3

I know that $1.000\dots$ and $0.999\dots$ are two different decimal representations of $1$, but how can I prove that they are unique?

I mean, there is no other decimal representation $a_n\dots a_0. a_{-1}a_{-2}\dots$ of $1$, different from $1.000\dots$ and $0.999\dots$

Blue
  • 75,673
Jorge S.
  • 782
  • 2
    Nice question!. – aarbee Jun 18 '22 at 17:52
  • 1
    I think it suffices to prove that there is a unique representation of $0$. – Deane Jun 18 '22 at 17:58
  • Shooting from hip: take any other representation $x$ and $|1 - x| < \varepsilon$ will fail for sufficiently small $\varepsilon >0$ – AlvinL Jun 18 '22 at 18:02
  • What is your definition of a decimal representation? There is no such thing as uniqueness in nature, for instance if you define it as any element of ${0,...,9}^\mathbb N$ – blamethelag Jun 18 '22 at 19:06

1 Answers1

3

There can be no other decimal representation of $1$.

Let $$p.s_1s_2...$$ be a decimal representation of $1$ where $p$ is the integer prefix and $s_1s_2...$ is the sequence of digits after the decimal place.

If $p\gt 1$ then $p.s_1s_2...\ge p\gt 1$ so $p\le 1$.

If $p=1$ then $1.s_1s_2...\text{ }-\text{ } 1.00....=.s_1s_2...=0$ so $s_1s_2...=0$ which only happens when all of the digits are zero.

If $p=0$ then consider $0.99...-0.s_1s_2...$

Since no digit can be greater than $9$ we can do this subtraction from left to right because no borrowing is required. If $s_1s_2...\ne 99...$ then there will be a first position $i$ such that $s_i\ne 9$. In that case $0.99... - 0.s_1s_2...\ge (9-s_i)\times 10^{-i}\ne 0$

Since the difference is not zero $p.s_1s_2...$ cannot represent $1$. But this is a representation of $1$ so $s_1s_2...$ must equal $99...$

John Douma
  • 11,426
  • 2
  • 23
  • 24