Or, assume 0.9999... < 1.
Then e = 1 - .9999... > 0.
Then the decimal representation of e is not all 0's.
Let the first non-zero entry in the decimal expansion of e be in place N.
But .9999... has way more than N places, so the Nth place must be 0.
Contradiction! Therefore .9999... DOES = 1.
Or..
.9999 means the LIMIT of 9 + .09 + .009 + ....
The limit equals 1. So by the definition of what .9999.... means, it's equal to 1.
No matter how you look at it, .99999... HAS to be the same as 1.
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Krishnamurthy say: 1
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roger say: 0.99999.... can be shown to be 1
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Chris P say: 0.999.... which is infinitely close to 1; and for most cases can be considered to be 1.
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Brainard say: 1
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Metalplanttag say: It is one, because there is no problems with partial fractions .
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D.W. say: 3/3 = 3×0.333... = 0.999... = 1
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Paul say: 3/3 = 0.99999... where the nines go on forever.
This is numerically the same as 1 (although not necessarily if you get to more advanced maths).
Here's the school leaver explanation.
0.999999... = x
9.999999... = 10x (remember the 9s go on forever so we never run out)
10x-x 9.999999.... - 0.999999....
9x = 9
x = 1.
Here's a video explaining this:
https://www.youtube.com/watch?v=4V5NJiGB...
Here's a more advanced video explaining why 0.999999... isn't necessarily one.
https://www.youtube.com/watch?v=aRUABAUc...
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