Where has this proof gone wrong? ∞= 1/0

[Mr Indeterminate]

Now I expect that most of you on this forum would be familiar with the equality between point nine reoccurring and one:

0.999...=1 ​

If your not familiar please review https://en.wikipedia.org/wiki/0.999...​

Now this equality can be used to imply something else, which is rather heterodox, consider the below:

Point nine reoccurring is only one infinith smaller than one
0. 999 ... +(1/∞)= 1

But as point nine reoccurring and one are equal, point nine reoccurring is one infinith smaller than itself
0. 999 ... +(1/∞)= 0. 999 ...

It is only logical then to conclude that one infinith is equal to zero
0. 999 ... − 0. 999 ... +(1/∞)= 0. 999 ... − 0. 999 ...

1/∞= 0

Which can in turn be inverted to reveal that infinity is equal to one divided by zero
∞ =1/0∎​

As the mathematical consensus is that division by zero is undefined, why is the proof incorrect?

 

[weirdoguy]

1/∞ is not a real number, so you cannot add it to other numbers.∞ is not a number, it's a symbol that means something only in a special context - limits. "one infinith smaller" does not make sense.

 

[mfb]

Point nine reoccurring is only one infinith smaller than one
0. 999 ... +(1/∞)= 1

There is no such thing. Whatever (1/∞) is supposed to mean, you are assuming (1/∞)=0 here, and that equation doesn't make sense in the real numbers.

 

[WWGD]

And the EDIT: Extended Reals (Standard Reals with a "Point at Infinity" appended to them) are not a ring, i.e., do not have a multiplication defined ( and, I believe, no "reasonable" version definable), so the operation $\infty \times 0$ is not defined. Still, if you had a ring, then $0 \times r =0$ for any r in the ring. So this operation is not definable within the Extended Reals either. Maybe you can try the Surreals or the Hyperreals where you have infinities as numbers. Still, a lot of the work on those numbers is pretty esoteric.

 

[Mr Indeterminate]

1/∞ is not a real number, so you cannot add it to other numbers.∞ is not a number, it's a symbol that means something only in a special context - limits. "one infinith smaller" does not make sense.

Why is infinity not a number, while zero is?

 

[weirdoguy]

It's not a real number. Why do you think it should be? Do you understand what "infinity" means in context of limits? Or other contexts (cardinality)? How would you define it as a real number? We have a few constructions of real numbers, in all of them 0 is a precisely defined object.

 

[jbriggs444]

Why is infinity not a number, while zero is?

If $\infty$ were a member of the field of real numbers and if one agreed that $\frac{1}{\infty}=0$ then one could proceed to prove nonsense. For example...

By definition, $\frac{a}{b}$ is that [unique] number which, when multiplied by $b$ produces $a$. So if $\infty$ is a real number and $\frac{1}{\infty}=0$ then, by definition:
$$0 \cdot \infty = 1$$
But it is also well known and provable that for all $a$, $0 \cdot a = 0$. In particular, if $\infty$ is a member of the field of real numbers then:
$$0 \cdot \infty = 0$$
Two things that are equal to the same thing are equal to each other. So it would immediately follow that:
$$0 = 1$$

There are some choices for how to deal with this problem. One way to go is to accept that $\infty$ is a legitimate number but to leave $0 \cdot \infty$ undefined. This leads to the extended reals as @WWGD has explained. Another way to go is to allow for multiple infinite numbers, each with its own infinitesimal inverse. This leads to Robinson's Non-standard Analysis and fields such as the "hyperreals" or "surreals" [as @WWGD has also suggested].

In the standard real numbers, there are no infinite numbers and only one infinitesimal number. Zero counts as the infinitesimal.
In the non-standard real numbers, there are many infinite numbers and many infinitesimal numbers but still only one zero (which still has no inverse).

 

[FactChecker]

There are many ways for something to be infinite and many different "sizes" of infinity (cardinality). So any time you try to use infinity as a number in a calculation, you are opening a can of worms that may be much more complicated than the original problem.

 

[fresh_42]

As the mathematical consensus is that division by zero is undefined, why is the proof incorrect?

Why is infinity not a number, while zero is?

With this point of view given, my argument will be:
Zero is not an element of the multiplicative group of the real number filed. Therefore the question, whether it has an inverse or even can be multiplied by something else simply does not exist. It is in a different set!

More elaborated: The connection between addition, where the zero belongs to, and multiplication, where inverses belong to, in a field is set by the distributive law only. It forces $0 \cdot a = 0$. As $\infty$ isn't an element of either of the groups, there is no chance to operate with it. You could object "... then let us define it as an element" of one or both of the groups, but then you will run into contradictions of all kind as already instanced in previous posts, including yours. You have to be rigorous from the start, not at the end and wonder why the sand castle breaks down.

 

[Carrock]

A problem here is the use of $\infty$ without clearly distinguishing it from the real numbers, which have a cardinality equal to $\aleph_1$, assuming the continuum hypothesis.

$\infty$ is defined here as "an abstract concept describing something without any bound or larger than any number." i.e. two different definitions, and later
"In real analysis, the symbol $\infty$, called "infinity", is used to denote an unbounded limit. ${n \rightarrow \infty}$ means that x grows without bound..."

i.e. $\infty$ as normally used refers to a large but not infinite number.

So $\lim_{x \rightarrow+\infty} {\frac {x+1} {2x+1}}=0.5$ should really be read as "as x increases without limit the expression becomes arbitrarily close to, but is never equal to $0.5$.

Using the '=' sign in this way is very convenient but also misleading.

Using x equal to the smallest infinity $\aleph_0$ you get ${\frac {\aleph_0+1} {2\aleph_0+1}}$ which is undefined.

Using '=' in a different way ...

Given that $0.999999...=1$
$\lim_{x \rightarrow+\infty}({0.999999...+\frac 1 x})>1$ since 1/(a positive number which can increase without limit, but is always finite) is always >0.

 

[I like Serena]

It is only logical then to conclude that one infinith is equal to zero
0. 999 ... − 0. 999 ... +(1/∞)= 0. 999 ... − 0. 999 ...

1/∞= 0

Which can in turn be inverted to reveal that infinity is equal to one divided by zero[/I]
∞ =1/0∎

As the mathematical consensus is that division by zero is undefined, why is the proof incorrect?

As said before, infinity is not defined for the real numbers, and neither is division by zero.

However, we can extend the real numbers to the Real Projective Line or to the Extended Real Number Line.
In both cases your proof is valid, and we have indeed $\frac 1\infty = 0$.
It's only on the Real Projective Line that we have $\frac 1 0=\infty$ though.
Have to be careful since not all of the usual operations are well defined.

For the record, the use of infinity in Projective Geometry allows us full equivalency between ellipses, parabolas, and hyperboles.

 

[Mr Indeterminate]

Hi All,

You've made quite clear to me that infinity isn't a real number because it hasn't been defined as one. However, what I really want to understand is why it wasn't included.

Obviously if something nonsensical results of the inclusion then there is a reason to exclude it. jbriggs444 follows this logic in his explanation:

By definition, $\frac{a}{b}$ is that [unique] number which, when multiplied by $b$ produces $a$. So if $\infty$ is a real number and $\frac{1}{\infty}=0$ then, by definition:
$$0 \cdot \infty = 1$$
But it is also well known and provable that for all $a$, $0 \cdot a = 0$. In particular, if $\infty$ is a member of the field of real numbers then:
$$0 \cdot \infty = 0$$
Two things that are equal to the same thing are equal to each other. So it would immediately follow that:
$$0 = 1$$

That being said, the 0=1 equality is only arrived at because it is assumed that:

If a=b and a=c then b=c

This is not always the case, consider Yx0=0

Y can be 1, 0 or 524 as all numbers work for Y. Just because Y=1 and Y=0 doesn't mean that 1=0

 

[jbriggs444]

This is not always the case, consider Yx0=0

Y can be 1, 0 or 524 as all numbers work for Y. Just because Y=1 and Y=0 doesn't mean that 1=0

This is the fallacy of equivocation. You are taking the single symbol Y and using it to denote two different numbers within the same context.

http://www.txstate.edu/philosophy/resources/fallacy-definitions/Equivocation.html

 

[Mr Indeterminate]

This is the fallacy of equivocation. You are taking the single symbol Y and using it to denote two different numbers within the same context.

http://www.txstate.edu/philosophy/resources/fallacy-definitions/Equivocation.html

(Y-2)(Y-3)=0

thus Y=2 and Y=3

2=3?

 

[jbriggs444]

(Y-2)(Y-3)=0

thus Y=2 and Y=3

2=3?

Let me try to translate to more coherent prose...

Let Y be the solution the equation, "(Y-2)(Y-3)=0"...
​

That's a violation of the rules of mathematical discourse. The word "the" carries with it an assertion of existence and uniqueness. You are not allowed to use the word unless you are prepared to provide those guarantees. In this case there is no guarantee of uniqueness.

Let Y be a solution the equation, "(Y-2)(Y-3)=0". Clearly then, either Y=2 or Y=3. We can conclude that both Y=2 and Y=3 are true.
​

That's just wrong. We can conclude one of the two possibilities (Y=2 or Y=3) must hold. But we do not know which. We most certainly know that both cannot simultaneously be true.

Edit: Let me try to put it in an entirely different way...

If we are given the equation "(Y-2)(Y-3)=0", that is an assertion about some state of affairs. We might have a state of affairs where Y=2, the sky is blue and water is wet. We might have a different state of affairs where Y=3, Pluto is not a planet and Elvis has left the building. Those are both consistent with the truth of the equation. [The equation rules out every state of affairs where Y is anything other than 2 or 3]

The transitive property: "if a=b and b=c then a=c" holds within a single state of affairs. You have to have decided on specific values for a, b and c. You are not allowed to equivocate, picking one state of affairs where a=b holds and a different state of affairs where b=c holds.

 

[Mr Indeterminate]

Let me try to translate to more coherent prose...

Let Y be the solution the equation, "(Y-2)(Y-3)=0"...
​

That's a violation of the rules of mathematical discourse. The word "the" carries with it an assertion of existence and uniqueness. You are not allowed to use the word unless you are prepared to provide those guarantees. In this case there is no guarantee of uniqueness.

Let Y be a solution the equation, "(Y-2)(Y-3)=0". Clearly then, either Y=2 or Y=3. We can conclude that both Y=2 and Y=3 are true.
​

That's just wrong. We can conclude one of the two possibilities (Y=2 or Y=3) must hold. But we do not know which. We most certainly know that both cannot simultaneously be true.

How would you approach plotting it on a Cartesian plane?

Flip a coin, if heads plot (0,2), if tails plot (0,3)?

That type of thinking seems very pre-Schrödinger's cat.

 

[jbriggs444]

How would you approach plotting it on a Cartesian plane?

Flip a coin, if heads plot (0,2), if tails plot (0,3)?

That type of thinking seems very pre-Schrödinger's cat.

Plotting what on a Cartesian plane.

You can plot the graph of f(Y) = (Y-2)(Y-3) and get a parabola which intersects the Y axis at Y=2 and Y=3.
You can plot the solution set to (Y-2)(Y-3)=0 on the number line by putting dots at 2 and 3.

None of that is relevant to the fallacy of equivocation.

 

[FactChecker]

It's not always true that ∞ - ∞ = 0. So including ∞ as a real number means that some fundamental rules of arithmetic no longer hold for the reals. That is too much to give up.

 

[Mark44]

(Y-2)(Y-3)=0

thus Y=2 and Y=3

2=3?

No. The correct conclusion is that EITHER Y = 2 OR Y = 3. It is incorrect to conclude that Y can be both numbers simultaneously, which is what the conjunction "and" implies.

 

[WWGD]

(Y-2)(Y-3)=0

thus Y=2 and Y=3

2=3?

Y=2 _Or_ Y=3 for product to be 0. Y=2 _and_ Y=3 does not follow.

EDIT: Still, per@ILikeSerena's last post, we can do arithmetic on $\infty$ in $P_1(K)$ ; arithmetic rules can be extended from K to work with the projective line.

 

[Mark44]

How would you approach plotting it on a Cartesian plane?
Flip a coin, if heads plot (0,2), if tails plot (0,3)?

Like this:

That type of thinking seems very pre-Schrödinger's cat.

No need for Schrödinger's cat -- this is pretty basic mathematics.

 

[FactChecker]

Y can be 1, 0 or 524 as all numbers work for Y. Just because Y=1 and Y=0 doesn't mean that 1=0

You changed the "or" to "and". That is wrong. No more can be said.

 

[thierrykauf]

Infinity is not a number. It can only be approached, in the limit, but never reached. Take a large number N. Add 1 to it. You obtain N+1 which is always greater than N. But in the limit N goes to infinity, lim N = lim N+1. Does that mean N=N+1? Not at all.

 

[Carrock]

Infinity is not a number. It can only be approached, in the limit, but never reached. Take a large number N. Add 1 to it. You obtain N+1 which is always greater than N. But in the limit N goes to infinity, lim N = lim N+1. Does that mean N=N+1? Not at all.

You're referring to $\infty$ which has been described as "not a cardinal number, but a description for going to as large a number as we like."
In the limit N goes to $\infty$, lim N < lim (N+1). Although "lim N = lim N+1" is often written, it is clear that the LHS is one smaller than the RHS. N is a finite number.
A highest number cannot be defined for $\infty$; similarly a truly infinite set such as $\aleph_0$ can be defined as a concept, which does not require each of its members to be enumerated. Such truly infinite sets are widely used in mathematics.

 

[jbriggs444]

lim N < lim (N+1)

In the topology where the limit exists, both limits are identical. However, that does not make the limit a real number.

 

[Carrock]

lim N < lim (N+1)

In the topology where the limit exists, both limits are identical. However, that does not make the limit a real number.

I made a rather careless late night post.
I should have confined myself to saying "N (or N+1) does not converge on a limit as N goes to infinity. lim N and lim (N+1) are not defined and can't be compared.

 

[sysprog]

... Just because Y=1 and Y=0 doesn't mean that 1=0 ...

I dislike the "just because ... doesn't mean ..." construction -- I rephrase your statement as:

"it is not the case that Y=1 and Y=0 entails that 1=0"

the statement [y=1 and y=0] is contradictory, and therefore false, and a false premise entails anything, including false statements, including the false statement that 1=0 ...

I disagree with your argument to the effect that [(y=1 and y=0) does not entail that (1=0)].

 

[Mr Indeterminate]

You changed the "or" to "and". That is wrong. No more can be said.

I concede that "and" was inappropriate wording.

However, I still think the 1=0 logic provided earlier in the thread is problematic.

Any chance we could cover off on this before moving back the original question?

 

[jbriggs444]

However, I still think the 1=0 logic provided earlier in the thread is problematic.

I stand behind the correctness of the logic in #7: If $\infty$ is an element in the field and has the property that $\frac{1}{\infty}$ = 0 then 1=0.

 

[Mr Indeterminate]

I stand behind the correctness of the logic in #7: If $\infty$ is an element in the field and has the property that $\frac{1}{\infty}$ = 0 then 1=0.

Then how you reason that x(x-1)=0 isn't proof of 1=0 while in #7 it is?