Fractions in Ascending OrderDate: 03/22/99 at 21:39:20 From: Anonymous Subject: Fractions in Ascending Order My question is: If a, b, c, and d are positive integers that satisfy a/b < c/d < 1, arrange the 5 quantities b/a, d/c, bd/ac, b+d/a+c, 1 in ascending order. Date: 08/04/2008 at 21:39:20 From: Doctor Ian Subject: Re: Fractions in Ascending Order Hi David, Here's the situation we're given: a/b < c/d < 1 Integer a, b, c, d > 0 To keep from getting lost, let's make a table where we can keep track of the comparisons we make. We'll give each fraction a label (I through V), and write the larger of the fractions at the intersection: bd b+d b/a d/c -- --- 1 ac a+c (I) b/a - - - - - (II) d/c - - - - bd (III) -- - - - ac b+d (IV) --- - - a+c (V) 1 - We can multiply our original inequality by (bd)/(ac) to get abd cbd --- < --- < bd bac dac which simplifies to d b bd --- < --- < -- c a ac II I III Let's add that information to our table: bd b+d b/a d/c -- --- 1 ac a+c (I) b/a - - - - - (II) d/c I - - - - bd (III) -- III III - - - ac b+d (IV) --- - - a+c (V) 1 - If a/b is less than 1, then b/a must be greater than 1. Similarly for d/c. And (bd)/(ac) is greater than both of those, so it must also be greater than 1: bd b+d b/a d/c -- --- 1 ac a+c (I) b/a - - - - - (II) d/c I - - - - bd (III) -- III III - - - ac b+d (IV) --- - - a+c (V) 1 I II III - How can we compare b/a and (b+d)/(a+c)? Well, we can look at the ratio, and see how it compares to 1. (If the ratio is less than 1, we know the numerator is smaller; if greater than 1, the numerator is larger.) b/a b a+c ab + bc ------- = - * --- = ------- b+d a b+d ab + ad --- a+c Now, we're golden if we know which is larger, bc or ad. Let's go back to our original condition: a c - < - < 1 b d We can multiply by bd, to get ad < bc < bd Since ad < bc, it must be true that ab + bc ------- < 1 ab + ad which means that b/a < (b+d)/(a+c). We can use similar reasoning to show that d/c < (b+d)/(a+c). And since both b/a and d/c are greater than 1, we can conclude that (b+d)/(a+c) is, too: bd b+d b/a d/c -- --- 1 ac a+c (I) b/a - - - - - (II) d/c I - - - - bd (III) -- III III - - - ac b+d (IV) --- IV IV - - a+c (V) 1 I II III IV - That leaves us with just one comparison to do. If we try some examples, e.g., bd 3*2 b+d 3+2 5 1/3 < 1/2 < 1 --> -- = --- = 6 --- = --- = - ac 1*1 a+c 1+1 2 bd 3*5 15 b+d 3+5 8 2/3 < 4/5 < 1 --> -- = --- = -- --- = --- = - ac 2*4 8 a+c 2+4 6 it suggests that bd b+d -- > --- ac a+c but suggestion isn't proof. A ratio is one way to compare two quantities. Another is to subtract them, and see if the difference is less than or greater than zero. Let's try that, and see if we can show that the difference must be positive: bd b+d 0 < -- - --- ac a+c bd(a+c) ac(b+d) 0 < ------- - ------- ac(a+c) ac(a+c) (abd + bcd) - (abc + acd) 0 < ------------------------- ac(a+c) 0 < (abd + bcd) - (abc + acd) 0 < abd + bcd - abc - acd 0 < ab(d - c) + cd(b - a) But we know that c < d, so (d - c) is positive; and similarly, (b - a) is positive. So this is always true, and that means that (bd)/(ac) > (b+d)/(a+c) So that completes our table: (I) (II) (III) (IV) (V) bd b+d b/a d/c -- --- 1 ac a+c (I) b/a - - - - - (II) d/c I - - - - bd (III) -- III III - - - ac b+d (IV) --- IV IV III - - a+c (V) 1 I II III IV - And now we have enough information to order everything: I > II, V II > V III > I, II, IV, V IV > I, II, V V > nothing So we have III > IV > I > II > V or bd b+d -- > --- > b/a > d/c > 1 ac a+c Just to bolster our confidence, let's check that with an example again: 2/3 < 4/5 < 1, a=2, b=3, c=4, d=5 That gives us 3*5 3+5 3 5 --- > --- > - > - > 1 2*4 2+4 2 4 15 8 3 5 -- > - > - > - > 1 8 6 2 4 15 8 3 5 -- > - > - > - > 1 8 6 2 4 1.875 > 1.333 > 1.5 > 1.25 > 1 And there ya go! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/