# Write # as a ratio of two integers

1. Apr 18, 2004

### noboost4you

Problem: Write the number 3.1415999999999... as a ratio of two integers.

In my book, they have a similar example, but using 2.3171717... And this is how they solved that problem.

2.3171717... = 2.3 + (17/10^3) + (17/10^5) + (17/10^7) + ...

After the first term we have a geometric series with a = (17/10^3) and r = (1/10^2). Therefore:

2.3171717... = 2.3 + [(17/10^3) / (1 - (1/10^2))] = 2.3 + [(17/1000)/(99/100)] = (23/10) + (17/990) = 1147/495 == 2.3171717...

Thinking I could follow the similar steps with a different number, I thought it would work, but it really isn't.

This is what I did:

3.1415999999999... = 3.1415 + (99/10^6) + (99/10^8) + (99/10^10)

a = (99/10^6) and r = (1/10^2)

3.1415 + [(99/10^6) / (1 - (1/10^2))] = 3.1415 + [(99/1000000)/(99/100) = (31415/10000) + (1/10000) = (31416/10000) = 3.1416 which isn't 3.1415999999999...

What am I doing wrong?

Thanks

2. Apr 18, 2004

### Hurkyl

Staff Emeritus
Actually, it is.

P.S. any particular reason you were grouping the nines in pairs?

3. Apr 18, 2004

### noboost4you

technically, it is, but is that correct though? and no, there was no reason i paired them up.

4. Apr 18, 2004

### master_coda

3.1416=3.141599999999... is very true. So any fractional representation of one is a representation of the other. In fact, that's how I would have solved this problem; I wouldn't have bothered with an infinite geometric series in this case.

5. Apr 20, 2004

### HallsofIvy

Staff Emeritus

Techically it's true but is it correct??? Is that what you are asking?

"True" is "true"- there is no "technically"! And if it's true, then it's correct.