# math axioms

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Are there certian postulated math axioms, that cannot be proven.

Do these axioms ( if they exist) , constitute the basis of all mathematics.

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All axioms cannot be proven.

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Mathematical theorems start from a set of axioms (i.e., assumptions) and proceed to prove the theorem via a set of rules of deduction/induction. You have been introduced to one set of axioms if you have taken secondary school geometry: Euclid's postulates.

There is not one set of axioms that constitute the basis of all mathematics. Geometry is a good example. Euclidean geometry postulates that given a plane containing a line and a point not on the line, exactly one line can be drawn through the point that does not intersect the line anywhere (Playfair's postulate). If you replace this axiom with something different you get a different geometry. Hyperbolic geometry says an infinite number of lines pass through the given point that do not intersect the line in question. Elliptical geometry says that all lines intersect somewhere.

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dh

"Mathematical theorems start from a set of axioms (i.e., assumptions) and proceed to prove the theorem via a set of rules of deduction/induction"

i see your point here ...

i was thinking in terms of pure mathematics ... for example one cannot prove 0 < 1 , we use this assumption though .

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for example one cannot prove 0 < 1

I'll use the standard definition of the natural numbers, starting with $0 \in \mathbb N$. The natural numbers are endowed with the operations '=','succ', '+', '<=', and '<' (there are others, I'll ignore them here).

1. Equality is defined to be

1. reflexive ($a=a$),
2. symmetric ($a=b\,\Rightarrow\,b=a$),
3. transitive ($a=b\,\textrm{and}\,b=c\,\Rightarrow\,a=c$), and
4. closed ($a \in \mathbb N\,\textrm{and}\, a=b\,\Rightarrow\,b \in \mathbb N$).

[*]The successor primitive recursively defines the natural numbers: $a \in \mathbb N \Rightarrow \textrm{succ}(a) \in \mathbb N$.

[*]A natural number and its successor are not equal to one another.

[*]Addition is defined as the successive application of the successor function:

$a+b \equiv \underbrace{\textrm{succ}(\cdots(\textrm{succ}(a)\cdots)}_{b\;\textrm{successors}}$.

[*]Less than or equal to is defined by given $a,b \in \mathbb N, a\le b \Rightarrow \exists c\in \mathbb N$ such that $a+c=b$.

[*]Less than is defined as $a<b \Rightarrow a\le b\,\textrm{and}\,a\ne b$.

Define $1=\textrm{succ}(0)$. $b=\textrm{succ}(a)\,\Rightarrow\,a<b$. Proof: $b=\textrm{succ}(a)\,\Rightarrow\,b=a+1\,\Rightarrow\,a\le b$. Since a number and its successor are not equal to one another, $a<b$ by the definition of '<'. Since $1=\textrm{succ}(0), 0<1$.

Edited by D H
Cleanup only. Nothing substantial.

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Most of classical mathematics is based on the axioms of set theory and in particular the axiom of choice.

You can prove $0<1$ using the Peano axioms as DH has done above.

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thanks ... sorry for the late response

i find this utterly facinating,

I hated discrete maths as a student , simply because there were to many symbols that i could not absorb and i could not grasp the logic or truth.

the above definitions just seem to simple to define classical mathematics.

I have a question

do you think these "natural" numbers we now have understood ,formed in earlier civilizations because of our own set of circumstances i.e social , economic and conciousness .

If so then after thinking about these numbers for a long time we found that this particular number set can be used by us to explain all the marvels that we extract from mathematics in our particular planet.

OR

Are there other number sets much more complicated that we will only develop in the future , depending on our enviroment.

If there are many number sets , irrational , imaginary etc .

Then is there one that encorporates all these sets as subsets of the " complete set" which is goverened by the " golden formula set". the ultimate definition of all mathematics. Anywhere in the universe.

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By "number sets" you mean rings and fields that are not subrings or subfields of the complex numbers?

If so, other things do exist.

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what i mean is ,

1.what is the most complete set of numbers and the axioms that define them.

2. Does this set of numbers apply anywhere in the universe?

i am trying to establish why mathematics evolved as it did on earth , and could it be different in other civilizations , because of their enviroment.

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Depends what you mean by numbers, maybe you mean division algebras?

I think mathematics is "universal", however notation is not.

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clearly im no mathematician , i mean any number that you can think of.

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I don't really follow your question. Are you asking if all elements in $\mathbb{C}$ or $\mathbb{R}$ are what?

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1.what is the most complete set of numbers and the axioms that define them.
Well the Peano axioms only go as far as the natural numbers (although really the axiomatic method gets tedious after that point anyway) and after that point I think ZFC gets you through the integer, rational, algebraic, real and complex spaces. Extended Reals, Hyperreals, Surreals, Nimbers, Quaternions etcetera etcetera all have their own axioms. I'm fairly sure it would be impossible to define a set of 'everything that's been called a number', bear in mind that the set of all sets is a contradiction which is a strong hint that the set all numbers would be shakey.

2. Does this set of numbers apply anywhere in the universe?
Hell no. Numbers as we've defined them are really just ideas in the platonic realm.

i am trying to establish why mathematics evolved as it did on earth , and could it be different in other civilizations , because of their enviroment.
Well the very basics of the natural numbers were derived from counting, so basically everywhere independently managed a concept of addition and multiplication, so I'd imagine that up to the rationals, any culture would have their own equivalent. But more advanced mathematics, in fact all mathematics that was developed for it's own sake varies massively in different cultures in terms of what people focus on, hence why the Chinese remainder theorem is called the Chinese remainder theorem.

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thank you ...your response is simple yet logical and i happen to agree with your views.

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