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

In summary, the use of infinity in mathematical calculations is a complex and nuanced topic. While it can be used to represent unbounded limits in real analysis, it cannot be treated as a real number with operations such as addition and multiplication. Attempts to do so can lead to contradictions and nonsensical results. Therefore, the mathematical consensus is that division by zero and the use of infinity as a number are undefined.
  • #1
Mr Indeterminate
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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?
 
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  • #2
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.
 
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  • #3
Mr Indeterminate said:
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.
 
  • #4
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.
 
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  • #5
weirdoguy said:
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?
 
  • #6
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.
 
  • #7
Mr Indeterminate said:
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).
 
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  • #8
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.
 
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  • #9
Mr Indeterminate said:
As the mathematical consensus is that division by zero is undefined, why is the proof incorrect?
Mr Indeterminate said:
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.
 
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  • #10
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.
 
  • #11
Mr Indeterminate said:
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.
 
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  • #12
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:

jbriggs444 said:
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
 
  • #13
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  • #15
Mr Indeterminate said:
(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.
 
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  • #16
jbriggs444 said:
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.
 
  • #17
Mr Indeterminate said:
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.
 
  • #18
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.
 
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  • #19
Mr Indeterminate said:
(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.
 
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  • #20
Mr Indeterminate said:
(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.
 
  • #21
Mr Indeterminate said:
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:
DotPlot.png

Mr Indeterminate said:
That type of thinking seems very pre-Schrödinger's cat.
No need for Schrödinger's cat -- this is pretty basic mathematics.
 
  • #22
Mr Indeterminate said:
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.
 
  • #23
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.
 
  • #24
thierrykauf said:
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.
 
  • #25
Carrock said:
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.
 
  • #26
jbriggs444 said:
Carrock said:
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.
 
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  • #27
Mr Indeterminate said:
... 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)].
 
  • #28
FactChecker said:
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?
 
  • #29
Mr Indeterminate said:
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.
 
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  • #30
jbriggs444 said:
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?
 
  • #31
Mr Indeterminate said:
Then how you reason that x(x-1)=0 isn't proof of 1=0 while in #7 it is?
#7 is a proof using the fundamental properties of arithmetic. In your "proof", you are confusing 'or' with 'and'. You should review the difference before continuing.
 
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  • #32
FactChecker said:
#7 is a proof using the fundamental properties of arithmetic. In your "proof", you are confusing 'or' with 'and'. You should review the difference before continuing.

Did you read #28?
 
  • #33
Mr Indeterminate said:
Did you read #28?
#28 made no argument that I can see.
 
  • #34
Mr Indeterminate 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?
To which 1=0 logic are you referring to exactly?

For the record, if we accept that ##\frac 1\infty=0##, we accept that we're not talking about the real numbers (or a field) any more, but for instance about the real projective line. In that case the implication that multiplying both sides by ##\infty## would yield ##1=0\cdot \infty## does not hold true. And that's a reason why infinity is not included in the real numbers.

To follow through with the real projective line, each number is represented as [x:w], which really means ##\frac x w##. With this definition, it becomes possible to calculate with infinity (represented as [1:0]) without problems. It does mean that any result that comes out as [0:0] must be treated as undefined.
 
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  • #35
Mr Indeterminate said:
Then how you reason that x(x-1)=0 isn't proof of 1=0 while in #7 it is?
When one writes down a variable name in mathematics saying something like...

"Let x be a solution to the equation: x(x-1)=0"​

... one is allowing the symbol "x" stand in for one of the two solutions of that equation. It does not mean that x=0 and x=1. It means that x=0 or x=1. Which of the two solutions it is remains unspecified.

An assertion that x(x-1)=0 is not strong enough to demonstrate that x=1. It might instead be 0.
An assertion that x(x-1)=0 is not strong enough to demonstrate that x=0. It might instead be 1.
So there is no basis on which to invoke the transitive property of equality and proceed to a conclusion that 1=0.

Edit to add:

Now let us go back to #7 and see if this flaw applies to the argument there...

The first line begins with a phrasing: "If ##\infty## were a member of the field of real numbers...". The mathematical meaning of this is that we are using the symbol ##\infty## to denote an arbitrary real number. A single number, albeit a number not yet fully specified.

That line proceeds with "if one agreed that ##\frac{1}{\infty} = 0##. This further restricts the set of real numbers that ##\infty## might denote. It still denotes a single real number, albeit one that still might not be completely specified.

In fact, it has been over-specified -- no such real number can exist. Which is demonstrated by the rest of the argument.

The next line argues that whatever ##\infty## is, the properties of arithmetic together with the premise that ##\frac{1}{\infty} = 0## means that ##0 \cdot \infty = 1##. Because ##\infty## denotes a single real number, ##0 \cdot \infty## evaluates to a single real number whose value must be 1. That is, no matter what value ##\infty## denotes, the expression ##0 \cdot \infty## must evaluate to 1.

The next line proceeds to demonstrate with equal force that no matter what value ##\infty## denotes, the expression ##0 \cdot \infty## must evaluate to 0.

Equality is transitive. Two things equal to the same thing are equal to each other. So one can conclude that 1=0.

Note well. We were able to demonstrate that ##0 \cdot \infty## = 0 and we were also able to demonstrate that ##0 \cdot \infty## = 1. That's and, not or.

This argument takes the form of a proof by contradiction. We have applied correct logic and concluded a falsehood. So at least one of the premises must be incorrect. The premises were:

"If ##\infty## were a member of the field of real numbers" and
"If one agreed that ##\frac{1}{\infty}= 0##"

At least one of those conditions must not hold.
 
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