Fraction help

[auville2000]

Find a fraction between each pair:

4/7 & 5/9

1/3 & 1/4

 

[Mrspi]

Find a fraction between each pair:

4/7 & 5/9 1/3 & 1/4

Sometimes it is helpful to write the fractions as equivalent fractions with the same denominator.

For example, if I were asked to find a fraction between 1/5 and 1/6, I might say, "What would be a common multiple of 5 and 6? Oh...I know! 30!"

And then, I would rewrite each fraction so that it has a denominator of 30.

To change 1/5 to an equivalent fraction with 30 for its denominator, I'd multiply both numerator and denominator by 6: (1/5)*(6/6) = 6/30

To change 1/6 to an equivalent fraction with 30 for its denominator, I'd multiply both numerator and denominator by 5: (1/6)*(5/5) = 5/30

Well....I'm still not sure what a fraction between those two might be. Maybe I should make the denominators larger. I could multiply the numerator and denominator of each fraction by 2:

(6/30)*(2/2) = 12/60
(5/30)*(2/2) = 10/60

Aha! Now I can see that a fraction between 12/60 and 10/60 would be 11/60! So, 11/60 is a fraction between 1/5 and 1/6.

You could try this approach on your problems.....

 

[lookagain]

Find a fraction between each pair:

4/7 & 5/9

1/3 & 1/4

auville2000,

other than understanding what Mrspi stated, if you want a "quick cheat,"
add the numerators together and add the denominators together:


$\displaystyle \dfrac{4 + 5}{7 + 9} \ = \ \dfrac{9}{16}$


This lies between 4/7 and 5/9.


---------------------------------------------------------


$\displaystyle \dfrac{1 + 1}{3 + 4} \ = \ \dfrac{2}{7}$


This lies between 1/3 and 1/4.




I have not proven my method here, though.

 

[soroban]

Hello, auville2000!

Here's another "quick cheat" ... though not as quick as lookagain's.

Find a fraction between each pair:

. . $\displaystyle \frac{4}{7}\text{ and }\frac{5}{9}$

. . $\displaystyle \frac{1}{3}\text{ and }\frac{1}{4}$

To find a number between any two numbers, find their average.
This produces a number exactly halfway between the two numbers.


For the first problem: .$\displaystyle \dfrac{\frac{4}{7} + \frac{5}{9}}{2} \;=\;\dfrac{\frac{36+35}{63}}{2} \;=\;\dfrac{\frac{71}{63}}{2} \;=\;\dfrac{71}{126}$

And we have: .$\displaystyle \displaystyle\frac{5}{9}\;<\:\frac{71}{126} \;<\;\frac{4}{7}$

 

[soroban]

Hello, all!

I just proved Lookagain's "quick cheat".


Suppose we have two distinct fractions: $\displaystyle \dfrac{a}{b}\text{ and }\dfrac{c}{d}$

. . $\displaystyle a,b,c,d$ are positive integers and $\displaystyle b\ne0,\:d\ne0$

We can assume that: .$\displaystyle \dfrac{a}{b}\:<\:\dfrac{c}{d}$

. . Hence: .$\displaystyle ad \:<\:bc$ .[1]


Add $\displaystyle ab$ to both sides of [1]: .$\displaystyle ab + ad \:<\:ab + bc$

. . and we have: .$\displaystyle a(b+d) \:<\:b(a+c) \quad\Rightarrow\quad \dfrac{a}{b} \:<\:\dfrac{a+c}{b+d}$ .[2]


Add $\displaystyle cd$ to both sides of [1]: .$\displaystyle ad + cd \:<\:bc+cd$

. . and we have: .$\displaystyle d(a+c) \:<\:c(b+d) \quad\Rightarrow\quad \dfrac{a+c}{b+d} \:<\:\dfrac{c}{d}$ .[3]


Combining [2] and [3], we have: .$\displaystyle \dfrac{a}{b} \:<\:\dfrac{a+c}{b+d} \:<\:\dfrac{c}{d}$

. . . . . . ta-DAA!