Decimal expansion of 1/2

Using the formula for a sum of a geometric series: $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$

$0.4999.... = 0.4 + 0.09 + 0.009 + 0.0009 + \ldots$

$= 0.4 + \frac{9}{100}\sum_{k=0}^{\infty} \left(\frac{1}{10} \right )^k$

$= 0.4 + \frac{9}{100} \times \frac{1}{1-(1/10)}$

$= 0.4 + \frac{9}{100} \times \frac{10}{9}$

$= 0.4 + 0.1 = 0.5$

Let $x=0.4\overline{9}$

So, $10x=4.\overline{9}$ and $100x=49.\overline{9}$ Subtracting second equation from first, $90x=49.\overline{9}-4.\overline{9}=45 \implies \boxed{x=0.5}$

Though the two numbers are nearly equal, they are not equal exactly.

Let us write $0.4999999...$ as $0.4+0.09+0.009+0.0009+...=0.4+\frac{9}{100}(1+\frac{1}{10}+\frac{1}{100}+...)=0.4+\frac{9}{100}\times \dfrac{1}{1-\frac{1}{10}}=0.4+\frac{9}{100}\times \frac{10}{9}=0.4+0.1=0.5$

In the image above we observe that the inequalities could go on infinitely. Let the right hand side be x. As x approaches 0, the inequality becomes 0.5(0) - 0.4(9) < 0 -> 0.5(0) < 0.4(9). Yet, we know that 0.5(0) > 0.4(9). We can observe how the conditions lead us to 0.5(0) = 0.4(9). Alternatively, an explanation by MIT Professor Agustín Rayo:

This means that the difference between 0.5(0) and 0.4(9) must be smaller than each of 0.01,0.001,0.0001, and so forth. Since the difference between real numbers must be a non-negative real number, and since the only non-negative real number smaller than each of 0.01,0.001,0.0001,… is 0, it follows that the difference between 0.5(0) and 0.4(9) must be 0.

$0.499...=0.4+0.0999...=0.4+0.999...\times10^{-1}=0.4+1\times10^{-1}=0.4+0.1=0.5=\frac{1}{2}$ See 0.999...=1