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**bossk171****Member**- Registered: 2007-07-16
- Posts: 303

Please bare with me, I've never taken a formal number theory class, but this is something I've been thinking about...

Any decimal can be represented by a fraction as long as the number is either repeating or terminating.

Terminating: Suppose you have the number 0.4 and wish to represent it as a fraction, it's quite easy, just take every thing after the decimal, and put it in the numerator. Now in the denominator put 10^k (where k = the number of decimal places after the decimal). In this case, .4 = 4/10 = 2/5. Pretty straight forward.

Repeating: Very similar process, take the number until the point it starts to repeat and put that over [(10^k)-1]. For example: .747474...= 74/99

The opposite can be done as well:

Assume a number:

By multiplying it by

You can easily turn it into a fraction b/(10^k) (where k = mn). I think it's kind of obvious even if I've done a poor job explaining it.

What I don't think it obvious is that any fraction NOT in the form:

can be written in the form:

Can anyone prove that with minimal technical jargon? I'd really like that.

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The k in your example with the terminating decimals should actually be m+n, not mn. It's also possible to use max(m,n) as your k instead, which means smaller numbers.

As for your question, I'd say it's easier to directly prove that any fraction's decimal form must repeat, rather than that what you're trying to do.

This isn't a formal proof, more of a (hopefully) convincing demonstration.

Consider how the decimal form of 1/7 is worked out.

0. 1 4 2 8 5 ...

7|1.[sup]1[/sup]0 [sup]3[/sup]0 [sup]2[/sup]0 [sup]6[/sup]0 [sup]4[/sup]0 [sup]5[/sup]0 0...

Hopefully you can see what I'm doing there. I've added zeroes after the decimal point and turned 1 into 1.0000000..., and the little numbers above each zero represent the remainder from the previous column.

1 goes no times into 7 and leaves a remainder of 1, which goes into the next column and becomes 10.

10 goes one time into 7 and leaves a remainder of 3, which moves and becomes 30.

etc...

We could keep going and until we find a repeat, but we don't need to. During this process, we've come across every remainder once, so the next stage will have to be a repeat of something we've already had. That will give the same answer and remainder as it did last time, which means the next stage will be same as the stage after the stage that just got repeated. Sorry for that horrible sentence, but hopefully you see what I mean.

Same applies to the stage after that, and so on. Therefore, 1/7 is a repeating decimal.

The numerator of the fraction didn't particularly matter, we'd just get the remainders in a different order, so to generalise we can say that n/7 is a repeating decimal.

(One note: we technically haven't had all possible remainders, because 0 is missing. But it's easy to see that if 0 ever popped up, then no remainder would get carried forward and we'd end up with a terminating decimal)

The important point here is that if any remainder ever repeats itself, the answer will recur.

I chose 7 because it exhausts all possible remainders before it starts repeating, which nicely demonstrates that a repeat *must* happen, but you don't necessarily need total exhaustion like that.

1/3, for example, just keeps throwing out remainders of 1 and you get 0.333... as the answer.

Now looking at a general denominator, it's fairly easy to see that the decimal expansion has to terminate or recur no matter what the denominator is. For a denominator n, the maximum number of decimal places you can go without repitition is (n-1), and after that a repeat will definitely happen. That or it'll have terminated by then.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Please bare with me, I've never taken a formal number theory class, but this is something I've been thinking about...

Number theory isn't much about fractions. It is however, about integers. There are some fractions involved at times such as continued fractions, but even those aren't actually fractions. In fact, most of the time you stay away from division as much as possible. You'd write m = k*n instead of m/n = k.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The proof would take four steps:

1. A rational number either repeats or terminates, but not both.

2. Any terminating rational number can be represented by a/(2^m5^n), a, m, and n being integers.

3. A non-terminating rational number can not be represented by a/(2^m5^n).

4. Any repeating rational number can be represented by b/(10^k - 1).

Most of the proof is outlined here. Mathsyperson did 1, you did 2 and 4 but they would need to be formalized, and all that remains is 3, which is of course rather easy.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I got the feeling that my thing was what he wanted to actually prove, and he was trying to get the thing he asked for because the other proof follows from it and he thought it would be easier.

Why did the vector cross the road?

It wanted to be normal.

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**infinitebrain****Member**- Registered: 2008-07-01
- Posts: 153

wow it works except for 0.9!

The best thing about life is you don't know what to expect

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**bossk171****Member**- Registered: 2007-07-16
- Posts: 303

I didn't forget about this thread, I just had to let it sink in for a while.

I don't really understand what you're saying, Ricky, but mathsyperson's demonstration was really nice. It makes a lot more sense now.

Infinitebrain: What? I'm guessing you're talking about .99999... which is a reoccurring discussion around here, but .9 = 9/10

A thought: can we consider a terminating decimal repeating if we say it's the "0" that's repeating? ex: 1/6=.16666666... and 1/5 = .200000000000...

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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