real number set and countablity

husseinshimal

the purpose of this post is an attempt tp show that real numbers set could be generated intensively,it also could be counted somehow by defining aspecial surjective or injectice function.i think the mathematical constructure of this post need to be fixed by an expert,thats why i need some help here.

Consider we express the tow positive real numbers ,A&B as,

A=Σam[(10)^(n-m)] B=Σbm[(10)^(n-m)] Where,( n,m=0,1.2,……) am,bm,positive integer Now if, am+bm=pm+10,pm<10 pm,positive integer Then we define the relationship R, ARB={pm*(10)^(n)}+{(pm+1)*(10)^(n-1)+..... Obviously, R; looks like adding backwards.e.g, 341R283=525 =(3+2)=5,(4+8)=12,(1+1+3)=5 Lets now pick up arbitrarily the infinite sequence

S0=Σn\(10)^(n),+Σn\(10)^(n+1) + Σn\(10)^(n+2)+.....

Where n=1to9 ,10to99,100to999 ,...etc.respectively

i.e, S0=0.123456789101112131415161718192021222324.... ,. In order to generate or count* the real numbers within the interval,e.g. (0,1),

We define the surjective function,F;

F:N→positive irrational numbers subset in(0,1)

Where ,
F(n)=SnR0.1,

Sn, the set of sequences,

S1=s0 R 0.1,

S2=s1 R 0.1,

Sn=Sn-1R0.1,

etc.

notice that (n-1) is suffix,

post(1)

There are an infinite sequences,s1,s2 that we can make S1RS2 Close enough to any real number.

post(2)
if the relation,R,has aseriouse mathematical use , can we solve equations of the form,
xRx=s,where,s=0.3,0.5,..etc ,or xR1=10? i mean can the relation,R,be generalized to involve such equations?or even negative numbers?

HallsofIvy

The fundamental problem is that your whole concept is flawed. The operations you define can only produce a countable set of numbers and the set of all real numbers is not countable. Or do you refuse to accept that? You define
"the surjective function,F;

F:N→positive irrational numbers subset in(0,1)"
which is, of course, impossible since that would imply the set of irrational numbers between 0 and 1 was countable.

husseinshimal

i dont get it,you said that The operations i defined can only produce a countable set of numbers and we know the set is countable if there exists acountable subset belong to the original set.

ramsey2879

The numbers that both HallsofIvy and I see here are only the rational numbers, but real numbers include the irrational numbers, so your set is incomplete. While you can make rational numbers close to irrational numbers they are still only the rational numbers and not the irrational numbers. Moreover for every rational number you say is close to an irrational number you can create an infinite number of irrational numbers still closer, so which of these irrational numbers does your counting number actually count?

HallsofIvy

No, that's non-sense. A set if countable it there exist a countable set containing it, not the other way around! Any set of real numbers, complex numbers, etc. contains the natural numbers- that does prove that any such set is countable!

husseinshimal

. first let me express my great apreciation for your being patient with me thank you all guys.I am just confused here alittle bit.i am not saying that real number set is countable.all what iam trying to understand is this,if the arbitrary sequence,S0,that i gave represents one irrational number in(0,1) and if the realtion,R,that i defined is not mathematically flawed.and if the definition of counable set which is,(G, is countable, i.e. there exists an injective function ,F:G→N.
Either, G, is empty or there exists a surjective function,F:N→G) is right.then dont you think that what i was trying to do might be right? all what iam saying is ,F(1)=S0R0.1=0.223456789101112....f(2)=S0R0.2=S1R0.1=0.323456....,i.e,F ,is asurjective function.but ,F,keeps all the irrational numbers within(0,1) as long as it just shifts the digits to the left as it defined in the relation,R,the question here guys,and iam sure that you are the experts,is , if there is no logical or mathematical objection about this then dont you think it is worth to study it?thank you

ramsey2879

If your function creates unique irrational numbers from a set of k "ordered" non-negative integers where k is a constant then I would say that your set is countable because all such sets of k integers can be sorted first by the sum of the k integers, then by ascending order, for instance from left to right of the k integers. Thus if k = 4, 1 = F(0,0,0,0), 2 =F(0,0,0,1), 3 = F(0,0,1,0), 4 = F(0,1,0,0), 5 = F(1,0,0,0), 6 = F(0,0,0,2), 7 = F(0,0,1,1), etc. When k itself is infinite this is not the case.