Is There a Number N That is Neither Rational nor Irrational?

[arbol]

The set N of natural numbers = {1, 2, 3, 4, ...}.

But there exists one (1) number N, such that

N = 12345678910111213... (where the unit's place is at infinity).

A good example of an irrational number then would be

1.234567891011121314...

 

[belliott4488]

N = 12345678910111213... (where unit's place is at infinity).

Doesn't that make N = infinity?

 

[mathwonk]

so, put a decimal in front of it.

oh he did that.

 

[HallsofIvy]

The set N of natural numbers = {1, 2, 3, 4, ...}.

But there exists one (1) number N, such that

N = 12345678910111213... (where the unit's place is at infinity).

No, there is no such number. All integers have only a finite number of digits. By the way, it is not at all a good idea by using "N" to represent the set of natural numbers and then say that "N" is a number.

A good example of an irrational number then would be

1.234567891011121314...

Now THAT is a perfectly good irrataional number.

 

[Hurkyl]

But there exists one (1) number N, such that

N = 12345678910111213... (where the unit's place is at infinity).

Are you sure that decimal string actually denotes a number? How can the unit's place be 'at infinity'? What digit is in the unit's place?

 

[CRGreathouse]

A good example of an irrational number then would be

1.234567891011121314...

That's 10 times Champernowne constant.

 

[HallsofIvy]

If it were possible to construct any irrational number by putting a decimal into some positive integer, that would imply that the set of irrational numbers is countable.

 

[arbol]

Are you sure that decimal string actually denotes a number? How can the unit's place be 'at infinity'? What digit is in the unit's place?

good question

 

[arbol]

No, there is no such number. All integers have only a finite number of digits. By the way, it is not at all a good idea by using "N" to represent the set of natural numbers and then say that "N" is a number.


Now THAT is a perfectly good irrataional number.

It is necessary that N is not an interger, but it is one number.

 

[arbol]

No, there is no such number. All integers have only a finite number of digits. By the way, it is not at all a good idea by using "N" to represent the set of natural numbers and then say that "N" is a number.


Now THAT is a perfectly good irrataional number.

you can call it anything you want

 

[arbol]

Doesn't that make N = infinity?

lim f(x) (as x approaches infinty) is infinity, but N is a single number (not a variable).

 

[arbol]

If it were possible to construct any irrational number by putting a decimal into some positive integer, that would imply that the set of irrational numbers is countable.

a definition of an irrational number is a number that cannot be expressed in the form m/n, where m and n are intergers and n not equal to zero. such numbers are infinite to right of the decimal point and do not repeat. for example,

1.234567891011...

 

[CRGreathouse]

a definition of an irrational number is a number that cannot be expressed in the form m/n, where m and n are intergers and n not equal to zero. such numbers are infinite to right of the decimal point and do not repeat.

Yes, so together with Hallsofivy's statement you know that 123456789101112... is not an integer.

 

[ramsey2879]

It is necessary that N is not an interger, but it is one number.

All Numbers must have a meaning such that a rational number can be found to approximate the number within a chosen value, a expression that is an infinite string of numbers without any fixed decimal point does not have any meaning and is not a number.

 

[HallsofIvy]

good question

Yes it was. Since it was about your post, do you have a good answer?

It is necessary that N is not an interger, but it is one number.

Okay, what do you mean by "number". And my criticism was simply about using the same symbol, N, with two different meanings.

you can call it anything you want

Thank you. But I do prefer to use standard terminology. If you did that, it might be easier to understand what you are trying to say.

lim f(x) (as x approaches infinty) is infinity, but N is a single number (not a variable).

??This is the first time you mentioned "f(x)". Where did that come from. Once again, the N you posit is NOT a "number" by any standard definition.

a definition of an irrational number is a number that cannot be expressed in the form m/n, where m and n are intergers and n not equal to zero. such numbers are infinite to right of the decimal point and do not repeat. for example,

1.234567891011...

Yes, we know that- it is not necessary to state the obvious.

 

[arbol]

Doesn't that make N = infinity?

Yes. I think it does.

If f(x) = x, then

lim (of f(x) as x approaches infinity) = infinity = N. (but the unit's place of N is at infinity.)

 

[CRGreathouse]

But since "infinity" is not an integer, you know that N isn't an integer.

 

[LorenzoMath]

1234567891011... is not a conventional way of representing real numbers, so unless you introduce your own convention, it doesn't mean anything. Whereas if you put a disimal point somewhere, it represents a real number in a conventional sense. Because, by convention, 1.234567... represents some real number to which the sequence, 1, 1.2, 1.23, 1.234, ... converges. This is what we call the completeness of R. If we agree to say that 1234567891011... represents where the sequence 1, 12, 123, 1234, ... go, then we may call it infinity, or more precisely, we introduce the concept of infinity.

 

[CRGreathouse]

or more precisely, we introduce the concept of infinity.

Nitpicking: that's "a concept of infinity", not "the concept of infinity".

 

[ramsey2879]

Yes. I think it does.

If f(x) = x, then

lim (of f(x) as x approaches infinity) = infinity = N. (but the unit's place of N is at infinity.)

No need to stay confused. Let go of your mindset which says it should be possible to put a decimal at a point of infinity of a string of numbers and have something meaningful. There can only be a finite string of numbers prior a decimal point to have anything resembling a number.

 

[arbol]

1. N = infinity = 1234567891011121314... (a single number, where the unit's place is at N).

2. If f(x) = x, then

lim (as x approaches N) of f(x) = N.

3. There must be a field, a set of elements having two operations, designated addition and multiplication, satisfying the conditions that multiplication is distributive over addition, that the set is a group under addition (where N is the unit of all the other elements in the set), and that the elements with the exception of an additive identity form a group under multiplication.

a. If X, Y, and Z are elements in the said set, then

1.) X + Y = Y + X.

2.) X*Y = Y*X.

3.) (X + Y) + Z = X + (Y + Z).

4.) (X*Y)*Z = X*(Y*Z).

5.) X*(Y + Z) = X*Y + X*Z.

6.) 0 + X = X.

7.) N*N^(-1) = 1.

4. (0.123456789101112...)*10^N = N.

 

[ramsey2879]

1. N = infinity = 1234567891011121314... (a single number, where the unit's place is at N).

7.) N*N^(-1) = 1.

This is nonsense $$\frac{\infty}{\infty}$$ is undefined and $$\infty$$ is not a number.

 

[HallsofIvy]

1. N = infinity = 1234567891011121314... (a single number, where the unit's place is at N).

Once again, that is nonsense. You have not defined any number space in which such a thing can exist.

2. If f(x) = x, then

lim (as x approaches N) of f(x) = N.

Since N is undefined, this is also nonsense.

3. There must be a field, a set of elements having two operations, designated addition and multiplication, satisfying the conditions that multiplication is distributive over addition, that the set is a group under addition (where N is the unit of all the other elements in the set), and that the elements with the exception of an additive identity form a group under multiplication.

Why must there exist such a field? Because you say so? You may be attempting to do what was suggested before and trying to define a number space in which N can exist. Unfortunately, you cannot define N first and then define the number space!

a. If X, Y, and Z are elements in the said set, then

1.) X + Y = Y + X.

2.) X*Y = Y*X.

3.) (X + Y) + Z = X + (Y + Z).

4.) (X*Y)*Z = X*(Y*Z).

5.) X*(Y + Z) = X*Y + X*Z.

6.) 0 + X = X.

7.) N*N^(-1) = 1.

4. (0.123456789101112...)*10^N = N.

What do you mean by "1" here? I thought you had said that N was the "unit" (multiplicative identity) for this field so "1" makes no sense.

 

[CRGreathouse]

I don't understand your point #1, the 'definition' of N. When you're defining your own field you can't just use ellipses that vaguely! But I can address your third point, assuming that N is some distinguished element of a set S over which your field lies:

3. There must be a field, a set of elements having two operations, designated addition and multiplication, satisfying the conditions that multiplication is distributive over addition, that the set is a group under addition (where N is the unit of all the other elements in the set), and that the elements with the exception of an additive identity form a group under multiplication.

a. If X, Y, and Z are elements in the said set, then

1.) X + Y = Y + X.

2.) X*Y = Y*X.

3.) (X + Y) + Z = X + (Y + Z).

4.) (X*Y)*Z = X*(Y*Z).

5.) X*(Y + Z) = X*Y + X*Z.

6.) 0 + X = X.

7.) N*N^(-1) = 1.

There does exist at least one such field: GF(2) suffices, for example. Here's the correspondence:
0 := 0
1 := 1
N := 1
N$^{-1}$ := 1

GF(3) also suffices. Here's one correspondence:
0 := 0
1 := 1
N := 2
N$^{-1}$ := 2

Heck, any field suffices, since there has to exist some invertible element, which is all 7 requires. (Of course all nonzero elements are invertible in a field.)

Now in the context of a field, let's examine the 'definition' for N:
"N = infinity = 1234567891011121314..."

Now "infinity" has no meaning in abstract algebra, so we ignore that part: clearly, that's just an alternate name for N. But "1234567891011121314..." seems to have meaning: let's examine that.

Definitions (over a generic field with additive identity 0 and multiplicative identity 1):
0 := 0
1 := 1
2 := 1 + 1
3 := 1 + 1 + 1
4 := 1 + 1 + 1 + 1
. . .
9 := 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
f(n) := n * (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)

Further, define g(n) to be the digit in the nth decimal place of the Champernowne constant, defined additively as above.

Let $$N_0 = 0$$ and $$N_k=f(N_{k-1})+g(k).$$

Now N is naturally defined as the limit of the $$N_k$$ if such limit is defined. But I can't think of any field in which it is defined, unless you consider the degenerate 0=1 (which convention does not consider a field). It's not defined for any Galois field, and it's not defined for the real or rational numbers.

 

[Bojan Keevill]

I have newbie question.
I have recently discovered that X^0 = 1
What happens if X = 0

 

[al-mahed]

0^0 makes no sense, is kind hard to explain, and x^0 = 1 for any x> 0 is a convention I think

 

[Bojan Keevill]

0^0 is kind hard to explain or impossible to explain? Please expand.
0^0 must equal something, either 0 or 1.
Might it exist in both states simultaneously, resolving to either 0 or 1 depending on the context in which it is placed?

 

[arbol]

What do you mean by "1" here? I thought you had said that N was the "unit" (multiplicative identity) for this field so "1" makes no sense.

Yes. You are right.

If X, Y, and Z are elements in the said set, then

X*X^(-1) = N.

I don't know myself what that really means.

 

[al-mahed]

0^0 is kind hard to explain or impossible to explain? Please expand.
0^0 must equal something, either 0 or 1.
Might it exist in both states simultaneously, resolving to either 0 or 1 depending on the context in which it is placed?

I thought a way to explain, I don't know if this is the best or most convincing...

x^0 = x^(n-n) for any number n, and x^(n-n) = $$\frac {x^n}{x^n}$$ = 1 if x<>0, so x^0= 1 if x > 0

but if x = 0, then we can write $$x^0 = 0^0 = \frac {x^n}{x^n} = \frac {0^n}{0^n}$$

as we know, 0^n for n > 0 is equal to 0, because 0^n = 0*0*0*0... n times

then $$0^0 = \frac {0}{0}$$ what is an undefined amount since infinity (1/0) times zero (what is the same of 0/0) is undefined because infinity is not a defined number or amunt

 

[CRGreathouse]

In most cases, $0^0=1$ is a sensible definition. A great many combinatorical identities rely on this.

In complex analysis the fact that $$\lim_{x\to0,y\to0}x^y$$ is ill-defined because different paths give different values means that it's best in that context to leave $0^0$ undefined.