What if 0 is special

[Azhar]

TL;DR

Special 0 rule

[Mentor Note: Cross-posted threads merged]

Rule:
0×0=0
Think like this,×n is time to type a number
example:
1×2=2 is "type number 1 2 times then sum"
So 0×0=0 is "type number 0 0 time then sum" the result is 0 because it's 0 time so you're not type to the input bar

0÷0=1:
Its like "how much time to type 0 so the result is 0" the result is 1 because 0 is already 0,if it's 0 then you're not type the number to input bar

0²÷0:
x²÷x,the ÷x neutrilize the ² so the result is x

Example:
x=0
Standard:
lim r->0,μ->0 μ²÷r,result is 0÷0=indeterminate->Hôpital thing->0
F(x)=2x÷x=(2×0)÷0=0÷0=indeterminate
G(x)=x²÷x:
0²÷0=indeterminate
Non standard(zero exclusion):
0²÷0,÷0 neutrilize the ²(x²÷x=x) so the result is 0
F(x)=2x÷x=2×0=0÷0=1
G(x)=x²÷x:
÷x neutrilize the ² like in 2²÷2=2,so 0²÷0=0

 

[bhobba]

There is some confusion on the part of the OP.

Becoming familiar with Peano's axioms and ZFC set theory, which most mathematicians accept, should clarify things.

Peano's axioms:
https://en.wikipedia.org/wiki/Peano_axioms

There is a strong connection with the Axiom of Infinity from the so-called ZFC axioms of set theory:
https://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory

Note that 0/0 is undefined, although intuitively it is related to the derivative.

Thanks
Bill

 

[Matterwave]

0 is already special. It is the additive identity in the Complex, Real, Integer, Natural, etc. number systems.

We use it a lot! So trying to redefine what it is is a very hard uphill struggle.

So I will just answer using the standard definition of numbers (and 0).

Rule:
0×0=0

This is correct.

0÷0=1:

This is incorrect. Division by 0 is undefined.

0²÷0:

Again, division by 0 is undefined.

 

[Azhar]

0 is already special. It is the additive identity in the Complex, Real, Integer, Natural, etc. number systems.

We use it a lot! So trying to redefine what it is is a very hard uphill struggle.

So I will just answer using the standard definition of numbers (and 0).

This is correct.


This is incorrect. Division by 0 is undefined.


Again, division by 0 is undefined.

But what if there's exception so the result is not indeterminate nor undefined

 

[FactChecker]

You might be interested in Abstract Algebra. They study the basic abstract properties of sets with an operation or two (usually called addition and multiplication). In that, the zero element is the element, 0, where (a+0 = a, for all a).
This would talk about the special properties of the 0 element in a very stripped-down context.

 

[FactChecker]

But what if there's exception so the result is not indeterminate nor undefined

Division is not defined arbitrarily. We start with two operations, addition and multiplication, that have very specific, desirable properties. Then division is defined as the inverse of multiplication. The result is that division by zero is undefined.

 

[Azhar]

Division is not defined arbitrarily. We start with two operations, addition and multiplication, that have very specific, desirable properties. Then division is defined as the inverse of multiplication. The result is that division by zero is undefined.

but there's exception in standard math too like positive x⁰=1,2² is for(i=0;i<=2;i++){2×2} so if it's <=0 then the multiplication is not even start then the result is 0

 

[Dale]

But what if there's exception so the result is not indeterminate nor undefined

You can make such an exception, but not with the real numbers.

The real numbers are a field and fields have an additive identity $0$ such that $a+0=a$ for all $a$ as well as a multiplicative identity $1\ne 0$ such that $a \cdot 1 = a$ for all $a$. These lead to an additive inverse $-a$ such that $a+(-a)=0$ for all $a$ and a multiplicative inverse $a^{-1}$ such that $a \cdot a^{-1}=1$ for all $a\ne 0$. Division is just shorthand for multiplication by the inverse $a/b=a \cdot b^{-1}$.

From these you can prove that $0 \cdot a = 0$ for all $a$:
$$0 \cdot a = (0+0) \cdot a$$$$0 \cdot a = 0 \cdot a + 0 \cdot a$$$$0 \cdot a + (- 0 \cdot a) = 0 \cdot a + 0 \cdot a + (- 0 \cdot a)$$$$ 0 = 0 \cdot a$$

Then the issue with assuming that $0$ has a multiplicative inverse is immediately apparent. Assume for proof by contradiction that $0^{-1}$ exists. By the multiplicative inverse definition $0 \cdot 0^{-1}=1$ but by the previous proof $0 \cdot 0^{-1} = 0$. So $0=1$ which is a contradiction. Therefore $0^{-1}$ does not exist for the real numbers (or any field).

There are other number systems (not real numbers) where $0^{-1}$ does exist, but those numbers are not fields. So they lose a lot of the desirable proofs and properties that rely on being a field.

 

[Azhar]

You can make such an exception, but not with the real numbers.

The real numbers are a field and fields have an additive identity $0$ such that $a+0=a$ for all $a$ as well as a multiplicative identity $1\ne 0$ such that $a \cdot 1 = a$ for all $a$. These lead to an additive inverse $-a$ such that $a+(-a)=0$ for all $a$ and a multiplicative inverse $a^{-1}$ such that $a \cdot a^{-1}=1$ for all $a\ne 0$. Division is just shorthand for multiplication by the inverse $a/b=a \cdot b^{-1}$.

From these you can prove that $0 \cdot a = 0$ for all $a$:
$$0 \cdot a = (0+0) \cdot a$$$$0 \cdot a = 0 \cdot a + 0 \cdot a$$$$0 \cdot a + (- 0 \cdot a) = 0 \cdot a + 0 \cdot a + (- 0 \cdot a)$$$$ 0 = 0 \cdot a$$

Then the issue with assuming that $0$ has a multiplicative inverse is immediately apparent. Assume for proof by contradiction that $0^{-1}$ exists. By the multiplicative inverse definition $0 \cdot 0^{-1}=1$ but by the previous proof $0 \cdot 0^{-1} = 0$. So $0=1$ which is a contradiction. Therefore $0^{-1}$ does not exist for the real numbers (or any field).

There are other number systems (not real numbers) where $0^{-1}$ does exist, but those numbers are not fields. So they lose a lot of the desirable proofs and properties that rely on being a field.

but the text is about custom rule not about standard math 0^-1

 

[Dale]

but the text is about custom rule not about standard math 0^-1

I understand that.

I showed why the custom rule you consider does not work for a field. So you can have either your custom rule or a field, not both.

 

[Azhar]

What if I combine both,what if the field axiom include exception for 0 but the exception is incomplete in field axiom so the custom rule complete it

I understand that.

I showed why the custom rule you consider does not work for a field. So you can have either your custom rule or a field, not both.

 

[FactChecker]

I don't think you paid enough attention to @Dale's post #8.
In your first post, you propose that $0/0 = 1$ and also that $0^2/0 = 0$. Since $0^2 = 0$, we have $0=0^2/0=0/0=1$.
This is using your special rules. $0=1$ is unacceptable.

 

[Azhar]

Wait yeah you're correct,0²÷0≠0÷0 0²÷0=0×0÷0,PEMDAS so multiplication first before division,0×0=0 then 0÷0=1 but 0≠1

But what if it's PNEMDAS so "neutralize" before ()

Neutralize definition:
Divide number with exponent so

(x³)÷(x²):
÷x neutralize the ² so the result is x,if it's (x³) then ÷(x²)

example:
(x³)÷(x²)
÷(x²) neutralize the ³ so the result is x

(2³)÷(2²)
8÷4=2

So 0²÷0:
÷0 neutralize the ² so the result is 0,0×0÷0 then ÷0 neutralize the ×0 so the result is 0

I don't think you paid enough attention to @Dale's post #8.
In your first post, you propose that $0/0 = 1$ and also that $0^2/0 = 0$. Since $0^2 = 0$, we have $0=0^2/0=0/0=1$.
This is using your special rules. $0=1$ is unacceptable.

 

[FactChecker]

Well, I don't know what you are trying to get at. There are an infinite variety of rules that can be thought up. The ones we use have many applications, are consistent, lead to a lot of theorems that are useful, etc.

There is nothing wrong with investigating your own ideas as long as you don't go too far down a "rabbit hole" for no reason. But this forum is not a good place for that because we have rules which forbid indulging in unpublished personal theories.
I will leave further discussion to others.

 

[PeroK]

What if I combine both,what if the field axiom include exception for 0 but the exception is incomplete in field axiom so the custom rule complete it

Then you end up with your own personal, probably inconsistent, mathematics.

 

[Azhar]

Then you end up with your own personal, probably inconsistent, mathematics.

More than 1 x×more than 1 x≠x

1×1=1,there's exception in standard math so there's exception for 0 too, same as custom rule,the exception is for certain number not for every number,if u force 2×2=2 then it's incorrect