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Dear Experts.

Do you know some literature on

representation of rational numbers by formulas? TIA.

Do you know some literature on

representation of rational numbers by formulas? TIA.

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- Thread starter nworm
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- #1

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Dear Experts.

Do you know some literature on

representation of rational numbers by formulas? TIA.

Do you know some literature on

representation of rational numbers by formulas? TIA.

- #2

arildno

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Eeh?

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For example, let B = {x + y, x - y, xy, x^(-1), 1},

B1 = {x + y, xy, x^(-1), 1}.

Constants 1 are formulas that realize number 1.

If F1, F2 are formulas in basis B (F1 realize number r1 and F2 realize number r2), and w from B is basic operation,

then F = w(F1,F2) is formula that realize number w(r1,r2) (by definition).

If operation w(x) = x^(-1) then w(F1) is formula, that realize number w(r1).

- #4

AKG

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None of this makes sense to me.

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then

7/5 = (1 + 1 + 1 + 1 + 1 + 1 + 1)*(1 + 1 + 1 + 1 + 1)^(-1)

1/3 = (1 + 1 + 1)^(-1).

Did you see this formulas somewhere?

- #6

arildno

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What nonsense is this???

- #7

Hurkyl

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I think nworm is asking something about constructing rational numbers out of the constant *1* and the functions +, -, *, and [itex](\cdot)^{-1}[/itex]. As to *what* he's asking, and for what purpose, I have absolutely no clue.

By the way, nworm, depending on what you're asking, you have forgotten to specify we can use parentheses in expressions.

By the way, nworm, depending on what you're asking, you have forgotten to specify we can use parentheses in expressions.

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- #8

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One day I read a book of arithmetical problems. I like the section that told about constructing of rational numbers by formulas (formulas are generated by superposition operations).

In this book "complexity of formula" is defined. It is number of symbols 1 in formula. "Complexity of number" r is minimal complexity of formula that realizes r.

(It is denoted as LB(r)).

Example of task:

Let B2 = {x + y, x^(-1), 1}, B3 = {x + y, (x^(-1)+y^(-1))^(-1), 1}.

Prove, that

LB3(r) = LB2(r), where r is any fraction.

I want to read about this else (exactly about rational numbers).

- #9

AKG

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Using no more 1's with B3, you can create x+y and (x

So getting back to the one we missed? Can we invert numbers using B3 without using any more 1's? Actually, I don't think it can be done. Using x

Actually, since the number of 1's required to make x is the same as the 1's required to make x

Maybe we can go back to assuming that x takes the same 1's with B2 or B3, and still prove that x

- #10

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I am thinking that we can't create x

I have found the solution of this task (on the other page)

As function (x

If a corollary formula of a formula in B2 is given by F

G=((...((F1

The formula F and G are equivalent. The complexity of G is the same as the complexity of F. By iterating this transformation we can find a formula H in B3. The complexity of H is the same as the complexity of F. Therefore LB3(r) <= LB2(r). Therefore LB3(r) = LB2(r).

I am repeating my question. Do you know some literature about this (about rational numbers)

I didn’t read a lot of books in the field of number theory (I read “A Course in Number Theory and Cryptography” by Neal Koblitz only).

- #11

AKG

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xnworm said:Thank you very much :rofl: These ideas are very interesting :tongue2:

I am thinking that we can't create x^{-1}by using B3.

Yes, that's correct.As function (x^{-1}+ y^{-1})^{-1}is expressed in B2 as noniterated superposition, then any formula in B3 can be transformed to formula in B2 without changing of complexity. This implies that LB3(r) >= LB2(r).

I see, very nice. So we are using induction. G (in B3) has the same complexity as F (in B2) as long as the corollary formulas can be made in B3 with the same complexity. But they can since each corollary formula FIf a corollary formula of a formula in B2 is given by F^{-1}, then this corollary formula is represented as (F1^{-1}+F2^{-1}+...+Fn^{-1})^{-1}, where Fi are corollary formulas of less complexity or constants 1. Write a new formula G.

G=((...((F1^{-1}+F2^{-1})^{-1})^{-1}+F3^{-1})^{-1})^{-1}+...+Fn^{-1})^{-1}= g(g(...g(g(F1,F2),F3)...),Fn), where g(x,y)=(x^{-1}+y^{-1})^{-1}.

The formula F and G are equivalent. The complexity of G is the same as the complexity of F. By iterating this transformation we can find a formula H in B3. The complexity of H is the same as the complexity of F. Therefore LB3(r) <= LB2(r). Therefore LB3(r) = LB2(r).

Sorry, I've never seen anything like this.I am repeating my question. Do you know some literature about this (about rational numbers)

I didn’t read a lot of books in the field of number theory (I read “A Course in Number Theory and Cryptography” by Neal Koblitz only).

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