# Are counting numbers or natural numbers a set within real numbers?

Discussion in 'Numerical Analysis' started by Martin, Jun 3, 2011.

1. ### MartinGuest

The definition of a real number is a quantity along a continuum. For
example -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. Counting numbers (orï»¿
natural numbers as they are called) are basically how we count out
loud and contain 1, 2, 3, 4, 5. Since 1, 2, 3, 4, 5 is a subset of -5,
-4, -3, -2, -1, 0, 1, 2, 3, 4, 5 counting numbers are a subset within
real numbers. Agree or disagree? Martin

Martin, Jun 3, 2011

2. ### calvinGuest

You're confusing integers and real numbers.

calvin, Jun 4, 2011

3. ### PubkeybreakerGuest

No, this is NOT the definition. I suggest that you should
actually learn this subject before making further ridiculous
claims. Take a course in real analysis. Look up "Cauchy sequence"
or "Dedekind cut".
This set does NOT form a "continuum". It is discrete.
Mathematics is not done by popular poll. However, in this
case, this is the first statement you have made that is correct.

Pubkeybreaker, Jun 4, 2011
4. ### MartinGuest

I understand the entire set of reals are not limited to integers or
whole numbers and did not confuse the two though I understand how it
may have appeared so.

Martin, Jun 4, 2011
5. ### MartinGuest

This is precisely the definition: In mathematics, a real number is a
value that represents a quantity along a continuum http://en.wikipedia.org/wiki/Real_number
I suggest that you should
I would like to.
Look up "Cauchy sequence"
Cauchy sequence convergant for R or converge in the reals. A sequence
a_1, a_2, ... metric d(a_m,a_n) satisfies lim_(min(m,n)-
&gt;infty)d(a_m,a_n)=0.
set partition of the rational numbers into two nonempty subsets S_1
and S_2 all members of S_1 are less than those of S_2 and S_1 has no
greatest member.
Agreed.
However, in this
How can a question of agree or disagree be correct or incorrect or
rather how can a question be correct or incorrect?

Martin, Jun 4, 2011
6. ### Arturo MagidinGuest

The very next paragraph in that page reads:

"These descriptions of the real numbers are not sufficiently
rigorous by the modern standards of pure mathematics. The discovery of
a suitably rigorous definition of the real numbers — indeed, the
realization that a better definition was needed — was one of the most
important developments of 19th century mathematics."

So, no, that is *not* "precisely the definition". It's not even a
definition, it's a nonrigorous description.

Arturo Magidin, Jun 4, 2011
7. ### nmm1Guest

Which is itself flawed! "Certainly, N and Z are entirely different
animals; set-theoretically, you can even show that they are disjoint."

The concept of disjointness applies only with two subsets of the
same superset. You can define sets where Z is a subset of R,
ones where they are disjoint, ones where they overlap, and even
more arcane possibilities.

Regards,
Nick Maclaren.

nmm1, Jun 4, 2011
8. ### nmm1Guest

I suggest that you look further into it.

Something that real mathematics (though not some of the computer
science that claims to be mathematics) is rather keen on is absolute
consistency, in that the statements A and not A should not both be
true. If you start applying the concept of disjointness to two
sets except in the context of being subsets of a superset, they
can be both disjoint and not disjoint. As that very reference
indicates!
And, without some superset to define the rules of doing that, the
union operator is undefined!

Regards,
Nick Maclaren.

nmm1, Jun 4, 2011
9. ### Aatu KoskensiltaGuest

There is no such requirement in ordinary set theoretic
mathematics. Even if we for some reason thought it prudent to talk of
two sets as disjoint only when they both are subsets of some set, this
would make no difference: given sets A and B they are both subsets of A
union B.
We can define all sorts of sets. What do you take to be the relevance
of this trivial observation?

Aatu Koskensilta, Jun 4, 2011
10. ### Aatu KoskensiltaGuest

There's nothing further to look into. If we inspect the definition of
disjointness in any standard set theory text we find no mention of any
supersets. We might find talk about some fixed superset U, and
operations on subsets of U, in introductory discrete mathematics texts,
of course.
Something real mathematicians are usually not particularly keen on is
contentless blather about "absolute consistency" and so on.

Aatu Koskensilta, Jun 4, 2011
11. ### Frederick WilliamsGuest

Are counting numbers or natural numbers a set within real numbers?

is "yes and no". One can construct the real numbers from the natural
numbers in various ways, and according to the ways that I know of the
answer is "no", but one identifies the natural number 6 (say) with the
real number 6. Or one can start with the real numbers which, being a
field, have 0 and 1 among them, and then define the natural numbers to
be

0, 0+1, 0+1+1, etc.

Frederick Williams, Jun 4, 2011
12. ### Frederick WilliamsGuest

That's not a problem. The usual notions of true and not are such that
if A is true, not A is false; and if A is false, not A is true.

Frederick Williams, Jun 4, 2011
13. ### Ronald BruckGuest

There's nothing wrong with this statement, since disjointness depends
only on the resolution of the statement "x = y", which is a primitive.
Reading the full paragraph, the authors define Z as something like N x
{0,1}, where a "0" corresponds to a "-" and a "1" to a "+" (removing
the extra representation of 0). Nothing of this form is in N, so
they're disjoint, AS SETS.

Of course, in the usual development this is followed by a redefinition
of N as N x {1}, and only then is N \subset Z.

-- Ron Bruck

Ronald Bruck, Jun 4, 2011
14. ### Arturo MagidinGuest

Okay, so what we actually have here is that you have absolutely no

That is complete, absolute, abject nonsense, without any basis
whatsoever in set theory. You have no idea what you are talking about.

Take a sentence out of context; add a dash of ignorance, and what you
get is Nick Maclaren's pronouncements on mathematics.

Arturo Magidin, Jun 4, 2011
15. ### Arturo MagidinGuest

The "very reference" indicates no such thing. Do stop projecting your
ignorance on others.

Two sets are disjoint if and only if their intersection is empty.
Period.

Sets cannot be both "disjoint" and "nondisjoint". Don't confuse the
*name* *you* give a set with the set itself.
The existence of unions is part of Set Theory. It's an axiom, in
fact.

In ZF set theory, the Axiom of Pairing asserts that for every sets A
and B, there exists a set Z such that A and B are elements of Z.
Applying then the Axiom of Separation, we conclude that for every sets
A and B, there is a set Z whose elements are *precisely* A and B.

The Axiom of Unions states that if W is any set, then there is a set Y
such that x is an element of Y if and only if there exists an element
w of W such that x is an element of W; that is, Y is the union of the
elements of W.

Combine the two, and you get that for *any* two sets A and B (with no
"superset to define the rules of doing" anything) there is a set which
is equal to the union of
A and B.

If you don't know the subject, perhaps you might want to ask instead
of issue nonsensical assertions?

Arturo Magidin, Jun 4, 2011
16. ### GuestGuest

Thank you. Martin

Guest, Jun 6, 2011
17. ### Tim Golden BandTech.comGuest

Hi. I like your question and prefer to use physical examples and
graphical analysis to consider it. Traditionally, real analysis does
build its continuum out of the natural number by moving along to
develop more types that fill in the line. When people bothered to name
a number 'real' shouldn't we admit that they took this name seriously?
So I justify my physical instances which will expose the arbitrary
choice of unity along the real line, whereas within the counting
numbers no such arbitrary choice is possible.

Simply compare one centimeter against one inch and we see that the
encoding of the continuum's address requires a definition of unity. We
name this thing the unit and denote it by the symbol
1
or
1.00000...
Likewise when we draw a line upon a piece of paper to address the
continuum we are forced to plop down two arbitrary positions; zero and
unity, at which point we feel confident that we can address the rest
of the representation.

Relativity theory further blurs the representation by allowing for
numerous references, the offsets from one reference to another being a
means of address which invokes relative offsets, and in the
multidimensional form taking on even more complexity.

I criticize the mathematician's real number firstly for its name. The
real world seems symbolically quantifiable, but only to some finite
precision. We might pull out a caliper and witness an instrument
capable of unencrypted continuous transcription abilities, whereas the
tape measure as it is used usually to produce an intermediate number
cannot provide this level of integrity. Still, as a matter of design
the numerically quantified piece is the most reproducible and
transferable design, presuming that the unit of the work was
accurately transferred in the first place.

I believe that the continuum can and should stand freely aside from
the discrete value. I believe that magnitude, as it lacks any
signature is more fundamental than the real number and so the problem
can be dissected slightly more. Under this dissection the ray becomes
more fundamental than the line, and in this way perhaps we can come to
view the vector as a more primitive format than it is built in today's
math. Let's not forget also that the term 'dimension' springs from the
real line as fundamental; the number of dimensions of a system being a
matter of the quantity of real lines(typically orthogonal) required to
represent that system. By splitting that real line in half what
becomes of these dimensional analyses? They do survive, though off by
one. Still, that one matters. The ray... the continuum... the discrete
dimension that none of us can deny... time as a mystery still belies
the fallacy of modern mathematics. That unidirectional ray mimics the
unidirectional time that we witness, while we stumble to quantify it
with a real number. Why? Because this is an old presumption, and the
human mind operates via mimicry. False belief propagates as readily
and as unchallenged as truth within the mimicry paradigm.

- Tim http://bandtech.com/polysigned

Tim Golden BandTech.com, Jun 6, 2011
18. ### Michael StemperGuest

Michael Stemper, Jun 6, 2011
19. ### Ronald BruckGuest

For what? His answer is WRONG.

To say that N is NOT a subset of R is to argue that the technicalities
of the proof are to build, from N, a set Z (which is set-theoretically
disjoint from N, but which contains a "copy" of N), then to build a set
Q which contains a "copy" of Z (but which is disjoint from Z), etc. To
leave it at that--where N, Z, Q and R are all disjoint--is a little
like hiring a contractor to build you a house, and when you arrive to
take possession, you find he's left all kinds of forms and wood scraps
and unpainted walls. He hasn't finished the job.

The final step is to REDEFINE Q, Z, and N, so that they ARE subsets of
R. OK, the original contractor argues that he PUT the walls in, but
just didn't paint them; but that's not what most people would call
finishing the job.

-- Ron Bruck

Ronald Bruck, Jun 8, 2011
20. ### fishfryGuest

Bad analogy. The construction you outlined is the standard way N, Z, Q,
and R are defined. Each set injects into the next one in a natural way.

The reason this is not the same as building a house, is that once the
construction of N, Z, Q, and R is accomplished, we can then freely use
these sets with their usual properties. Nobody ever cares or mentions
that Z is not really a subset of Q. It's irrelevant once we've shown
that we could build Q out of Z if we wanted to.

If you want an analogy from the field of buildings. it's like the sign
in an elevator that the inspection certificate is on file at the
superintendent's office. Nobody cares that the certificate might be
torn, taped up, old, yellowed, and hard to find. We don't care about it
except to know that it exists; and even then, we really don't care about
it one way or another.

I believe xkcd made this point recently ...

http://xkcd.com/897/

fishfry, Jun 8, 2011