Repeating decimals.

Number Theory Level 2

Is it always true that $\displaystyle a.999999999\cdots = a.\dot 9 = \frac { a9 - a } 9 = a + 1$ ?

Please don't calculate it as calculating as normal fractions calculations.

Like $\displaystyle \frac { a9 - a } 9 = \frac { 9a - a } 9 = \frac { 8a } 9$ . It is false calculation.

Warn that $a$ must not be less than zero. So omly $a \ge 0$ .

And $a9$ means the first digit of the 2-digited number $a$ , and the last digit of 2-digited number 9.

Not $a \times 9$ .

It is always true. Maybe so. It is true, but not always. It is always false. It is false(not true), but not always.

2 solutions

Elijah Frank
Mar 10, 2021

0.999... = 1 = 9/9 (a=0)

. .
Mar 10, 2021

It is the easiest question of mine.

The proof is very simple.

Like $\displaystyle a.0$ , substituting it into the above question, then $\displaystyle 0.9999\cdots = 0.\dot9 = \frac 99 = 1$ .

We get $\displaystyle a = 0$ , and $\displaystyle 0.9999\cdots = 1 = 0 + 1$ .

And working on it to $\displaystyle a = 1$ , then $\displaystyle 1.99999\cdots = \frac { 19 - 1 } { 9 } = \frac { 18 } 9 = 2$ .

Then, $\displaystyle 1.999999\cdots = 2 = 1 + 1$ .

If I work on it for every integer, then I can verify it as $\displaystyle \boxed { \text { True } }!!!$ easily.