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 December 8th, 2014, 11:41 AM #1 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 Interesting Proof Exercise Let a, b, c, and d be any positive integers. Then: 1. If $\displaystyle ad  December 8th, 2014, 02:37 PM #2 Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 1) Is a straightforward division. 2) Multiply top and bottom by$d$, apply 1) and simplify. Repeat with$b$. December 8th, 2014, 08:19 PM #3 Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Hello, Timios! Quote:  Let$a, b, c, d$be any positive integers. Then: (1) If$ad

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

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

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

Quote:
 (2) If $\,\dfrac {a}{b} < \dfrac {c}{d}$, prove that $\,\dfrac {a}{b} < \dfrac {a+c}{b+d}<\dfrac {c}{d}$

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

 December 9th, 2014, 07:10 PM #4 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 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 $\displaystyle \frac{4}{5}$
 December 9th, 2014, 08:59 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra $$\frac{7}{12} \lt \frac{3}{4} \lt \frac{4}{5}$$ I suspect you meant $$\frac{7}{9}$$
December 9th, 2014, 10:05 PM   #6
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, Timios!

Quote:
 1. Quickly determine which is smaller: $\:\dfrac{5}{7}\,\text{ or}\,\dfrac{3}{4}$

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: $\:\dfrac{5}{7}\,\color{red}{<}\,\dfrac{3}{4}$

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