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

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.