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December 8th, 2014, 11:41 AM  #1 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 371 Thanks: 68  Interesting Proof Exercise
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 $\displaystyle \frac {a}{b} < \frac {a+c}{b+d}<\frac {c}{d}$ 
December 8th, 2014, 02:37 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 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: 407  Hello, Timios! Quote:
We have: $\:ad \:<\:bc$ Divide by $bd\!:\;\;\dfrac{ad}{bd}\:<\:\dfrac{bc}{bd}$ Therefore: $\:\dfrac{a}{b} \:<\:\dfrac{c}{d}$ Quote:
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: 371 Thanks: 68 
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: 7,327 Thanks: 2451 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: 407  Hello, Timios! Quote:
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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