Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**UltraV****Guest**

I'm trying to solve a question related to recurring decimals and I've just found out I need to represent them algebraically.

For example,

If I have the recurring decimal 0.11798... (798 being repeated), I get an answer of 3929/33300.

But, I have the problem

0. ... 251 (251 being repeated and there being an unknown number of digits before this).

Can anyone help me?

Thank you!

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

Hi UltraV

Welcome to the forum.

I agree with 3929/33300.

But it sounds like an impossible task if the number of missing digits is unknown.

0.251 recurring ;. 0.0251251 recurring; and 0.6789251251251 recurring will have very different fractional forms. You have to know what all the digits are. It would

be like trying to solve an equation when you don't know what it is.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

Can you solve

0.251251251251...

first?

1000/x -> 251

x - >4 ?

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

suppose we have A/B = 0.251251251...

A*10//B = 2 remainder A1

A1*10//B = 5 remainder A2

A2*10//B = 1 remainder A3

and A3==A?

Let us check an example of 2/11

A = 2 & B =11

A*10 // B = 1 remainder 9

90 // B = 8 remainder 2=A

so 2/11 = 0.181818......

And you may try primes as A and B and solve 0.251251... out.

**X'(y-Xβ)=0**

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

The OP has not suggested what digits come before the 251 repeats, so I'll make up an example. It is always possible to get a fraction.

example. Convert 0.3251251251 recurring into a fraction.

step 1. Split off the non recurring part.

0.3 + 0.0251251251...…. note 0.3 = 3/10, so that part is now a fraction.

step 2. Call the recurring part a/b, where a and b are integers and multiply the recurring part by a suitable power of ten. There are three recurring digits so 10^3 is suitable. For n recurring digits use 10^n.

a/b = 0.0251251251.... 1000a/b = 25.1251251251251.....

step 3. Subtract to eliminate the recurring part.

999a/b = 25.1

step 4. Multiply by a suitable power of ten to make this a ratio of integers. There was one non-recurring digit so 10^1 is suitable. If there had been m non-recurring digits then use 10^m.

9990a/b = 251 => a/b = 251/9990

step 5. Combine the two fractions to complete the conversion.

3/10 + 251/9990 = (2997 +251)/9990 = 3248/9990

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

Great, the solution of the rational seems to utilize the 0.999...=1 'proof'.

**X'(y-Xβ)=0**

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

hi George,

Yes it does in a way.

I know from many posts on the subject of 0.9999..... that some folk are unhappy with playing around with 'infinite' series like this but I think they fail to understand something fundamental about maths. The job of maths is just to help make 'models' of real situations. The test of whether it does this job is 'does it work?'

In the above example if you evaluate 3248/9990 you do get that recurring decimal so I claim that makes the method valid.

I have met an alternative definition of the real number system that says that between every pair of different real numbers there is another. So as not to cause a contradiction with 0.99999..... such sets of recurring digits are disallowed in the definition. This also works but requires a fresh approach when considering 'infinite' series.

If all terminating decimals and all recurring decimals can be converted to fractions, this means that the irrationals are the decimals that neither terminate nor recur. At university I was shown a proof that 'e' and 'pi' are such numbers but I didn't really follow it then so don't expect me to be able to say what it is now. I expect Mr Google knows though

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

bob bundy wrote:

hi George,

Yes it does in a way.

Bob

Hi Bob, I think both views have their rational.

By the way, you seem to be fond of pentagons. Is it because it contains the golden ratio phi = (√5 -1) /2 ?

**X'(y-Xβ)=0**

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

hi George,

That's as good a reason as any I suppose. At school, geometry was my best topic and I rather like that gif as it has some interesting properties as you say. From what I can remember I used some static geometry software to make the diagram and then animated it in flash. That's about the extent of my flash programming though. Once I made it I decided to stick with it.

There's a related thread here: http://www.mathisfunforum.com/viewtopic … 87#p378587

As a 'lets do some exploratory maths for fun rather than just following the text book' I used to do the following.

Start with Pascal's triangle. By adding diagonally discover the Fibonacci sequence. Divide each term by the previous one to generate a converging sequence leading to the golden ratio. Draw a rectangle with sides in the golden ratio. Cut off a square. Check out the ratio of the new rectangular remainder and discover it's in the golden ratio too. Find out why some seed heads form golden ratio spirals; why snowdrops multiply in a Fibonacci sequence etc.

So, thinking about it, yes, that must be the reason .

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

bob bundy wrote:

hi George,

That's as good a reason as any I suppose. At school, geometry was my best topic and I rather like that gif as it has some interesting properties as you say. From what I can remember I used some static geometry software to make the diagram and then animated it in flash. That's about the extent of my flash programming though. Once I made it I decided to stick with it.

I felt strange about it. A famous Chinese mathematician Hua Luogeng preached 0.618 rule, orginated by Jack Kiefer, to substitute bisection method.

Do you know why it improves efficiency? Or does it improve efficiency at finding roots?

**X'(y-Xβ)=0**

Offline