# Set of ALL numbers?

1. Jan 15, 2012

### 1MileCrash

If I want to denote the set of ALL numbers, is the set of complex numbers fitting? a + bi, where b is 0, allows for reals, and where a is 0, allows for imaginary.

2. Jan 15, 2012

### mathman

What are you looking for? The set of all complex numbers includes all reals and all pure imaginary.

3. Jan 15, 2012

### pwsnafu

Define number. Seriously, all you have done is written down the definition of the complex numbers. There are numbers which are outside the set of complex numbers.

4. Jan 15, 2012

### Char. Limit

That's not sufficient. You've neglected to mention quaternions, octonions, vectors, tensors, the like.

5. Jan 16, 2012

### micromass

If you consider transfinite numbers as actual numbers, then there is no set of all numbers. The collection will be too large. It will be a proper class.

6. Jan 17, 2012

### MrAnchovy

The problem is that '(the collection of) all numbers' doesn't actually mean anything - what we have here is a bunch of different interpretations of this, all of which are equally correct (and equally incorrect).

If you are looking for the set of all numbers which are the solutions of equations involving arithmetic operations (including exponentiation) on the integers, you want the set of complex numbers as in the original question.

I think that in all ordinary usage, the word 'number' implies a scalar quantity, so Char. Limit's post misses the mark for me.

Similarly, transfinites such as $\aleph_0$ do not fit most people's definition of a number since they do not behave the same way as all finite numbers do (e.g. $x + 1 > x$ is not true when $x =\aleph_0$).

Working from the other end, the whole numbers can be considered to be 'all numbers' if you infer that a number is a number of things.

Many people would add 0 to that to get the set of integers, so we can now have 'no things'.

But that leaves the simple equation of $x = 1 - 2$ without a solution, so the negative numbers are surely numbers too.

Next come the rationals as the solutions to $x = a \div b$.

The next one is a bit more of a leap because most of the time we accept that the four arithmetical operations of addition, subtraction, multiplication and division are all we need to deal with the 'real' world. But the length of the diagonal of a square with sides $1m$ long is $\sqrt 2 m$, and $\sqrt 2$ cannot be expressed in terms of rational numbers so I want irrational numbers in my set of all numbers. Now if I mark out a circle with radius $1m$ it will have a circumference of $2 \pi m$ so we need to add the transcendentals too and now we have now got to the reals. If we think of numbers as points on a line that extends infinitely in two directions, each real number is represented by a point on the line and vice versa. So for me, this is the set of 'all numbers'.

But what about $\sqrt {-1}$? Taking square roots is not an unreasonable thing to do with a number - the diagonal of a square is an obvious example. So perhaps the number that is the solution to the equation $x = \sqrt {-1}$ which we call $i$ belongs in the set of all numbers too - we are then going to end up admitting all the complex numbers to the 'set of all numbers' and have the answer proposed in the original post.

For me, 'all numbers' are represented by points on a line so this is a step too far - you cannot have a square with sides of length $-1{m}$, so the concept of its diagonal is meaningless (or if you like, imaginary) in this context, so I'll stick with the reals. But I'll admit that the complex plane is quite tempting - after all, the rules of arithmetic hold true for all complex numbers, so they certainly behave like the rest of the numbers - they just don't (all) represent the distance between two points like all the real (and 'real') numbers do.

Last edited: Jan 17, 2012
7. Jan 17, 2012

### pwsnafu

The transcendentals are defined as real numbers that are not algebraic.
You can't take the algebraic numbers, and union it with the set "real numbers not algebraic" and claim "we now got to the reals".

Edit: nothing you post argues we should use reals over p-adics.

Last edited: Jan 17, 2012
8. Jan 18, 2012

### MrAnchovy

Perhaps my ninth paragraph could be improved (I erroneously imply that the irrationals do not include $\pi$):

The next one is a bit more of a leap because most of the time we accept that the four arithmetical operations of addition, subtraction, multiplication and division are all we need to deal with the 'real' world. But the length of the diagonal of a square with sides $1m$ long is $\sqrt 2 m$, and $\sqrt 2$ cannot be expressed in terms of rational numbers. Now if I mark out a circle with radius $1m$ it will have a circumference of $2 \pi m$ and again $\pi$ is not a rational number so we need to add irrational numbers too and now we have now got to the reals.

I don't think that anyone could argue that the union of the rationals and the irrationals is not the reals.

Nothing I post argues that you should use anything. My point was that 'all numbers' has no universal meaning: for me it implies the reals, for someone with a better knowledge of number theory than mine it may mean something else.

However, the mapping of the reals to the points on a line is the clincher for me - the p-adics cannot do this.

9. Jan 18, 2012

### jgens

You realize that there are one-to-one mappings from Rn onto R, right? So all n-tuples of real numbers can be identified with points on a line too. Granted these maps do not preserve algebraic structure, but you never indicated that this was an important point to you.

If the fact that R can be identified with points on a line in a way that preserves algebraic structure is what the real clincher is, then R is no more special than Q or R∪{_∞}∪{-∞} or the hyperreal numbers for that matter.

10. Jan 18, 2012

### MrAnchovy

I did rule out n-tuples as not fitting any definition of 'number' that is satisfactory to me (although I did acnowledge at the end that number pairs are a difficulty because a + ib certainly behaves like a number).

Er, yes it is, Q cannot do that. However if cardinality equal to that of the continuum is all I am interested in, I may as well pick [ 0 , 1 ] as my set of all numbers! Clearly I also want my set to be a complete metric space and R is (I believe) the simplest set that satisfies both of those conditions.

11. Jan 18, 2012

### jgens

No. The embedding QR identifies points of Q with the line in a way the preserves algebraic structure. If you want completeness, then you need to state that separately.

12. Jan 18, 2012

### MrAnchovy

Yes you are right. My original statement was "each real number is represented by a point on the line and vice versa", but I ommitted the reverse mapping in my later comment.

13. Jan 20, 2012

### Jamma

What a ridiculous post.

14. Jan 20, 2012

### Jamma

What do you mean by "a point on the line".

Do you mean "a real number"?

15. Jan 20, 2012

### MrAnchovy

I'm not exactly sure whether you are asking one question or two.

If it is one question, no I do not mean 'a real number' because then my statement would indeed have been a ridiculous tautology.

If it is two questions, (i) I am not sure if I can add to the clarity of 'a point on the line'. I am not trying to construct a formal argument, I am trying to explain what the phrase 'all numbers' means to me. And (ii) yes I do mean a real number.

I do appreciate the element of tautology, essentially I am saying that the elements of a continuum map to the elements of a continuum. This is straying from my point however which is the sufficiency (and necessity) of the reals to satisfy my concept of 'all numbers'. I am perfectly willing to accept that you find that ridiculous.

16. Jan 20, 2012

### checkitagain

pwsnafu,

aside from the choice of words and phrases of MrAnchovy,

you are not disagreeing with that the set of algebraic numbers

unioned with the set of transcendental numbers equals the

set of Real numbers?

17. Jan 20, 2012

### pwsnafu

The algebraic numbers unioned with transcendentals is equal to the set of all reals.
That is not in contention.

MrAnchovy's used the terms "we got the reals". The connotation is that there's this magical process that let's you go from the algebraics -> real by adding on "numbers which are not algebraic".

That's not how it works. We define the rationals, then use that to construct the reals (using Dedekind or Cauchy). After that we consider the algebraic and transcendentals.

This isn't just theoretical. I was taught in highschool "the real numbers are numbers like square root 2 and transcendental numbers like pi". As if that explains anything. :uhh:

18. Jan 20, 2012

### micromass

To be fair, you can define an irrational number as a number whose decimal expansion is non repeating. That is: you can treat real numbers as decimal expansions. That is a third way to construct the reals and it's probably the most honest way. But it's terribly tedious!!!!

19. Jan 20, 2012

### Deveno

but...but...to be fair, if you are going to start with decimals, as one's basis for understanding numbers, you don't even have the full set of rationals to begin with, you just have the ring of all terminating decimals (that is, we might just as well regard 1/3 as the Cauchy sequence:

0.3
0.33
0.333
0.3333
0.33333
etc.

mirroring in theory, what we do in practice).

is there even a well-established name for this ring?

it seems to me this ring is dense in the reals (isn't this obvious?), and that it's completion gives us the reals. the fact that some integers aren't units in this ring (4 is, but 3 isn't) is, um, inconvenient, but if you really want to use decimals as your jumping-off point, there is a piper to be paid.

by the way, your definition needs a bit of work. for example, 1/12 isn't a "repeating decimal" (it's "eventually repeating"), but it is hardly irrational.

20. Jan 20, 2012

### micromass

I don't really see why this must be true.