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I am not a mathematician but I have a question.

I found

  • $1 / 3 = 0.33333....$
  • $2 / 3 = 0.666666...$
  • $3 / 3 = 1$ while it should be $0.999999....$

Same as dividing by $9$

  • $1/9=0.11111..$
  • $2/9= 0.2222..$
  • $3/9=0.3333..$
  • $9/9=1$ while it should be $0.99999..$

Divided by $7$ is little bit different. As

  • $1/7= 0.142857 \, 142857 \, 142857$
  • $2/7= 0.285714 \, 285714\, 285714$
  • $3/7= 0.428571\, 428571\, 428571$

The same pattern of number repeated in the same arrangement. But with $7/7$ the answer is $1$

I can understand the approximation but with the number increases the error increased.

To explain it:

                        Should be
1/3=0.3333.            1/3=0.3333
2/3=0.6666.            2/3=0.6666
3/3=1.                 3/3=0.9999
4/3=1.333.             4/3=1.2222
5/3=1.666.             5/3=1.55555
6/3=2                  6/3=1.88888

Explanation is appreciated

Henry
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Ehab Mohamed
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    Cf. https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-dots-1 – Travis Willse Apr 19 '19 at 23:52
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    Uhm... Provided that there is no such thing as digit-by-digit operations for numbers with an infintite decimal expansion, what makes you assume that $0.\overline 9\times 2=1.\overline 8$? In my book, for instance, $0.9999\times 2=1.9998$. –  Apr 19 '19 at 23:56
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    Also, $142857\cdot 7=999999$, so by your (incorrect) idea, $\frac{7}{7}=0.\dot{9}$ too. – Rhys Hughes Apr 20 '19 at 00:06
  • You forgot to include the carries. –  Apr 20 '19 at 00:13

2 Answers2

2

$$0.9999999...=\frac{9}{10}(1+\frac{1}{10}+\frac{1}{10^2}+.....)=\frac{9}{10}\frac{1}{1-\frac{1}{10}}=1$$

E.H.E
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There is a bijection between infinite ultimately periodic decimal expansions and rational numbers, except when the period consists only of the digit $9$. In this case,, one proves that deleting the final $9$s and replacing last digit $d\ne 9$ with the digit $d+1$ defines a decimal number which is equal to the given infinite decimal expansion.

Bernard
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