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

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.

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)].

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
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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.

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?

FactChecker
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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.

#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.

jbriggs444
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#28 made no argument that I can see.

I like Serena
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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.

jbriggs444
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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.

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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Math_QED
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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.
Indeed, I agree as well to this. There is even a field axiom, called the non-triviality axiom, that forbids that $0 = 1$.

Mark44
Mentor
From post #7 (with underlining added by me):
jbriggs444 said:
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...
In the post above jbriggs444 continues to the nonsense result that 1 = 0.

Then how you reason that x(x-1)=0 isn't proof of 1=0 while in #7 it is?
A shorter version of what jbriggs444 said in post #35.
If x(x - 1) = 0, then we can conclude that x = 0 or x = 1. x can be one of these values but not both. There's no way we can conclude that 0 = 1 from this.

Since we seem to be beating a dead horse here, I'm closing this thread.