if you have 8/9 then you would divide 8 by9 and it would be .8888888 repeating. To turn that into a fraction again would it be 8/9, 88/99, 888/999, 8888/9999, ect. what would it be because if it's anyone but the first one it wouldn't equal 8/9 like ot originally was?
2007-06-15 06:56:30 · 3 answers · asked by gigglewabbit 1 in Science & Mathematics ➔ Mathematics
88/99 does equal 8/9 exactly:
8/9 * 1 = 8/9
8/9 * 11/11 = 8/9
(8*11)/(9*11) = 8/9
88/99 = 8/9
The same is true of 888/999 (multiply by 111/111), 8888/9999 (multiply by 1111/1111), etc.
No matter how many digits you pick for the length of the repeating sequence (using the sequence of digits over an equal length sequence of 9s), you get the exact same fraction back. It may not be in simplest terms, but it's the same fraction.
The same is true for repeating decimals of other lengths. Consider 10/11:
0.909090909090909090...
90/99 = 10/11 * 9/9
9090/9999 = 10/11 * 909/909
909090/999999 = 10/11 * 90909/90909
etc.
2007-06-15 07:00:25 · answer #1 · answered by McFate 7 · 2⤊ 0⤋
IT WOULD BE 8/9 AGAIN BECAUSE THE 9 IS IN THE TENTH PLACE . one digit in the bottom of the fraction always remain the same if written again.
2007-06-15 15:31:10 · answer #2 · answered by Kandice F 4 · 0⤊ 0⤋
Let x = .8888888888888...........
10x = 8.888888888888.........
Subtract rhe 1st from the second getting:
9x =8
x = 8/9
2007-06-15 14:10:27 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋