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May 23rd, 2015, 04:31 AM  #1 
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0  Need Help : to show 2 almost equal fraction is not equal using contradiction??
This question is taken from : https://brilliant.org/problems/same...p=N36CSB8k5Wpz Below shows two numbers accurate up to 4 significant figures. Which of these numbers is larger? $\displaystyle \large \frac{122}{353} \approx 0.3456, \ \ \ \ \ \frac{75}{217} \approx 0.3456 $ Solution 1 : $\displaystyle 353 \times 75 = \boxed {26475} \ge 122 \times 217= \boxed {26474}$ therefore $\displaystyle \frac{75}{217}$ is larger However , in the solution page , the question master lay down a challenge. 'Bonus question: Since you already know the answer, can you prove that $\displaystyle \frac{75}{217} > \frac{122}{353}$ by contradiction? I have read the sqrt 2 contradiction on wiki but i still not really have a good idea on how to start it. Anyway this is my working so far : $\displaystyle let \ \ \frac{75}{217} > \frac{122}{353} \ \ be \ \ \frac{odd_{a}}{odd_{b}} > \frac{even_{a}}{odd_{c}}$ if $\displaystyle \frac{add_{a}}{odd_{b}} \ \ has \ \ an \ \ integer \ \ solution$ the integer must be $\displaystyle odd_{d}$ otherwise $\displaystyle odd_{b} \times even \ \ = \ \ even \ \ ≠ \ \ odd_{a}$ if $\displaystyle \frac{even_{a}}{odd_{c}} \ \ has \ \ an \ \ integer \ \ solution$ the integer must be $\displaystyle even_{b}$ otherwise $\displaystyle odd_{c} \times even \ \ = \ \ odd \ \ ≠ \ \ even_{a}$ therefore $\displaystyle \frac{75}{217} ≠ \frac{122}{353}$ as they are odd and even , so one has to be bigger than the other. Finally we know that the bigger the differences between the numerator and denomenator , the smaller the number. 217  75 = 142 (smaller differences) 353  122 = 231 (bigger differences) therefore , $\displaystyle \frac{75}{217}$ must be larger than $\displaystyle \frac{122}{353}$ i am not sure whether this count as contradiction or not , if otherwise....i hope someone can show me and i pray it is not rocket maths proving.... 
May 23rd, 2015, 03:51 PM  #2 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85  That is only certainly true if two fractions have the same denominator and cannot be used generally. For example, 6/10 > 4/7, but 6/10 has the greater difference. I used proof by contradiction to prove some of your proof wrong.
Last edited by EvanJ; May 23rd, 2015 at 03:56 PM. 
May 23rd, 2015, 04:31 PM  #3 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  Sorry to nitpick, but that's not a proof by contradiction. It's a counterexample. Proof by contradiction requires you to assume something is true, then use that assumption to obtain an impossible result. Then, you may say that your original assumption is false and the opposite must be true. For example, in the 'root 2 is irrational' proof, the assumption is that $\sqrt{2}$ is rational, or $\sqrt{2} = \dfrac{m}{n}$ where $m$ and $n$ have no common factors. They then use that assumption to show that both $m$ and $n$ are even and thus have a common factor of 2. This is the impossible result. After that, they say that $\sqrt{2}$ cannot be rational and thus must be irrational. 
May 23rd, 2015, 07:21 PM  #4  
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0 
Thank you EvanJ and Azzajazz Senior , i wouldnt mind that as nitpick at all. Wrong is wrong , in my field of study(industrial science) wrong maths cost lives , pls dont hesitate on mistake. Quote:
but that would make it seems like proof 2 > 1 by contradiction , which doesnt seems to make much sense........( i wonder if the question itself is flawed? )  
May 23rd, 2015, 08:34 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,559 Thanks: 2558 Math Focus: Mainly analysis and algebra 
I think it's a slightly silly question. But, Suppose that $${122 \over 353} \ge {75 \over 217}$$ Then $$\begin{aligned} {122 \over 353}  {75 \over 217} &\ge 0 \\ {217 \times 122  353 \times 75 \over 353 \times 217 } &\ge 0 \\ 26474  26475 &\ge 0 \\ 1 &\ge 0 \end{aligned}$$ 
May 23rd, 2015, 08:59 PM  #6 
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0 
=.= , it sure does look a bit silly.... anyway, v8archie, I'll post what you wrote with credit; I'll reply here if the questionmaster replies. Last edited by skipjack; May 23rd, 2015 at 09:28 PM. 
May 23rd, 2015, 10:28 PM  #7 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 
By contradiction, suppose 122/353 > 75/217. Then we can write 122/353 = 75/217 + x, x > 0. Solving for x obtains x < 0. Contradiction. 
May 24th, 2015, 02:15 AM  #8 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
Alternatively, exploit $\displaystyle 0 < a < b$ gives (and is equivalent with) $\displaystyle 0 < \frac{1}{b} < \frac{1}{a}$ (1) and $\displaystyle a < b$ gives $\displaystyle a + c < b + c$ (0 might be in a different place!) Suppose $\displaystyle \frac{75}{217} \le \frac{122}{353}$ As $\displaystyle \frac{75}{217} = \frac{122}{353}$ gives 75 * 353 = 217 * 122 where the RHS is a multiple of 2 and the LHS isn't by the fundamental theorem of arithmetic, they aren't equal (you might not need it but I like to use it) So we suppose $\displaystyle \frac{75}{217} < \frac{122}{353}$ This also gives by (1) $\displaystyle \frac{217}{75} = 2 + \frac{67}{75} > 2 + \frac{109}{122} = \frac{353}{122}$ Subtracting 2 from both sides, by (2) we get that it's equivalent to saying that: $\displaystyle \frac{67}{75} > \frac{109}{122}$ Similarily, $\displaystyle 1 + \frac{8}{67} < 1 + \frac{13}{109}$ and $\displaystyle 8 + \frac{3}{8} > 8 + \frac{5}{13}$ (*) and $\displaystyle 2 + \frac{2}{3} < 2 + \frac{3}{5}$ $\displaystyle 1 + \frac{1}{2} > 1 + \frac{2}{3}$ (**) Finally: 2 < 1 + 1/2 A contradiction. So $\displaystyle \frac{75}{217} \le \frac{122}{353}$ is false and therefore $\displaystyle \frac{75}{217} > \frac{122}{353}$. (*) We might recognize 3/8 and 5/13 as division of Fibonacci numbers of the form $\displaystyle \frac{F_i}{F_{i+2}}$ which alternates. $\displaystyle \frac{F_{2k + 1}}{F_{2k+3}} > \frac{F_{2k}}{F_{2k+2}}$ But checking that might be harder then proceding. (**) Here we might see in the fractions the numerator and denominator have an equal difference (denominator  numerator) of 1 and both are > 0 which gives the fraction with the larger numerator and denominator is larger. 
May 24th, 2015, 03:52 AM  #9 
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0 
the post master / moderator replied , v8archie solutions onwards are all correct. (can close thread) Still i didnt expect it to be able reduce to 2 < 1+ 1/2 ..... it looks really beautiful. could it be done without reciprocal both side? i know it sounds weird tho.(sry i replied in a rush , didnt have much time to think about it) 
May 24th, 2015, 04:50 AM  #10 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
Maybe as follows (rough sketch) : As the approximation to 4 sf of the numbers is the same, the numbers are within 1/10^4 from each other. Now, 122 * 217 ends in 4 and 353 * 75 ends in 5. The difference is of the form 10k + 1 which must be as close to 0 as possible. This happence when k = 0 so the difference 353 * 75  122 * 217 is 1. Together this yields that 122 * 217 < 353 * 75. Therefore, 122 / 353 < 75 / 217. This might need more elaboration and work to a contradiction. Maybe suppose k < 0 or so. 

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