raw proof

Proof that 0.9999... is equal to 1

Robert Eisele

The fact that \(1=0.99999...\) may seem counterintuitive at first glance, but upon closer examination, it becomes clear that these two expressions indeed represent the same value.

For a first example lets say you have \(1\) and divide it by \(3\):

\[\frac{1}{3} = 0.\overline{3}\]

Where \(0.\overline{3}\) means \(0.33333...\). If we now multiply this equation back by 3, we get

\[\underbrace{\frac{1}{3}\cdot 3}_{=1} = 0.\overline{9}\]

If you don't trust this simple example, lets first examine what \(0.\overline{9}\) actually means for a better proof:

\[0.\overline{9} = \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+... = 9\cdot\sum\limits_{i=1}^\infty10^{-i}\]

When we now call \(x=0.\overline{9}\), we get

\[\begin{array}{rl} x &= \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...\\ \end{array} \]

If we multiply both sides by 10, we just move the decimal power by one place:

\[\begin{array}{rl} 10x &= 9+\frac{9}{10}+\frac{9}{100}+...\\ \end{array} \]

Now subtracting the definition of \(x\) from this equation yields

\[\begin{array}{rrl} &10x &= 9+\frac{9}{10}+\frac{9}{100}+...\\ -(&x &= \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...)\\ \end{array}\]

Which of course gives

\[\begin{array}{rl} 9x &= 9\\ x &= 1 \end{array}\]

And since we initially said that \(x=0.\overline{9}\), we showed that \(1=0.\overline{9}\checkmark\)