addition is order preserving, so if x>0, for any y you get that x+y>y. if you start adding positive numbers, it will always be a bigger number, and therefore you won’t come back.
Outline of a proof:
Let P(n+1)>P(n), and P(0)=1>0.
Now, through induction on n, we can prove that if P(n)>0, then P(n+1)>0.
If a>0, and b>a, then b>0, so since P(n+1)>P(n), P(n+1)>0.
Thus, if we ''move on'' to any bigger number, the newer number will still be bigger than 0. Since -infinity is smaller than 0, we can never reach -infinity.
---
If we want we can decide to glue the ends of the number line together at infinity, but then we're talking about a new mathemathical object, (e.g. the riemann sphere https://en.m.wikipedia.org/wiki/Riemann_sphere ), and no longer actually talking about the number line.
The fact that natural numbers aren’t “circular” is one of Peanos axioms, so you can’t prove it. There’s probably a simple proof that the integers don’t wrap around based on that axiom, but I’m not familiar with one.
Which axiom of PA are you referring to? I’m not aware of any formulation that explicitly describes any geometric or topological structure of the standard model of PA.
I’m don’t think that works by itself. One could have an extension or maybe like a direct limit of cyclic orderings where 0 is not a successor, but is a limit. You could maybe think of the set {e^(2πi\(1-1/\(n+1\)\)):n∈ℕ} as a visualization of this.
My interpretation of the original question would mean in order for the natural numbers to be cyclic, 0 has to be a successor If you have a set that lies on a circle and is order isomorphic to the natural numbers then I would say that doesn’t count. There may be a counter example that I haven’t considered, but for a basic question that explanation would still be “good enough” in the style of “you can’t take a square root of a negative number.”
Ok that’s reasonable. My contention is that the definition of the ordering is not part of PA itself and the relation is not (does not *need* to) even included in the basic language. It’s definable from the successor function. The divisibility order is definable from the multiplication function.
A cyclic order C is definable from the standard <, itself being definable from the successor function^(\[1\]), as a ternary relation on ℕ satisfying (a,b,c)∈C iff 0≤a
I think your intuition for the successor function is fine, but I’m claiming that no specific axiom of PA says that ℕ cannot be cyclic. In fact, PA says nothing at all about order or topology. The standard ordering is defined from the successor function, but one can also define the divisibility ordering which gives ℕ the order structure of a bounded infinite-dimensional atomic lattice.
We could just as well define a cyclic ordering C as a ternary extension of the standard ordering < by (a,b,c)∈C iff 0
If you think of moving on the number line as successively passing through each number based on their order, then it can't be circular since every number would be both greater than and less than all other numbers.
If you construct the natural numbers (successor of n is n+1 = n∪{n}, so n ∈ n+1) with the foundation axiom, this cannot happen. Lets say for an example that 4 + 1 = 2 (circularity at that point), consider A = { 2,3,4 }, so 4 ∈ 2, 2 ∈ 3, 3 ∈ 4 with the definition of successors, so 4 ∈ A∩2, 2 ∈ A∩3, 3 ∈ A∩4 since A contains 2, 3, 4, which contradicts the axiom of foundation that says that there is at least one x ∈ A such that x∩A=∅ since A is non-empty.
Not in the reals, but there are some systems where positive and negative infinities are the same point and others where an infinite string of nines added to one makes zero.
The number line isn't a physical object like a table. You can't hold a measuring tape up to it and find out that it's really one thing or another. It's basically an idea, and has the properties we assign to it. We can design arithmetic that works like you describe, but it would no longer be the real number line.
If the reals looped around, then there should be some n where n is positive and n+1 is negative, so there should be some n such that n>0 and 0>n+1. Adding these inequalities together yields n>n+1. Subtracting n yields 0>1. However, 0<1, so the reals don't loop back around. Pretty simple.
Not really. It's an axiom. It only requires there to be a successor function for it to be true, so not a very controversial one. Basically you'd need do not believe some quite basic things about finite sets to not have it.
Actually if you plot all points in the Euler circle on a computer using an approximation of i (-0.225f) you will find that numbers go in circles via imaginary multiplication and in lines via addition and subtraction.
https://wiki.betterexplained.com/assets/img/Eulers_Identity.5e5ba91c.png
Multiplying by i to go from -n to +n is all about going in the imaginary circle which is just a side effect generated by the pure possibility of numbers to exist. That's a proof because it is using the Principle of Sufficient Reason (Leibniz's).
The Euler Circle along with Euler's formula and Euler's identity is what you are after.
There is an experimental question which says that by adding 1, eventually you run out of numbers and go back to -infinity, but this simply isn't possible. It isn't possible (A) for any abstract number, such as infinity and -infinity, to exist in reality or (B) for number to be limited, since the radius of any Euler Circle (not 1 or anything arbitrary) can be anything at all no matter how high number goes which is just a case of building a bigger number that generates bigger numbers inside that radius (another proof: that numbers beget bigger numbers, meaning since there was no cap on radius=1's numbers inside its Euler circle, there can be no limit any higher no matter how high we go).
Numbers aren't sequential in 1 dimension. They have 1 dimensional application but frequencies do not have dimensionality. This is all made clear if you understand wave mathematics via something like audio processing and DSP.
Fun fact: the soul is 2 dimensional since all wave mathematics is 100% guaranteed to be 2 dimensional. The 3d world is given by quantum mechanics which operates on mathematical expressions as its input, in an uncertain order, across a range of minds using their certainty to determine where to draw from next.
As others have noted, yes it is trivially true that in standard arithmetic adding one to an integer never produces a smaller integer.
You may be interested in [modular arithmetic](https://en.m.wikipedia.org/wiki/Modular_arithmetic), a system of arithmetic which does behave as you describe, and has applications in computer programming, where most numbers are a fixed length i.e. cannot be incremented indefinitely.
If it were circular then the concept of Positive and Negative would mean something entirely different and that's just one consequence, another would be the basis of Imaginary Numbers would be destroyed since they are based on the Square Root of Negative 1, even things like simple arithmetic would be radically different because for example Addition relies on progression through the Number Line e.g. 2+3 progresses +3 from +2 to +5
addition is order preserving, so if x>0, for any y you get that x+y>y. if you start adding positive numbers, it will always be a bigger number, and therefore you won’t come back.
can you define your ‘move’ step? i don’t even think you can move to infinity…
Outline of a proof: Let P(n+1)>P(n), and P(0)=1>0. Now, through induction on n, we can prove that if P(n)>0, then P(n+1)>0. If a>0, and b>a, then b>0, so since P(n+1)>P(n), P(n+1)>0. Thus, if we ''move on'' to any bigger number, the newer number will still be bigger than 0. Since -infinity is smaller than 0, we can never reach -infinity. --- If we want we can decide to glue the ends of the number line together at infinity, but then we're talking about a new mathemathical object, (e.g. the riemann sphere https://en.m.wikipedia.org/wiki/Riemann_sphere ), and no longer actually talking about the number line.
The fact that natural numbers aren’t “circular” is one of Peanos axioms, so you can’t prove it. There’s probably a simple proof that the integers don’t wrap around based on that axiom, but I’m not familiar with one.
Which axiom of PA are you referring to? I’m not aware of any formulation that explicitly describes any geometric or topological structure of the standard model of PA.
One of the axioms is that 0 is not the successor of any natural number, so I think that covers it. Maybe it is a bit more complicated though
I’m don’t think that works by itself. One could have an extension or maybe like a direct limit of cyclic orderings where 0 is not a successor, but is a limit. You could maybe think of the set {e^(2πi\(1-1/\(n+1\)\)):n∈ℕ} as a visualization of this.
My interpretation of the original question would mean in order for the natural numbers to be cyclic, 0 has to be a successor If you have a set that lies on a circle and is order isomorphic to the natural numbers then I would say that doesn’t count. There may be a counter example that I haven’t considered, but for a basic question that explanation would still be “good enough” in the style of “you can’t take a square root of a negative number.”
Ok that’s reasonable. My contention is that the definition of the ordering is not part of PA itself and the relation is not (does not *need* to) even included in the basic language. It’s definable from the successor function. The divisibility order is definable from the multiplication function. A cyclic order C is definable from the standard <, itself being definable from the successor function^(\[1\]), as a ternary relation on ℕ satisfying (a,b,c)∈C iff 0≤a
the definition of the Successor function means that the number line will never loop back onto itself that it my intuition, i might be wrong
I think your intuition for the successor function is fine, but I’m claiming that no specific axiom of PA says that ℕ cannot be cyclic. In fact, PA says nothing at all about order or topology. The standard ordering is defined from the successor function, but one can also define the divisibility ordering which gives ℕ the order structure of a bounded infinite-dimensional atomic lattice. We could just as well define a cyclic ordering C as a ternary extension of the standard ordering < by (a,b,c)∈C iff 0
Projective geometry go brrr
If you think of moving on the number line as successively passing through each number based on their order, then it can't be circular since every number would be both greater than and less than all other numbers.
You can't "reach" negative infinity.
The projective real number line is in fact circular. The conventional real number line is not circular. Which number line are you talking about?
Infinity is not a number, it is representing a Process of constant increase
Tell me your axioms and I'll come up with a proof.
If you construct the natural numbers (successor of n is n+1 = n∪{n}, so n ∈ n+1) with the foundation axiom, this cannot happen. Lets say for an example that 4 + 1 = 2 (circularity at that point), consider A = { 2,3,4 }, so 4 ∈ 2, 2 ∈ 3, 3 ∈ 4 with the definition of successors, so 4 ∈ A∩2, 2 ∈ A∩3, 3 ∈ A∩4 since A contains 2, 3, 4, which contradicts the axiom of foundation that says that there is at least one x ∈ A such that x∩A=∅ since A is non-empty.
Not in the reals, but there are some systems where positive and negative infinities are the same point and others where an infinite string of nines added to one makes zero.
The number line isn't a physical object like a table. You can't hold a measuring tape up to it and find out that it's really one thing or another. It's basically an idea, and has the properties we assign to it. We can design arithmetic that works like you describe, but it would no longer be the real number line. If the reals looped around, then there should be some n where n is positive and n+1 is negative, so there should be some n such that n>0 and 0>n+1. Adding these inequalities together yields n>n+1. Subtracting n yields 0>1. However, 0<1, so the reals don't loop back around. Pretty simple.
Not really. It's an axiom. It only requires there to be a successor function for it to be true, so not a very controversial one. Basically you'd need do not believe some quite basic things about finite sets to not have it.
Actually if you plot all points in the Euler circle on a computer using an approximation of i (-0.225f) you will find that numbers go in circles via imaginary multiplication and in lines via addition and subtraction. https://wiki.betterexplained.com/assets/img/Eulers_Identity.5e5ba91c.png Multiplying by i to go from -n to +n is all about going in the imaginary circle which is just a side effect generated by the pure possibility of numbers to exist. That's a proof because it is using the Principle of Sufficient Reason (Leibniz's). The Euler Circle along with Euler's formula and Euler's identity is what you are after. There is an experimental question which says that by adding 1, eventually you run out of numbers and go back to -infinity, but this simply isn't possible. It isn't possible (A) for any abstract number, such as infinity and -infinity, to exist in reality or (B) for number to be limited, since the radius of any Euler Circle (not 1 or anything arbitrary) can be anything at all no matter how high number goes which is just a case of building a bigger number that generates bigger numbers inside that radius (another proof: that numbers beget bigger numbers, meaning since there was no cap on radius=1's numbers inside its Euler circle, there can be no limit any higher no matter how high we go). Numbers aren't sequential in 1 dimension. They have 1 dimensional application but frequencies do not have dimensionality. This is all made clear if you understand wave mathematics via something like audio processing and DSP. Fun fact: the soul is 2 dimensional since all wave mathematics is 100% guaranteed to be 2 dimensional. The 3d world is given by quantum mechanics which operates on mathematical expressions as its input, in an uncertain order, across a range of minds using their certainty to determine where to draw from next.
As others have noted, yes it is trivially true that in standard arithmetic adding one to an integer never produces a smaller integer. You may be interested in [modular arithmetic](https://en.m.wikipedia.org/wiki/Modular_arithmetic), a system of arithmetic which does behave as you describe, and has applications in computer programming, where most numbers are a fixed length i.e. cannot be incremented indefinitely.
If it were circular then the concept of Positive and Negative would mean something entirely different and that's just one consequence, another would be the basis of Imaginary Numbers would be destroyed since they are based on the Square Root of Negative 1, even things like simple arithmetic would be radically different because for example Addition relies on progression through the Number Line e.g. 2+3 progresses +3 from +2 to +5