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I am reading the following exercise:

Use mediants to compare $\frac{13}{21}$ and $\frac{17}{27}$

My approach:
Since the fractions given are already reduced, if we do the "addition" with $\frac{4}{6}$ to get $\frac{17}{27}$ the fraction $\frac{4}{6}$ we would use is not reduced so:
$$\frac{13}{21} + \frac{17}{27} = \frac{30}{38}$$
This is correct since in my understanding the mediant has to be between reduced fractions but the mediant itself does not have to be in reduced form.
Now we have either:
a) $$\frac{13}{21} \lt \frac{30}{38} \lt \frac{17}{27}$$ or

b) $$\frac{17}{27} \lt \frac{30}{38} \lt \frac{13}{21}$$

I can see that $\frac{13}{21} \lt \frac{30}{38}$ but the way I do that is by comparing $13 \cdot 38 \lt 21\cdot 30 \equiv 494 \lt 630 \implies true$

Because:
I can see that $\frac{17}{27} \lt \frac{30}{38}$ but the way I do that is by comparing $17 \cdot 38 \lt 27\cdot 30 \equiv 646 \lt 810 \implies true$

So $\frac{30}{38}$ is not the mediant after all!

What am I messing up here?
I guess my assumption that the mediant is allowed to be in non-reduced form is wrong. But:
$$\frac{13}{21} + \frac{4}{6} = \frac{17}{27}$$ the $\frac{4}{6}$ is not reduced.
And $$\frac{13}{21} + \frac{17}{27} = \frac{30}{38}$$
does not lead to a solution, so I am stuck on what other approaches there are

And is it wrong to compare the fractions the way I do in the context of the specific problem?

JMP
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Jim
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  • The mediant is $\frac{30}{48}=\frac{5}{8}$. – JMP Jul 05 '21 at 16:38
  • @JMP: Am I allowed to reduce the mediant then? In that case how would I have preferred that over reducing $\frac{4}{6} = \frac{2}{3}$ without doing direct calculation which I think goes against the idea of the problem – Jim Jul 05 '21 at 16:41
  • As an inequality, yes. Note that $\frac23+\frac12=\frac35$, but $\frac46+\frac12=\frac58\ne\frac35$, so you need to take care if this is your intent. – JMP Jul 05 '21 at 16:43
  • @JMP: Yes I was aware of what you point out i.e. that the original fraction and the reduced fraction lead to different rational numbers, that is why I was not sure if I was supposed to reduce $\frac{30}{48}$ or not. Additionally I suspect the way I do the comparison i.e. $a \cdot d < c \cdot b$ to compare $\frac{a}{ b}, \frac{c}{d}$ should not be the way, because I could have used it to begin with instead of using mediants right? So how could I have compared to determine the order of the boundaries of the mediant? – Jim Jul 05 '21 at 17:08
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    The mediant provides a quick fraction between two fractions, another method being $\frac{ad+bc}{2bd}$, and your comparison test is fine. The problem is you have $38$ as your denominator, which is why your inequalities aren't right. – JMP Jul 05 '21 at 17:40
  • It seems like you already know that the mediant always lies between the two original fractions. Now, notice that $\frac{17}{27}=\frac{13+4}{21+6}$. Since $\frac46 =\frac23$, can you now figure it out? – arbashn Jul 05 '21 at 18:23
  • See this simple comparison alhorithm – Bill Dubuque Jul 05 '21 at 18:53
  • @arbashn: I don't understand what you are hinting towards. I already excluded $\frac{4}{6}$ as I stated in the post because the boundary has to be in reduced form and $\frac{13+2}{21+3} \ne \frac{17}{27}$ – Jim Jul 05 '21 at 21:00
  • It doesn't have to be in reduced form. The only condition to find the mediant of $\frac ab$ and $\frac cd$ is that $a,b,c$ and $d$ are positive integers. So we know that $\frac{17}{27}$ lies between $\frac{13}{21}$ and $\frac 46$. And since, $\frac 46 = \frac 23 = \frac{14}{21} \gt \frac{13}{21}$, we get that $\frac{17}{27} \gt \frac{13}{21}$. – arbashn Jul 06 '21 at 05:17
  • @BillDubuque: Is that an alternative to mediants or is it for my question on how to determine which boundary is larger? – Jim Jul 06 '21 at 08:18

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