Why do I get 0.098765432098765432... when I divide 8 by 81?I got this remarkable thing when I divided 16 by 162, or, in a simplified version, 8 by 81. It's 0.098765432098765432 ⋯ , or more commonly known as 0. 098765432 ¯ with all the one-digit numbers going backwards...except for 1. Yeah, it's missing the 1. One, how do I get this remarkable outcome and two, why is it missing the 1?

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Why do I get   0.098765432098765432... when I divide 8 by 81?I got this remarkable thing when I divided 16 by 162, or, in a simplified version, 8 by 81. It&#039;s   0.098765432098765432  ⋯ , or more commonly known as   0.      098765432    ¯    with all the one-digit numbers going backwards...except for 1. Yeah, it&#039;s missing the 1. One, how do I get this remarkable outcome and two, why is it missing the 1?

Leah Conley · Accepted Answer

Maybe you can also see it by the following calculations  x  =  0.      098765432    ¯        10    9    x  =  98765432.      098765432    ¯    (      10    9    −  1  )  x  =  98765432  x  =  8      /    81We could actually find the number   y  =  0.      0987654321    ¯   by similar logic.      10          10        y  =  987654321.      0987654321    ¯    (      10          10        −  1  )  y  =  987654321  y  =      109739369    1111111111  So maybe the reason why you&#039;re number is cool is that it has a easy fractional representation and a decimal representation with interesting properties. Any interesting repeating decimal representation is of course a fraction and can be calculated, but it may not have an easy fractional representation as shown with y.

hughy46u · Accepted Answer

More generally, in base b (any integer   &amp;gt;  1)                    b        −        1                    b        2              +                    b        −        2                    b        3              +  …  +            2              b                  b          −          1                      =                    (        b        −        2        )        (                  b                      b            −            1                          −        1        )                              b                      b            −            1                          (        b        −        1                  )          2                    so that                                                        b              −              1                                      b              2                                                  +                                            b              −              2                                      b              3                                      +        …        +                              2                          b                              b                −                1                                                    +                              0                          b                              b                                                    +                                            b              −              1                                      b                              b                +                1                                                    +                                            b              −              2                                      b                              b                +                3                                                    +        …        =                  ∑                      j            =            0                    ∞                          b                      −            j            (            b            −            1            )                                                              (              b              −              2              )              (                              b                                  b                  −                  1                                            −              1              )                                                      b                                  b                  −                  1                                            (              b              −              1                              )                2                                                                              =                                            b                              b                −                1                                                                    b                                  b                  −                  1                                            −              1                                                                          (              b              −              2              )              (                              b                                  b                  −                  1                                            −              1              )                                                      b                                  b                  −                  1                                            (              b              −              1                              )                2                                                    =                                            (              b              −              2              )                                      (              b              −              1                              )                2                                                        What you have is the case    b  =  10

Why do I get 0.098765432098765432... when I divide 8 by 81? I got this remarkable thing when

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