Get the solution to "If 9.999⋯=10, then is there a general proof..."

NepanitaNesg3a

Answered question

2022-04-24

9.999 \dots = 10

, then is there a general proof for any number that has infinite trailing 9s?

Patricia Stanley

Beginner2022-04-25Added 10 answers

Step 1
First of all, assume that

x \in (0, 1)

. Then

x = 0 . d_{1} d_{2} \dots d_{n} 99999 \dots

, with

d_{n} \neq 9

. Therefore

10^{n} x = a + 0.999 \dots

, where

a = 10^{n - 1} a_{1} + 10^{n - 2} a_{2} + \dots + a_{0}

So,

10^{n} x = a + 1

, and therefore

x = 0 . d_{1} d_{2} \dots d_{n - 1} (d_{n} + 1)

In the general case, x can be written as

m + x^{'}

, with

m \in Z

and where

x^{'} \in (0, 1)

has trailing 9's. So, the previous argument aples to

x^{'}

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