Rational Number and its Reciprocal
Date: 03/14/2002 at 21:21:57 From: Nelson Phu Subject: Algebra word problem Dear Dr. Math, I recently ran into this problem in a math team meeting and got stuck and don't even know where to start Problem: A rational number greater than one and its reciprocal have a sum of 2 1/6. What is this number? Express your answer as an improper fraction in lowest terms. I'd appreciate it if there were some type of general formula or way to solve all problems like this. Thank you! Sincerely, Nelson
Date: 03/14/2002 at 21:36:56 From: Doctor Ian Subject: Re: Algebra word problem Hi Nelson, I'm guessing you won't ever come across a problem like this again in your whole life, so it's not clear how helpful a formula would be. But maybe we can find a way to solve the problem, and learn something from the solution. Any time you see the phrase 'a rational number' with no specific value, you should consider representing it as two integers, e.g., p/q. In this case, if we do that, we get A rational number (p/q) greater than 1 (p>q) and its reciprocal (q/p) have a sum of 2 1/6. So now we know two things: 1) p/q + q/p = 2 1/6 = 13/6 2) p > q Note that the second piece of information isn't all that useful, since if p/q isn't greater than 1, q/p will be. (Unless they're equal, in which case it's immediately clear that the sum in question can't be a fraction.) Anyway, we can simplify the first equation: p q 13 - + - = -- q p 6 p^2 q^2 13 --- + --- = -- pq pq 6 p^2 + q^2 13 --------- = -- pq 6 Now, at this point, you may be able to find values of p and q by inspection. (Hint: What would you have to multiply to get a denominator of 6?) Or, you could multiply both sides by 6pq to get rid of the fractions. That would give you a quadratic equation, which you could solve in any of the usual ways. But if you're doing this in a contest, the sucker's way would be to work it out. The quick way would be to realize that you're going to add two fractions, and you need to get a common denominator of 6. That tells you most of what you need to know to make an intelligent guess and quickly confirm it, which is the essence of succeeding in timed contests. I know that if _I_ ever see a problem like this again, that's the way I'm going to handle it. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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