Proof to find fraction inbetween to fraction

[Taylor_1989]

I keep getting slightly confused with the algebraic method of finding a fraction between, two other fraction. Here is an example question, I have been doing

Find the fraction between 13/15 and 14/15? I personally convert both to percentages and find the average between the two, the convert back to a fraction. So in this case I did:

13/15 = 0.8666666 = 87%
14/15 = 0.933333 = 93%
87+93 = 180/2= 90
90/100 = 9/10: which I believe is correct

The method I do not understand is the proof behind this method: 13/15+14/15 = 27/30 / 3 top and bottom you get 9/10. But its the proof that confuse me. I will show where I get confused:

I understand this part: a/b < c/d cross multiply ad < bc

This is the part I don't understand: Add ab to both sides ab+ad < ab+bc

why add ab to both sides, where dose this come from?

then factor : a(b+d)<b(a+c)⇒ a/b < a+c/a+b

Then you add cd to both sides, once again why? then you factor out again.

 

[Vorde]

Think of it this way:

You start with 13/15 and 14/15

13/15 is equal to 26/30
and
14/15 is equal to 28/30

This is done by a simple multiplication of two on both the numerator and the denominator.

Looking at your new fractions, it is obvious that 27/30 is in between the two of them, and 27/30 simplifies to 9/10

 

[Mentallic]

This is the part I don't understand: Add ab to both sides ab+ad < ab+bc

why add ab to both sides, where dose this come from?

Because it works. Notice in the next line the factoring worked out such that we can get a/b on its own on the left side.

then factor : a(b+d)<b(a+c)⇒ a/b < a+c/a+b

Just a typo but it should be a/b < (a+c)/(b+d)

Then you add cd to both sides, once again why? then you factor out again.

I guess when you say you add cd to both sides you're talking about

$$ad<bc$$

$$cd+ad<cd+bc$$

$$d(a+c)<c(b+d)$$

$$\frac{a+c}{b+d}<\frac{c}{d}$$

Which again work exactly the way we want it to. We've now just shown that $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$

By using algebraic manipulations that were cleverly used to give us the answer we were looking for.
But keep in mind that this value $$x=\frac{a+c}{b+d}$$ is not always exactly in the middle of a/b and c/d. When b and d are different, it doesn't turn out to be the average of the two fractions.

If you wanted the average of a/b and c/d as your in-between fraction, then you'd have

$$x=\frac{\frac{a}{b}+\frac{c}{d}}{2}$$
$$=\frac{ad+bc}{2bd}$$

Which is a lot more calculations than the value of x obtained from the proof above.

 

[pwsnafu]

Trivia: the fraction $\frac{a+c}{b+d}$ is called the mediant of $\frac{a}{b}$ and $\frac{c}{d}$.

Further the mediant does not preserve order. Suppose that $\frac{a}{b} < \frac{A}{B}$ and $\frac{c}{d} < \frac{C}{D}$, but it is possible to have $\frac{a+c}{b+d} > \frac{A+C}{B+D}$. This is known as Simpson's paradox.

 

[CellCoree]

First of all, it is important to note that there are multiple ways to find a fraction between two other fractions. The method you are using, which involves converting to percentages and finding the average, is a valid approach. However, there is also a more algebraic method, which is what the question is asking for.

The proof you are referring to is based on the concept of comparing fractions using cross multiplication. This concept states that if a/b < c/d, then ad < bc. This is because when we cross multiply, we are essentially comparing the same number (a) to two different numbers (bc and ad) and seeing which one is larger.

Now, let's apply this concept to the fractions 13/15 and 14/15. We can rewrite these fractions as 13/15 = 13/15 and 14/15 = 14/15. Since 13/15 < 14/15, we can use cross multiplication to compare the fractions. This gives us 13*15 < 14*15, which simplifies to 195 < 210. We can then divide both sides by 15 to get 13 < 14, which we already know is true.

To find a fraction between 13/15 and 14/15, we need to find a fraction that is greater than 13/15 and less than 14/15. This means that we need to find a fraction that is greater than 13/15 and has a smaller denominator, or less than 14/15 and has a larger denominator.

This is where the proof you mentioned comes in. We can add ab to both sides of the inequality ad < bc. This gives us ad + ab < bc + ab. Now, we can factor out a from the left side and b from the right side, giving us a(d + b) < b(a + c). We can then divide both sides by b(d + b) to get a/b < (a + c)/(b + d).

In simpler terms, by adding ab to both sides, we are essentially finding a common denominator for the fractions ad and bc. This allows us to compare the fractions more easily and find a fraction between them.

Similarly, by adding cd to both sides and factoring, we are finding a common denominator for the fractions ad and bc, but in this case, the denominator is larger than both a and b. This allows us to find a fraction