# How does $\frac29 + \frac{2}{9^2}+\frac{2}{9^3}+\cdots=\frac14$ imply that the base 3 expansion of $1/4$ is $0.020202…$?

I found in this response that

$$\frac29 + \frac{2}{9^2}+\frac{2}{9^3}+\cdots=\frac14$$ implies that the base-$$3$$ expansion of $$1/4$$ is $$0.020202...$$. I am very new to the concept of base-$$3$$ expansions. Could someone please explain how this statement is true?

• Compare your expression with $0\cdot \frac{1}{3}+2\cdot \frac{1}{3^2}+0\cdot \frac{1}{3^3}+2\cdot\frac{1}{3^4}+0\cdot \frac{1}{3^5}+2\cdot\frac{1}{3^6}+\dots$ and then look at the definition of base-n expansions. – JMoravitz Apr 2 at 3:11
• Build intuition by remembering how base-10 expansions work. $0.51042$ for instance corresponding to $5$ tenths and $1$ hundredth, zero thousandths, four ten-thousandths, and two hundred-thousandths... that is $0.51042 = 5\cdot 10^{-1}+1\cdot 10^{-2}+0\cdot 10^{-3}+4\cdot 10^{-4}+2\cdot 10^{-5}$ – JMoravitz Apr 2 at 3:13
• @JMoravitz, this is good, and you should copy it into an answer. – Sammy Black Apr 2 at 3:14
• It's exactly the same as base $10$. we have $0.020202020.....$ is defined in any base $b$ as $0\cdot \frac 1b + 2\times \frac 1{b^2} + 0\cdot \frac 1{b^3} + 2\times \frac 1{b^4} + ....$. That is the definition of base $b$. So if $b= 3$ the we have $0.02020202... = 2\frac 1{3^2} + 2\frac 1{3^4} + 2\frac 1{3^6} = \frac 2{9} + \frac 2{9^2} + \frac 2{9^3} + .... = \frac 14$. .... Of course the significant part is knowing $\frac 2{9} + \frac 2{9^2} + \frac 2{9^3} + .... = \frac 14$ – fleablood Apr 2 at 3:19

By definition $$0.a_1a_2a_3a_4.....$$ in base $$b$$ is by definition equal to

$$\frac {a_1}{b} + \frac {a_2}{b^2} + \frac {a_3}{b^3}+\frac {a_4}{b^4} + .....$$

So in base $$3$$ we will have $$0.020202020..... =$$

$$\frac 03 + \frac 2{3^2} + \frac 0{3^3} + \frac 0{3^4} + ....$$.

As only the even spaces have digits and the odd spaces are all $$0$$ then this is equal to

\$\frac 2{3^{2\cdot 1}} + \frac 2{3^{2\cdot 2}} + \frac 2{3^{2\cdot 3}}+ ....=

$$\frac 2{9} + \frac 2{9^2} + \frac 2{9^3}+....$$

And that's that.

We can use geometric series that $$\frac 29 + \frac 2{9^2} + ... = 2(\sum \frac 1{9^k}) = 2\frac {1}{9-1} = \frac 14$$.

Alternatively $$0.0202020202020......= k$$.

So $$3^2k = 2.02020202020..... = 2 + k$$.

So $$9k = 2+k$$ and $$8k = 2$$ and $$k = \frac 14$$.

This is the exact same thing as in base $$10$$ that $$0.020202020... = \frac 2{100} + \frac 2{100^2} + \frac 2{100^3} +.....= \frac 2{100-1} =\frac 2{99}$$.

And that if $$0.02020202020.... = m$$ then

$$100m =2.0202020202.... = 2 + m$$

$$100m = 2+m$$

$$99m = 2$$ and $$m= \frac 1{99}$$.

Converting between base $$b$$ and base $$b^k$$ for $$k\ge2$$ is extremely simple: one base-$$b^k$$ digit corresponds one-for-one with $$k$$ base-$$b$$ digits.

The given result shows that the base-$$9$$ expansion of $$\frac14$$ is $$0.\overline2$$. Each digit $$2$$ becomes $$02$$ in base $$3$$, yielding the desired result $$\frac14=0.\overline{02}_3$$.