User talk:Leland Prior

0.999...
I have reverted your addition to this page as, apart from being badly formatted and argumentative and so unencyclopaedic, it was completely wrong. If you do not understand the proofs you could ask about them at the reference desk, Reference desk/Mathematics. -- JohnBlackburne wordsdeeds 20:34, 23 March 2013 (UTC)

Show me how to shift digits left, in multiplying the endlessly repeating decimal 0.999... by 10, without using the carry. The carry cannot be used because there is no rightmost digit in 0.999... . Leland Prior (talk) 01:04, 24 March 2013 (UTC)

Let x = 0.999... = 0.9...99 be a repeating decimal where "..." represents an infinite number of nines.

*    0.9...99     3.9...96      6.9...93 *    +0.9...99    +0.9...99     +0.9...99 *    -    -     -  *     1.9...98     4.9...95      7.9...92 *    +0.9...99    +0.9...99     +0.9...99 *    -    -     - *     2.9...97     5.9...94      8.9...91 *    +0.9...99    +0.9...99     +0.9...99 *    -    -     - *     3.9...96     6.9...93      9.9...90 = 10(0.9...99) = 10(0.999...) = 10x. *    *     10x-x = 9x = 9.9...90 - 0.9...99 = 8.9...91. *        x = 8.9...91/9 = 0.9...99 = 0.999... ≠ 1. *    0.999... ≠ 1.   *      Disproves the Digit Manipulation "proof" on page 0.999... . Leland Prior (talk) 15:14, 1 April 2013 (UTC)

Digit manipulation

When a number in decimal notation is multiplied by 10, the digits do not change but each digit moves one place to the left. Thus 10 × 0.999... equals 9.999..., which is 9 greater than the original number. To see this, consider that in subtracting 0.999... from 9.999..., each of the digits after the decimal separator cancels, i.e. the result is 9 − 9 = 0 for each such digit. The final step uses algebra: *           X = 0.999...           *          10X = 9.999...           *          10X - X = 9X = 9.999... - 0.999... = 9          *            X = 9/9 = 1 *  :What exact mistake is being made in the "Digit manipulation" proof, immediately above? Its * :author is confusing a sum where addition has the commutative property with the ordered sum * :representing a repeating decimal. Each term's relative "place" is fixed, in the ordered sum. User talk:Leland Prior (talk) 16:06, 7 April 2013 (UTC) Leland Prior (talk) 16:08, 7 April 2013 (UTC) Leland Prior (talk) 18:43, 7 April 2013 (UTC)
 * :   When "..each digit moves one place to the left.", the rightmost place must be vacated.  The
 * :ellipsis in 9.999... cannot, then, represent an infinite number of 9's. The actual number is more
 * :accurately represented as 9.9...90 . In subtracting, then, it is not true that ".. each of the
 * :digits after the decimal separator cancels,.."