There exists one number N

  • Thread starter arbol
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In summary: You can NOT put a decimal at a point. All you can do is to put a decimal in some place. That place is the "unit's" place. But since there is no point, there is no unit's place. So your "number" N has no meaning.In summary, the conversation discusses the concept of irrational numbers and how they are defined as numbers that cannot be expressed in the form of m/n, where m and n are integers and n is not equal to zero. The conversation also mentions the example of 1.234567891011... as an irrational number and how it can be represented as an infinite sequence without a fixed decimal point. However, the idea
  • #1
arbol
50
0
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...
 
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  • #2
arbol said:
N = 12345678910111213... (where unit's place is at infinity).

Doesn't that make N = infinity?
 
  • #3
so, put a decimal in front of it.

oh he did that.
 
  • #4
arbol said:
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.
 
  • #5
arbol said:
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?
 
  • #6
arbol said:
A good example of an irrational number then would be

1.234567891011121314...

That's 10 times Champernowne constant.
 
  • #7
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.
 
  • #8
Hurkyl said:
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
 
  • #9
HallsofIvy said:
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.
 
  • #10
HallsofIvy said:
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
 
  • #11
belliott4488 said:
Doesn't that make N = infinity?

lim f(x) (as x approaches infinty) is infinity, but N is a single number (not a variable).
 
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  • #12
HallsofIvy said:
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...
 
  • #13
arbol said:
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.
 
  • #14
arbol said:
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.
 
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  • #15
arbol said:
good question
Yes it was. Since it was about your post, do you have a good answer?

arbol said:
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.

arbol said:
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.

arbol said:
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.

arbol said:
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.
 
  • #16
belliott4488 said:
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.)


:confused:
 
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  • #17
But since "infinity" is not an integer, you know that N isn't an integer.
 
  • #18
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.
 
  • #19
LorenzoMath said:
or more precisely, we introduce the concept of infinity.

Nitpicking: that's "a concept of infinity", not "the concept of infinity".
 
  • #20
arbol said:
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.)


:confused:
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.
 
  • #21
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.
 
  • #22
arbol said:
1. N = infinity = 1234567891011121314... (a single number, where the unit's place is at N).

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

This is nonsense [tex]\frac{\infty}{\infty}[/tex] is undefined and [tex]\infty[/tex] is not a number.
 
  • #23
arbol said:
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.
 
  • #24
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:

arbol said:
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[itex]^{-1}[/itex] := 1

GF(3) also suffices. Here's one correspondence:
0 := 0
1 := 1
N := 2
N[itex]^{-1}[/itex] := 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 [tex]N_0 = 0[/tex] and [tex]N_k=f(N_{k-1})+g(k).[/tex]

Now N is naturally defined as the limit of the [tex]N_k[/tex] 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.
 
  • #25
I have newbie question.
I have recently discovered that X^0 = 1
What happens if X = 0
 
  • #26
0^0 makes no sense, is kind hard to explain, and x^0 = 1 for any x> 0 is a convention I think
 
  • #27
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?
 
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  • #28
HallsofIvy said:
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.
 
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  • #29
Bojan Keevill said:
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) = [tex]\frac {x^n}{x^n}[/tex] = 1 if x<>0, so x^0= 1 if x > 0

but if x = 0, then we can write [tex]x^0 = 0^0 = \frac {x^n}{x^n} = \frac {0^n}{0^n} [/tex]

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

then [tex]0^0 = \frac {0}{0}[/tex] 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
 
  • #30
In most cases, [itex]0^0=1[/itex] is a sensible definition. A great many combinatorical identities rely on this.

In complex analysis the fact that [tex]\lim_{x\to0,y\to0}x^y[/tex] is ill-defined because different paths give different values means that it's best in that context to leave [itex]0^0[/itex] undefined.
 
  • #31
CRGreathouse said:
In most cases, [itex]0^0=1[/itex] is a sensible definition. A great many combinatorical identities rely on this.

Sorry, I didn't follow you...

I tought 0^0 didn't make any sense...

I know that if we take x^n such that x AND n --> 0, then x^n --> 1, and the two lateral limits converges to 1 in an ideal situation, but what of that since the "speed" of how x and n --> 0 is what matters?
 
  • #32
al-mahed said:
I know that if we take x^n such that x AND n --> 0, then x^n --> 1, and the two lateral limits converges to 1 in an ideal situation, but what of that since the "speed" of how x and n --> 0 is what matters?

The rate of convergence has nothing to do with the final value.
 
  • #33
It is generally convenient to let the notation x^0 denote the constant polynomail 1. In fact, in terms of polynomial arithmetic, raising the polynomial x to the integer exponent 0 really does yield you the constant polynomial 1.

Since we are often interested in the functions associated to a polynomial or to a power series, the correct answer to "plug 0 into the polynomial x^0" is, in fact, 1. The order of operations is subtle here:

This is what happens:
Compute the polynomial exponent x^0 = 1
Evaluate the polynomial 1 at 0, getting 1.

This is not what happens:
Plug 0 in for x, getting the expression 0^0
Evaluate the exponentation 0^0, getting undefined.


In every case where I've seen it useful, that is what is meant by the convention "let 0^0 = 1".
 
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  • #34
Hurkyl said:
It is generally convenient to let the notation x^0 denote the constant polynomail 1. In fact, in terms of polynomial arithmetic, raising the polynomial x to the integer exponent 0 really does yield you the constant polynomial 1.

Since we are often interested in the functions associated to a polynomial or to a power series, the correct answer to "plug 0 into the polynomial x^0" is, in fact, 1. The order of operations is subtle here:

This is what happens:
Compute the polynomial exponent x^0 = 1
Evaluate the polynomial 1 at 0, getting 1.

This is not what happens:
Plug 0 in for x, getting the expression 0^0
Evaluate the exponentation 0^0, getting undefined.


In every case where I've seen it useful, that is what is meant by the convention "let 0^0 = 1".


I think is best to leave 0^0 undefined since division by zero is undefined (unless you apply the limit concept).
 
  • #35
arbol said:
I think is best to leave 0^0 undefined since division by zero is undefined (unless you apply the limit concept).

No, it is best to leave it undefined in many cases precisely BECAUSE if you apply the limit concept you get that x^y cannot be extended continuously so that 0^0 has any value.
 
<h2>What does it mean when someone says "There exists one number N"?</h2><p>When someone says "There exists one number N", they are stating that there is at least one specific number, denoted as N, that satisfies a given condition or set of conditions. This statement is often used in mathematical proofs to show that a solution or value exists for a particular problem.</p><h2>How do you determine the value of N in the statement "There exists one number N"?</h2><p>The value of N can be determined by analyzing the given conditions or constraints. In some cases, N may be explicitly stated or can be found by solving a mathematical equation or inequality. In other cases, N may be unknown and must be found through further investigation or experimentation.</p><h2>What is the significance of the statement "There exists one number N" in scientific research?</h2><p>The statement "There exists one number N" is often used in scientific research to show that a particular phenomenon or relationship exists. It can also be used to prove the existence of a solution or value in a mathematical or scientific problem. This statement is crucial in forming hypotheses and making conclusions based on data and evidence.</p><h2>Can there be more than one value of N that satisfies the statement "There exists one number N"?</h2><p>Yes, there can be more than one value of N that satisfies the statement "There exists one number N". This is because the statement only guarantees the existence of at least one value, but it does not limit the number of possible values that can satisfy the given conditions or constraints.</p><h2>What is the difference between "There exists one number N" and "For all numbers N"?</h2><p>The statement "There exists one number N" means that there is at least one specific number, denoted as N, that satisfies a given condition or set of conditions. On the other hand, the statement "For all numbers N" means that the given condition or set of conditions is true for every possible number, including N. In other words, "There exists one number N" is a more specific statement than "For all numbers N".</p>

What does it mean when someone says "There exists one number N"?

When someone says "There exists one number N", they are stating that there is at least one specific number, denoted as N, that satisfies a given condition or set of conditions. This statement is often used in mathematical proofs to show that a solution or value exists for a particular problem.

How do you determine the value of N in the statement "There exists one number N"?

The value of N can be determined by analyzing the given conditions or constraints. In some cases, N may be explicitly stated or can be found by solving a mathematical equation or inequality. In other cases, N may be unknown and must be found through further investigation or experimentation.

What is the significance of the statement "There exists one number N" in scientific research?

The statement "There exists one number N" is often used in scientific research to show that a particular phenomenon or relationship exists. It can also be used to prove the existence of a solution or value in a mathematical or scientific problem. This statement is crucial in forming hypotheses and making conclusions based on data and evidence.

Can there be more than one value of N that satisfies the statement "There exists one number N"?

Yes, there can be more than one value of N that satisfies the statement "There exists one number N". This is because the statement only guarantees the existence of at least one value, but it does not limit the number of possible values that can satisfy the given conditions or constraints.

What is the difference between "There exists one number N" and "For all numbers N"?

The statement "There exists one number N" means that there is at least one specific number, denoted as N, that satisfies a given condition or set of conditions. On the other hand, the statement "For all numbers N" means that the given condition or set of conditions is true for every possible number, including N. In other words, "There exists one number N" is a more specific statement than "For all numbers N".

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