FMM20
From charlesreid1
Friday Morning Math Problem
Arrange in Order
Given the positive integers a, b, c, d, with (a/b) < (c/d) < 1, arrange the following quantities in order of increasing magnitude:
(b/a)
(d/c)
(bd)/(ac)
(b+d)/(a+c)
1
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Since (c/d) < 1, we know (d/c) > 1.
Since (a/b) < (c/d), we know (b/a) > (d/c). Hence, 1 < (d/c) < (b/a). If we know (d/c) > 1 and (b/a) > 1, then their product will be greater than both. Hence, 1 < (d/c) < (b/a) < (bd)/(ac). Next, we can show that the fraction (b+d)/(a+c) is greater than (d/c) and less than (b/a). Since (d/c) < (b/a), we know (ad < bc), and (cd + ad) < (cd + bc). Rearranging gives (d/c) < (b+d)/(a+c). Likewise, since (bc > ad), we know (ab + bc) > (ab + ad), which can be rearranged to (b+d)/(a+c) < (b/a). The order is therefore 1 < (d/c) < (b+d)/(a+c) < (b/a) < (bd)/(ac) |
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Source: Problem 2-5, Challenging Problems in Algebra
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