## You are here

Homedefinition of vector space needs no commutativity

## Primary tabs

# definition of vector space needs no commutativity

In the definition of vector space one
usually lists the needed properties of the vectoral addition^{}
and the multiplication of vectors by scalars as eight axioms,
one of them the commutative law

$u+v=v+u.$ |

The latter is however not necessary, because it may be proved
to be a consequence of the other seven axioms. The proof can
be based on the fact that in defining the group,
it suffices to postulate only the existence of a right identity^{}
element and the right inverses of the elements (see the article
“redundancy of two-sidedness in definition of group”).

Now, suppose the validity of the seven other axioms, but not necessarily the above commutative law of addition. We will show that the commutative law is in force.

We need the identity^{} $(-1)v=-v$ which is easily justified
(we have $\vec{0}=0v=(1+(-1))v=\ldots$). Then we can
calculate as follows:

$\displaystyle v+u$ | $\displaystyle=(v+u)+\vec{0}=(v+u)+[-(u+v)+(u+v)]$ | ||

$\displaystyle=\;[(v+u)+(-(u+v))]+(u+v)=[(v+u)+(-1)(u+v)]+(u+v)$ | |||

$\displaystyle=[(v+u)+((-1)u+(-1)v)]+(u+v)=[((v+u)+(-u))+(-v)]+(u+v)$ | |||

$\displaystyle=[(v+(u+(-u)))+(-v)]+(u+v)=[(v+\vec{0})+(-v)]+(u+v)$ | |||

$\displaystyle=[v+(-v)]+(u+v)=\vec{0}+(u+v)$ | |||

$\displaystyle=u+v$ |

Q.E.D.

This proof by Y. Chemiavsky and A. Mouftakhov is
found in the 2012 March issue of The American Mathematical
Monthly.

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## Commutativity sum on the vector spaces

All it is based upon the transitivity property embedded on the the couple equations (-v)+v=0, v+(-v)=0, where v and (-v) are members of some vector space V, where v is arbitrary and (-v) enjoys uniqueness.

## transitivity

Dear perucho, I don’t clearly see whether it it a question of the transitivity (of the equality of vectors?). Can you please explain elementarily?

Jussi

## A polynomial that never fails

A polynomial that nnever fails in generating a valid base for pseudoprimality of 105: 41^n + 41^(n-1) + 104. When we program this polynomial we get a sequence of valid bases for pseudoprimality of 105.

## A polynomial that never fails

A polynomial that nnever fails in generating a valid base for pseudoprimality of 105: 41^n + 41^(n-1) + 104. When we program this polynomial we get a sequence of valid bases for pseudoprimality of 105.

## A monomial that never fails

41^n*29^(n-1) is a monomial that never fails to generate a valid base for Fermat pseudoprimality of 105.

## A constructed ring

Let S be the set generated by 29^n(14 + 15i) where n belongs to N. Let S’ be the set generated by (-29)^(2n-1).(14+15i). Let S” be the set generated by (-29)^(2n).(-14-15i). Let S_1 be the set (0, S,S’, S”). Then R= ( S_1, +, . ) is a ring as it satisfies the 6 basic axioms pertaining to a ring. Note: 1) Excepting 0,each member of the ring is a valid base for pseudoprimality of 105 in Z(i). 2) R is not a subring of Z(i).

## Good news

Glad to note message section working again!

## conjecture

Any irreducible quadratic polynomial (where the variable belongs to Z ) generates an infinite set of prime numbers. For example f(x) = x^2 + x +1 will generate an infinite set of prime numbers: 3, 7, 13, 31, 43, 73. . . . A sketch proof, using failure functions, will be furnished seperately.

## conjecture

Hi Deva, can you give the proof without failure funtions which ones I don’t understand?

BTW, see the example http://planetmath.org/exampleofgcd

Jussi

## conjecture

Hi, Jussi; Sorry am unable to give the sketch proof without using the mathematical tool viz. ”failure functions ”.

## Conjecture - sketch proof

Before I give the sketch proof I request members to read my article ” Failure functions ” and the the following message : Application of failure functions - indirect primality test ( both on planetmath ).

## Conjecture- sketch proof

Conjecture: any irreducible quadratic polynomial, in which the variable belongs to Z, generates an infinite set of prime numbers. For illustration I am taking the irreducible quadratic polynomial: f(x) = x^2+x +1. Definitions:1) failure: a composite number. 2) failure function: x = x_0 + k*x_0 (here a belongs to N and is fixed; k belongs to Z and x_0 is a specific value of x. In other words x is a function of x_0. When we substitute the values of x generated by the failure function in f(x) we get ONLY failures ( composites ). Example: x = 1 +3*k generates 4,7, 10, 13. . . Any of this infinite set, when substituted in f(x) results in a failure (composite). Indirect primality test: in the sketch proof I am developing I will be testing only whether a particular value of x, say x_1, is is covered by one or more failure functions or not. If it is covered, then f(x) is a failure (composite ); otherwise f(x _1) is prime which need not be tested for primality. We do not directly test whether f(x_1) is prime or composite; that is why this is an indirect primality test. (to be continued).

## Conjecture- sketch proof (contd.)

Briefly the sketch proof consists of the following steps: 1) Iteration: Let p_0 be the largest known prime with shape x^2 + x +1. Let x_0 be the relevant value of x i.e. x_0^2 +x_0 +1 = p_0. Consider the interval x_0, x_0 + p_0.( this interval is chosen because f(x_0 + k*f(x_0)) is congruent to 0 (mod (f(x_0)). ). In each iteration there is a percentage of values of x not covered by the relevant failure functions - these are such that the relevant f(x)s are prime (which need not be tested for primality). Do the relevant failure functions cover the whole interval? If so the iteration has come to an end which means p_0 is the largest prime of the shape x^2 + x + 1 and that there are only a finite number of primes of this shape. If not we go to the second iteration, i_2.: let x_1 be the largest value of x not covered by the relevant failure functions; f(x_1) is prime - let this be p_1. Consider the interval x_1, x_1 + p_1. Do the relevant failure functions cover the this interval completely? If so iteration has come to an end and p_1 is the largest prime of this shape; this also means there are only a finite number of primes of this shape. If not we go to iteration, i_3. My conjecture: the percentage of xs not covered by the relevant failure functions will decrease from iteration to iteration progressively; however the decrease of percentage is asymptotic to 3. i.e. it never reaches 3. Hence the iteration is perpetural. Therefore the infinitude of primes of shape x^2 + x + 1 is proved. (to be continued ).

## Conjecture - sketch proof - concluding message

Note 1) p_0 less than p_1 less than p_2…….. Hence the intervals get progressively larger; even a small percentage of xs not covered by the relevant failure functions means a sizable number of candidates for initiating the next iteration. However we need only one x not covered by any failure function to go to iteration i + 1. This is the largest of the xs not covered. 2) Only a competent programmer can program the failure functions and perform the iteration. 3) For the sake of demonstration I have selected x = 12 and the interval x = 12, 12 + 157 (12^2 + 12 +1 =157 ). The iteration is done as follows: I make a list of xs starting with 12 and ending with 169. Which are the relevant failure functions? 1+2k, 2 + 7k, 3 + 13k, 4 + 7k, 5 +31k, 6 +43k……..ending with 47 + 61k. These are relevant because these cover the interval chosen. Next I circle the xs covered by each failure function. For example 1 + 3k generates 4, 7, 10, …After performing the same procedure with all the relevant failure functions I found I was left with 37 unmarked xs. The percentage of unmarked xs is 24The list of 37 unmarked xs: 14, 15, 17, 20, 24, 27, 33,38, 41, 50…ending with 167. 3) What are the implications of the iteration coming to an end after the ith iteration? p_(i-1) is the largest prime with with shape x^2 + x+1. This means x_(i-1) is the largest uncovered x; all subsequent xs are such that f(x) are composite having shape 3^q7^m13^n…..x_(i-1)^z ( here q,m,n…z are exponents belonging to N). This is highly improbable, perhaps impossible.

## conjecture - three points

1) Carl Pomerance has just pointed out an exception to my generalisation.However I stand by my message on x^2 +x +1. 2) A minor correction -second last line - shape 3^q….. (p_(i-1)^z. 3) I ran the program p(n) = factorint( n^2+n+1) from n=1 to n=1000. I could not find a single exponent greater than 2.

## Conjecture - clarification

Sorry I misunderstood Pomerance; he has clarified that it has not yet been proved.

## conjecture - a fewpoints

1) I double checked my computations - I am convinced that a value of x, say x’, if generated by one or more of the relevant failure functions then f(x’) is a failure (composite). If not f(x’) is prime and it need not be tested for primality. 2) in the demonstration sample the largest uncovered value of x in the interval chosen is 167; f(167) = 28057 and the second interval to be considered is 167, 28224. Obviously this cannnot be dealt with manually.

## conjecture - a fewpoints (contd)

3) Bonus: when a value of x, say x’, is generated by a failure function we not only know that f(x’) is composite- we also know what f(x’) is a multiple of. For example let us take the failure function x = 14 + 211k; when k =1, x’ = 225. f(14+ 211) is a composite and it is a multiple 211. Needless to say this bonus aspect is not important for application of failure functions in proving the infinitude of primes os shape f(x).

## Random thoughts on proofs

1) Perhaps some conjectures cannot be proved without collaboration with a programmer - unless the mathematician himself happens to be a gifted programmer 2) What is the value of alternate proofs? For example suppose someone comes up with an alternate proof of Fermat’s last theorem - who is going to pay attention even if the latter is simpler?

## failure functions - another example

Let our definition of a failure be a non-Carmichael number. Let the parent function be the polynomial f(x) = x^2 + x + 9. When x = 23 we get the Carmichael number 561. However x = 8 + 81k ( k belongs to Z ) is a failure function since when we substitute a value of x generated by this failure function we get only failures ( non-Carmichael numbers ).

## sketch proof

Attention: Pahio and others. Many feel that my terminology is not comprehensible. I therefore propose to develop the sketch proof, step by step, using only commmonly understood terminology. After each step I will give a break for a day. Is this ok?

## It would be desirable.

It would be desirable.

## Please show the proposition

Please show the proposition you want to prove. Then give its proof in a concise and clear form.

Regards,

Jussi

## sketch proof - in commonly understood terminology

Thank you, Jussi. Will commence soon.

## sketch proof - in commonly understood terminology

It is known that a sketch-proof is a plan for a proof; if things go as per plan we may get a proof. To begin: Aim- to prove that they are infinitely many primes of shape n^2 + n +1. Here n belongs to N. As f(n) = n^2+n+1 we get f(1) =3, f(2) = 7, f(3) = 13, f(4) = 21…… (to be contd.)

## sketch proof - in commonly understood terminology

It is known that a sketch-proof is a plan for a proof; if things go as per plan we may get a proof. To begin: Aim- to prove that they are infinitely many primes of shape n^2 + n +1. Here n belongs to N. As f(n) = n^2+n+1 we get f(1) =3, f(2) = 7, f(3) = 13, f(4) = 21…… (to be contd.)

## sketch proof - in commonly understood terminology

It is known that a sketch-proof is a plan for a proof; if things go as per plan we may get a proof. To begin: Aim- to prove that they are infinitely many primes of shape n^2 + n +1. Here n belongs to N. As f(n) = n^2+n+1 we get f(1) =3, f(2) = 7, f(3) = 13, f(4) = 21…… (to be contd.)

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

## sketch proof - in commonly understood terminology

n=1+3k is an arithmetic progression.Any value of n generated by this progression is such that it is congruent to 1 (mod3). Similarly any n generated by n=2+7k (here k belongs to Z) is congruent to 2 (mod7). n, thus generated, is such that f(n) is composite. Generally f(n_0+kf(n_0)) is congruent to 0 (mod(f(n_0)).

## sketch proof - in commonly understood terminology

Any value of n not covered by any of the arithmetic progressions like 1+3k, 2+7k, 3+13k…. is such that f(n) is prime.