# A question on Q

1. Jun 9, 2004

### Shemesh

1/2 = 2/4 or 4/8 or 8/16 or 37/74... and so on.

It means that in this case 1/2 is a basic Q member, which can be represented by infinitely many non-basic Q members.

In short 1/2 = n/2n.

What is the name of the Q members which their ratio is the basis of inifinely many non-basic Q members (as I show above)?

Is there a general way to define them?

What common properties we have to these basic Q members?

What is the connaction with the prime numbers and the basic Q members?
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A clearer version of my question:

1/2 = 2/4 or 4/8 or 8/16 or 37/74... and so on.

It means that in this case 1/2 is a basic Q member, which can be represented by infinitely many non-basic Q members.

In short 1/2 = n/2n.

By based on the "Fundamental Theorem of Arithmetic" http://mathworld.wolfram.com/Fundam...Arithmetic.html [Broken]
how can we find these irreducible Q members?

Last edited by a moderator: May 1, 2017
2. Jun 9, 2004

### matt grime

You are talking about the field of fractions of Z, these are the equivalence classes of elements in the localization. Your other questions are too vague for me to answer. You also sound like organic/www/dialog/doron shadmi and I hesitate to answer. Hopefully your next post will clear that up (sorry, if you aren't he, but I get paranoid) and make your other questions more focussed.

3. Jun 9, 2004

### Shemesh

Is there a general way to define any basic Q member?

For example, there is no arithmatical oparation that can define a prime number in a single step, and we need an algorithm for this.

Is this problem of not having a single-step formula, also related to the basic Q members?

Last edited: Jun 9, 2004
4. Jun 9, 2004

### matt grime

From your reply I'm even more convinced you're he, so I'll just tell you that the words you're looking for are LOWEST and TERMS, with the DENOMINATOR positive.

5. Jun 9, 2004

### Shemesh

Can you give an example of how I have to write the questions in my previous post, in a standard way?

Please refresh screen because i updated them, thank you.

6. Jun 9, 2004

### matt grime

Oh hell, not the prime number thing again? Firstly,you've not defined what YOU mean by operation, arithmetical or otherwise, and step is not defined either. So whilst we have a rough idea of what you mean it isn't formally understood to be meaningful mathematically. In particular why use the word 'problem'? A prime is defined as generating a maximalr ideal in Z. Arithmetic, not algorithmic. Though from the way you speak it sounds as if everything has to be algorithmic.

7. Jun 9, 2004

### Shemesh

By "single step" I mean that there exist some formula that your desired result can be found by it in single input/output iteration.

8. Jun 9, 2004

### matt grime

Then there is, the function has domain Z, and range 0,1, the definition of f, say, is f(x)=0 if x is composite, or a unit, or zero, and 1 if z is a prime. Voila.

9. Jun 9, 2004

### Shemesh

Matt, do you want to tell us that there exists a formula that get some n as an input and by a single iteration give us a 100% indication that this n is a prime number?

And i am not talking about Rabin-Miller probabilistic test http://mathworld.wolfram.com/PrimalityTest.html

10. Jun 9, 2004

### matt grime

The function I gave is exactly that. That you don't understand what a function is is your fault.

11. Jun 9, 2004

### Shemesh

By Rabin-Miller function you do not get a 100% indication thet your single iteration result is really a prime.

12. Jun 9, 2004

### matt grime

Doron, you do understand that the definition of a function as I gave, and its difficulty to evaluate for any arbitrary input are not related? The function I gave is a functional definition that distinguishes between composites and primes. You are once more confusing the calculation of "something" with the "something".

13. Jun 9, 2004

### Shemesh

Can we be 100% sure that our tested number is a prime?

14. Jun 9, 2004

### matt grime

Using my function, if the output is 1, then by DEFINITION of the function the input is prime, so I believe the answer is YES.

I am not talking about the probablitistic algorithm you mention.

Now, it's clear once more that you aren't going to learn from the responses, or even try to understand the mathematics, and are only here to peddle your crackpot theories, so don't bu shocked when the thread is moved. And don't expect another reply from me in this thread.

15. Jun 9, 2004

### Shemesh

Why do we have to believe that the answer is YES?

Please can you give an address when we can find the function that can give us 100% indication that our abitrary large n is really a prime?

16. Jun 9, 2004

### Shemesh

1/2 = 2/4 or 4/8 or 8/16 or 37/74... and so on.

It means that in this case 1/2 is a basic Q member, which can be represented by infinitely many non-basic Q members.

In short 1/2 = n/2n.

By based on the "Fundamental Theorem of Arithmetic" http://mathworld.wolfram.com/FundamentalTheoremofArithmetic.html
how can we find these irreducible Q members?

17. Jun 9, 2004

### hello3719

well for that you have to use a prime generating function, I think there is one , but it is a very big one and not at all efficient to find primes(it uses some results of wilson's theorem).

18. Jun 9, 2004

### maverick280857

Hi

You'll have to forgive me if my post is not germane to the topic...

Using multiple applications of the identity,

$$\frac{1}{k} = \frac{1}{k+1} + \frac{1}{k(k+1)}$$

the fraction,

$$\frac{1}{2}$$

can be written in an infinite number of ways.

Cheers
Vivek

19. Jun 9, 2004

### Hurkyl

Staff Emeritus
There's a trivial one; simply try to divide n by every number smaller than it. It's not quick, but it works.

And, I believe, there exist deterministic algorithms that can test primality in polynomial time (in the number of digits).

Use Euclid's algorithm to find the GCD of the numerator and denominator.

Last edited by a moderator: May 1, 2017
20. Jun 9, 2004

### master_coda

Yes. The AKS primality test can run in about O(log^6 n) time, I believe.