Thinking about the Mediant of Two FractionsDate: 12/10/2009 at 23:24:08 From: Teresa Subject: new fraction formula Students were trying to find a fraction between two given fractions with 1 as the numerators (such as 1/2 and 1/3). After working on several problems, one person noticed that when given 1/a and 1/b, the fraction 2/(a+b) always seems to be in between 1/a and 1/b. Is this a new fraction rule, or has this been discovered and proved before? I created the following "proof" but would like it to be checked, since it has been many years since I've tried to prove a new idea: If a and b are positive integers and a > b, then would 1/a < 2/(a+b) < 1/b? [b(a+b)]/[ab(a+b)] < 2ab / [ab(a+b)] < [a(a+b)]/[ab(a+b)], b(a+b) < 2ab < a(a+b) b(a+b) < 2ab and 2ab < a(a+b) a+b < 2a and 2b < a+b b < a and b < a, so it is true? Date: 12/11/2009 at 10:46:42 From: Doctor Peterson Subject: Re: new fraction formula Hi, Teresa. Yes, this is true. In fact, more generally, given ANY two fractions a/b and c/d (with a, b, c, and d positive numbers), the fraction (a+c)/(b+d), sometimes called the mediant, is between a/b and c/d. Your proof contains all the right elements, but just needs to be rearranged or explained more fully. What you've done is to show that IF 2/(a+b) is between 1/a and 1/b, THEN b < a. You want to show the converse of this. Since all the statements are equivalent to one another, you could essentially just reverse the order of your statements and it constitutes the proof you want. It would read a little oddly that way, though, so it needs to be explained carefully. Let's see if I can write a cleaner version of your proof for the general case. Given positive integers a, b, c, and d, such that a/b < c/d, we want to prove that a/b < (a+c)/(b+d) < c/d. The given condition, that a/b < c/d, is equivalent to ad bc -- < --, and therefore to ad < bc bd bd What we want to show, writing all the fractions with a common denominator, is ad(b+d) bd(a+c) bc(b+d) ------- < ------- < ------- bd(b+d) bd(b+d) bd(b+d) This is equivalent to showing that the numerators fall in the same order: ad(b+d) < bd(a+c) < bc(b+d) or abd + ad^2 < abd + bcd < b^2c + bcd We will take this in two parts: abd + ad^2 < abd + bcd, and abd + bcd < b^2c + bcd The first desired inequality is true because ad < bc ad^2 < bcd abd + ad^2 < abd + bcd The second is true because ad < bc abd < b^2c abd + bcd < b^2c + bcd We're done. (This could be written without working backward from the goal, but it would be very unclear why we did what we did.) There is also an interesting visual "proof" of this. Interpret the fractions as slopes, and consider any two points A (b,a) and D (d,c) in the first quadrant: ^ | a+c| ____---C | ____---- . / c| B--- . / | / . / | / . / | / . / | / . / a| / . ____---A |/. ____---- O-----------------------------> d b b+d Here the slope of OA, a/b, is less than the slope of OB, c/d. When we add the two vectors A and B (that is, make a parallelogram OACB so that AC and OB have the same slope and length), C has coordinates (b+d, a+c), so that the slope of the diagonal OC is (a+c)/(b+d). Clearly this diagonal goes through the middle of the parallelogram, so that its slope is between those of OA and OB. For more on this, including a more subtle proof, see: Wikipedia: Mediant (Mathematics) http://en.wikipedia.org/wiki/Mediant_(mathematics) If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 12/11/2009 at 15:37:03 From: Teresa Subject: Thank you (new fraction formula) Thank you so much! I am going to share your answer with the person who noticed the rule (who usually doesn't feel very good at math) and tell him that he can name it. That way we can always call it by his name in our class! Thanks again! Teresa Date: 12/11/2009 at 20:58:54 From: Doctor Peterson Subject: Re: Thank you (new fraction formula) Hi, Teresa. I just realized that, although I focused on a more general rule that I find interesting, your student's rule can be explained much more simply, which may even be what the student had in mind. We're looking at two unit fractions (fractions with numerator 1), and trying to find a fraction that is between them. We know that if a > b, then 1/a < 1/b. If we pick any number c BETWEEN a and b, then 1/a < 1/c < 1/b In particular, if c is the average (mean) of a and b, this will be true. For example, if a is 5 and b is 3, the average, c, is 4, and we know that 1/5 < 1/4 < 1/3 But in the case you were looking at, 1/2 and 1/3, there is no WHOLE number between 2 and 3. Their average is (2+3)/2 = 5/2. If we ignore the fact that this is not whole and just take its reciprocal (1/c), we get the fraction 2/5. The reasoning that said that 1/c is between the others is still valid, so we do know that 1/3 < 2/5 < 1/2 as desired. The general rule is that we can take c = (a+b)/2 and use its reciprocal, 2/(a+b): 1/a < 2/(a+b) < 1/b So this is a very reasonable choice to have made, and its truth follows from two elementary facts, namely that the average is between the two numbers, and that increasing the denominator decreases the fraction. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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