Binomial Expansion: Find 1/3 Decimal Expansion & Repeating Digits

[fk378]

Homework Statement

Find the binary decimal expansion of the fraction 1/3. Identify the repeating decimals of digits.

The Attempt at a Solution

I have that 1/3=0.0101111... and so the repeating digit is 1.

Is this right? It's the first time I've been exposed to binary expansion.

 

[Citan Uzuki]

No. In binary, 0.010111111... = 0.011 = 11/1000 = 3/8_dec. What method are you using to find the binary expansion?

 

[fk378]

1/3=x(1/2) + x(1/2)^2 + x(1/2)^3...

where each x is either 0 or 1. The first x cannot be 1 because 1/2 > 1/3
I used the same reasoning for the other x's.

How do you do it correctly?

 

[Citan Uzuki]

As a general method, I'd just do the long division in binary. See, for instance:

http://mathforum.org/library/drmath/view/55944.html
http://mathforum.org/library/drmath/view/55951.html

 

[fk378]

What am I supposed to divide by though? And what does this mean from the website you linked:
"33 / 3 (i.e. 100001 / 11)"

 

[Citan Uzuki]

You divide 1_bin = 1_dec by 11_bin = 3_dec.

"(i.e. 100001 / 11)" is simply stating that the binary equivalent of 33_dec is 100001_bin, and that 3_dec = 11_bin, so 33/3_dec = 100001/11_bin

 

[HallsofIvy]

I would do this by saying: 1/3 is less than 1/2 but larger than 1/4: 1/3- 1/4= 1/12 so 1/3= 1/4+ 1/12. 1/12 is less than 1/8 but larger than 1/16: 1/12- 1/16= 1/48 so 1/3= 1/4+ 1/16+ 1/48. 1/48 is less than 1/32 but larger than 1/64: 1/48- 1/64= 1/192 so 1/3= 1/4+ 1/16+ 1/64+ 1/192. You should be able to see the pattern now.

 

[Citan Uzuki]

I would do this by saying: 1/3 is less than 1/2 but larger than 1/4: 1/3- 1/4= 1/12 so 1/3= 1/4+ 1/12. 1/12 is less than 1/8 but larger than 1/16: 1/12- 1/16= 1/48 so 1/3= 1/4+ 1/16+ 1/48. 1/48 is less than 1/32 but larger than 1/64: 1/48- 1/64= 1/192 so 1/3= 1/4+ 1/16+ 1/64+ 1/192. You should be able to see the pattern now.

And once you see the pattern, you could always check its correctness by summing the geometric series. I really would try to get a handle on doing arithmetic in binary though, since I think it provides more insight than just looking at the problem in longhand. After all, if the problem was "find the denary decimal expansion of 1/3", would you really set it up by saying

$$\frac{1}{3} = \frac{d_1}{10} + \frac{d_2}{10^2} + \frac{d_3}{10^3} ...$$

and trying to find the $$d_i$$, or would you take advantage of the compact notation provided by the division algorithm?