Admin at 19:42, 7 February 2020

2020-02-07T19:42:43Z

← Older revision		Revision as of 19:42, 7 February 2020
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	1 < (d/c) < (b+d)/(a+c) < (b/a) < (bd/ac)		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)
			}}

Admin: Created page with "{{FMM |title=Arrange in Order |problem= Given the positive integers a, b, c, d, with (a/b) < (c/d) < 1, arrange the following quantities in order of increasing magnitude: (b..."

2020-02-07T19:36:12Z

Created page with "{{FMM |title=Arrange in Order |problem= Given the positive integers a, b, c, d, with (a/b) < (c/d) < 1, arrange the following quantities in order of increasing magnitude: (b..."

New page

{{FMM
|title=Arrange in Order
|problem=

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

|solution=

1 < (d/c) < (b+d)/(a+c) < (b/a) < (bd/ac)

Source: Problem 2-5, Challenging Problems in Algebra

}}

FMM20 - Revision history

Admin at 19:42, 7 February 2020

Admin: Created page with "{{FMM |title=Arrange in Order |problem= Given the positive integers a, b, c, d, with (a/b) < (c/d) < 1, arrange the following quantities in order of increasing magnitude: (b..."