i dunno how many times this has been asked, but does .9 repeated equal 1 or an infinitively close number? this has been giving me a headache the past few days.
2006-11-05 09:34:19 · 11 answers · asked by Toby 1 in Science & Mathematics ➔ Mathematics
Yes, 0.999... = 1. (And yes it gets asked a lot!) :)
The proof:
1/3 + 1/3 + 1/3 = 1
Rewrite 1/3 as 0.333...
0.333... + 0.333... + 0.333... = 1
0.999... = 1
2006-11-05 09:34:51 · answer #1 · answered by Anonymous · 2⤊ 0⤋
yes, it's a unique problem with the base 10 math system we use. 0.9 repeated does equal 1. It's the same thing for any case where the 9 repeats to the right of the decimal. 0.19999999999999999999999999... equals .2 and so on. I can't think of how to explain it very well other than to say yes it does.
2006-11-05 17:38:12 · answer #2 · answered by C D 3 · 2⤊ 0⤋
It is INFINITELY CLOSE to 1
RESPONSE to "DAVE" (he did a great job)
If you have
x=.9999999999.....
then 10x would be ALMOST EQUAL to 9.999999999...
Why?
If you have a number with THREE digits, when you multiply by 10 you'd now have a number with FOUR digits, with an added 0 at the right end. (999 x 10 = 9990).
Therefore, multiplying a number of infinite digits would result in a number with (as kids say it) "infinite plus one" amount of digits.
Imagine that IF you could see the end, would this number look like?
It would be like this:
0.9999999999... ...99999 = x
9.9999999999... ...99990 = 10x
9x would be:
8.9999999999... ...99991 = 9x
Divide by 9 and you'd get the original number:
0.9999999999... ...99999 = x
=)
RESPONSE to "YUPCHAGEE"
9/9 = 1 (exactly one) thus breaking the pattern you were trying to establish. Why? x/x=1.
Also, in Calculus you have to understand that "approach" 1 is not the same as "equal" one. Just try to graph y=1/x. You'll get close (aka "approach"), but you'll never ever get a value of y=0.
Also, BY DEFINITION of Equality, 1. = 0.999... is mathematically incorrect.
2006-11-05 18:49:33 · answer #3 · answered by chinaman 3 · 0⤊ 3⤋
It is definately 1. Another way to prove it is to look at it as a geometric series.9+.09+.009+.0009+.......Use the formula a/1-r to see what this series converges to: Since a=.9 and r=.1,a/1-r=.9/.9=1.
2006-11-05 17:36:32 · answer #4 · answered by bruinfan 7 · 2⤊ 0⤋
Here's how I see it, and how probably most math teachers would too.
.9 repeating does not equal one. One equals one. 0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 can get infinatly close to one, but it will never equal one.
For all practical purposes (like in the real world) you either aren't going to have something that is .9 repeating, or everything that is 1 is slightly off of one, like .9 repeating or 1.0000 with a million zeroes and then a one. But I could be wrong about that.
2006-11-05 17:39:14 · answer #5 · answered by T'Vral 3 · 0⤊ 4⤋
It equals 1. Here's how you show it:
x = 0.999999999.....
10x = 9.99999999....
If you subtract the first from the second you get:
9x = 9
x = 1
2006-11-05 17:36:22 · answer #6 · answered by Dave 6 · 3⤊ 0⤋
1/9=.1111111111111111111111111111...
2/9=.2222222222222222222.....................
etc. 9/9=.99999999999999999999999999999....
in the limit that the # of digits approaches infinity, it approaches 1.
2006-11-05 17:37:43 · answer #7 · answered by yupchagee 7 · 2⤊ 0⤋
No, .99999... repeating does not equal 1 for the simple reason that .33333... repeating times 3 equals 1, not .99999...
2006-11-05 17:39:46 · answer #8 · answered by SomeGuy 6 · 0⤊ 4⤋
it is the closest you can get to 1 without actually being 1. But generaly it is rounded to 1
2006-11-05 17:36:14 · answer #9 · answered by Sarah 4 · 0⤊ 4⤋
yes
2006-11-05 17:36:03 · answer #10 · answered by Melanie Smith 2 · 1⤊ 0⤋