Timios

Let a, b, c, and d be any positive integers. Then:

1. If $\displaystyle ad<bc$ prove that $\displaystyle \frac {a}{b} < \frac {c}{d}$

2. If $\displaystyle \frac {a}{b} < \frac {c}{d}$ prove that $\displaystyle \frac {a}{b} < \frac {a+c}{b+d}<\frac {c}{d}$

v8archie

1) Is a straightforward division.
2) Multiply top and bottom by $d$, apply 1) and simplify. Repeat with $b$.

soroban

Hello, Timios!


We have: $\:ad \:<\:bc$

Divide by $bd\!:\;\;\dfrac{ad}{bd}\:<\:\dfrac{bc}{bd}$

Therefore: $\:\dfrac{a}{b} \:<\:\dfrac{c}{d}$




From (1), we have: $\:ad \:<\:bc$

Add $ab$ to both sides: $\:ab+ad \:<\:ab+bc$

Factor: $\:a(b+d) \:<\:b(a+c)$

Divide by $b(b+d)\!:\;\;\dfrac{a}{b}\:<\:\dfrac{a+c}{b+d}$


From (1), we have: $\:ad \:<\:bc$

Add $cd$ to both sides: $\:ad+cd \:<\:bc+cd$

Factor: $\:d(a+c) \:<\:c(b+d)$

Divide by $d(b+d)\!:\;\;\dfrac{a+c}{b+d} \:<\:\dfrac{c}{d}$


Therefore: $\:\dfrac{a}{b} \:<\:\dfrac{a+c}{b+d} \:<\:\dfrac{c}{d}$

Timios

Yes, Soroban. Those are the precise proofs with which I came up.
It's also a useful bit of knowledge.

1. Quickly determine which is smaller:

$\displaystyle \frac {5}{7} or \frac{3}{4}$

Draw a downstroke from the 5 to the 4. Multiply.
Is the product smaller or greater than the product of 3 and 7?
Since the downstroke product is 20 and is less than the other product, 21
then $\displaystyle \frac {5}{7} < \frac{3}{4}$

2. Find a fraction between $\displaystyle \frac{3}{4} and \frac{4}{5}$
Just add the numerators and denominators to find a fraction between.

$\displaystyle \frac{7}{12}$ is a fraction between $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{4}{5}$

v8archie

$$\frac{7}{12} \lt \frac{3}{4} \lt \frac{4}{5}$$
I suspect you meant $$\frac{7}{9}$$

soroban

Hello, Timios!


I was taught to "cross multiply", but always upward.

We have: $\;\dfrac{5}{7}\!\! \rlap{\;\;\nearrow} \nwarrow\,\dfrac{3}{4}$

And we get: $\:5\cdot4\:\gtrless\:7\cdot3$

Since $20\,\color{red}{<}\,21$, we have: $\:\dfrac{5}{7}\,\color{red}{<}\,\dfrac{3}{4}$