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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: $\:\dfrac{5}{7}\,\color{red}{<}\,\dfrac{3}{4}$ Tags exercise, interesting, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ManosG Real Analysis 3 March 26th, 2013 02:11 PM challenge_accepted Advanced Statistics 3 November 23rd, 2012 03:08 PM icemanfan Number Theory 2 March 15th, 2012 04:58 PM john616 Applied Math 4 February 16th, 2012 10:02 AM Noob1 Advanced Statistics 7 April 26th, 2010 05:55 AM

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