{l}.\) In other words, we will prove: \(\displaystyle \frac{m}{n} = \frac{cm}{cn}\).
The fraction \(\displaystyle \frac{m}{n}\) is the \(\displaystyle m\)-th point in the sequence of \(\displaystyle n\)-ths. Now divide each of the segments between consecutive points in the sequence of \(\displaystyle n\)-ths into \(\displaystyle c\) equal parts. Thus each of \(\displaystyle [0,1], [1,2], [2,3]\), ... is now divided into \(\displaystyle cn\) equal parts. This the sequence of \(\displaystyle n\)-ths together with the new division points become the sequence of \(\displaystyle cn\)-ths. A simple reasoning shows that the \(\displaystyle m\)-th point in the sequence of \(\displaystyle n\)-ths must be the \(\displaystyle cm\)-th point in the sequence of \(\displaystyle cn\)-ths. This is another way of saying \(\displaystyle \frac{m}{n} = \frac{cm}{cn}\). Thus the proof is complete.
Recall that two fractions are equal if they are the same point on the number line. We obvserved that for all nonzero whole numbers \(\displaystyle n\) and \(\displaystyle k\), \(\displaystyle \frac{nk}{n} =\frac{k}{1}\), as both are equal to \(\displaystyle k\). The following generalizes this fact.
Theorem. Given two fractions \(\displaystyle \frac{m}{n}\) and \(\displaystyle \frac{k}{l}\), suppose there is a nonzero whole number \(\displaystyle c\) so that \(\displaystyle k=cm\) and \(\displaystyle l = cn\). Then \(\displaystyle \frac{m}{n} = \frac{k}{l}\)
Proof. Let \(\displaystyle k = cm\) and \(\displaystyle l = cn\) for whole numbers \(\displaystyle c\), \(\displaystyle k\), \(\displaystyle l\), \(\displaystyle m\), and \(\displaystyle n\). We will prove that \(\displaystyle \frac{m}{n} = \frac{k}{l}.\) In other words, we will prove: \(\displaystyle \frac{m}{n} = \frac{cm}{cn}\).
The fraction \(\displaystyle \frac{m}{n}\) is the \(\displaystyle m\)-th point in the sequence of \(\displaystyle n\)-ths. Now divide each of the segments between consecutive points in the sequence of \(\displaystyle n\)-ths into \(\displaystyle c\) equal parts. Thus each of \(\displaystyle [0,1], [1,2], [2,3]\), ... is now divided into \(\displaystyle cn\) equal parts. This the sequence of \(\displaystyle n\)-ths together with the new division points become the sequence of \(\displaystyle cn\)-ths. A simple reasoning shows that the \(\displaystyle m\)-th point in the sequence of \(\displaystyle n\)-ths must be the \(\displaystyle cm\)-th point in the sequence of \(\displaystyle cn\)-ths. This is another way of saying \(\displaystyle \frac{m}{n} = \frac{cm}{cn}\). Thus the proof is complete.
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