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Suppose we have following logical statement: if and only if x=666 is a true equality, then x is the devil's number (we assume that there is only one devil's number in existence).

Now let's use 1/0 as x. Of course it's undefined. The question is, would it be correct to conclude that 1/0 is NOT the devil's number? Or would we be more correct by saying that we don't know if it's the devil's number?

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    1/0=666 is undefined. Thus it has no truth value. If so, what is the meaning of "x=666 is a true equality" when the equality is undefined ? Oct 19 '20 at 14:05
  • Having said that "if and only if x=666 is a true equality, then x is the devil's number " is wrongly written; it must be "if x=666 is a true equality, then x is the devil's number " Oct 19 '20 at 14:06
  • Perhaps you are unaware that there are distinct types of truths. One type of truth is a universal truth also called an objective truth. This statement has a constant truth value that never changes. So it will always be true or always be false.There are no mixtures of true false in the same column. So all dogs are animals happens to be always true with no false cases. All women are reptiles is always false with no true cases. These are absolute then there are contingent truths. These truth are mixed.True & false would appear in the same column. You seem to think awareness is truth. It isn't.
    – Logikal
    Oct 19 '20 at 14:08
  • What happens if someone makes a claim that you are unaware of? Does that statement have a truth value? Well people into science & math may say NO. There are no unicorns until we become aware of them. We are not aware so the claim is false. This is the sense verification type of proof which is contingent. The mixed colum of true & false values. Objective truth has one value in the entire colum not both values. So if scientists are correct the universe is over 100 million years old BUT humans were unaware for centuries of this truth. It was true with no human awareness, Notice that.
    – Logikal
    Oct 19 '20 at 14:15
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    @HotLicks Mathematicians closed my question as off-topic, it's why I'm here =))
    – user161005
    Oct 20 '20 at 3:39
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One solution would be to apply Russell's analysis of definite descriptions. Statements with n-place functions, f, can be rewritten as more complicated statements involving (n+1)-place functional relations, F (see NOTE 1).

For 1-place functions this looks like:

f(x) = y iff Fxy ∧ ∀z(Fxz → z = y))

For 2-place functions (like division) this looks like:

f(x,y) = z iff Fxyz ∧ ∀w(Fxyw → w = z))

For division in particular, this looks like:

x/y = z iff div(xyz) ∧ ∀w(div(xyw) → w = z))

Since the first conjunct, div(xyz) is false for every choice of z, no matter the choice of x and y, the entire statement will always be false. Intuitively, if a statement that uses a functional relation (like "the present king of France") implies the existence of something that doesn't actually exist, then (since any statement that implies a falsehood is false) such statements will be false.

"The present king of France is bald" is false, because there is no present king of France, and there would need to be in order for that statement to be true. "1/0=666" can be read as "the value of 1/0 is 666" which is also false, because there is no value of 1/0 and there would need to be for that statement to be true.

Of course, this is only one solution (one that uses "classical" logic). You can also use a three valued logic or one that allows statements with no truth values.


NOTE 1: The way this works in general is n-place functions f(x₁,x₂,...,xₙ) are replaced with (n+1)-place relations F(x₁,x₂,...,xₙ,y) and an axiom that guarantees existence and uniqueness for each combination of x₁,x₂,...,xₙ if f is a function (or only a uniqueness axiom if f is a partial function on the domain like division, or in discourse like philosophy where the domain is unclear):

Uniqueness: ∀x₁∀x₂...∀xₙ∀y∀z((F(x₁,x₂,...,xₙ,z) ∧ F(x₁,x₂,...,xₙ,z)) → y = z)

Existence: ∀x₁∀x₂...∀xₙ∃y(F(x₁,x₂,...,xₙ,y))

Then, every statement that refers to the value of f gets rewritten to state the existence of a y determined by F, which is guaranteed to be unique, and to refer to that y instead:

P(... f(x₁,x₂,...,xₙ) ...) becomes ∃y(F(x₁,x₂,...,xₙ,y) ∧ P(... y ...))

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Assuming all the symbols used in your expression have the meaning commonly assigned to them,

1/0 = 666 implies (*) 1 = 666 * 0 and 1 = 0, which is false. Therefore, even if the status of 1/0 as a number is dubious, 1/0 = 666 is unambiguously false.

(*) I previously said "is equivalent to", but it was rightfully pointed to me that the inverse operation being unlawful, it is only a one way implication.

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  • "1/0 = 666 is equivalent to 1 = 666 * 0 and 1 = 0" It's not. 1/0 is undefined. It's like saying that "Socrates=666 is equal to 2*Socrates=666", where Socrates isn't even a word, but a person.
    – user161005
    Oct 20 '20 at 3:42
  • @user161005 If you know that 1/0 is undefined, then shouldn't it be obvious to you that something undefined is not equal to something defined? Oct 20 '20 at 3:47
  • @user161005: That is where you are mistaken, IMO. a/b=c is equivalent to a = b * c, independently of the values of a, b or c, because of the very definition of multiplication and division. Pry tell me why transferring the 0 from one side of = to another is not legit ? On the other hand, multiplying one side of the equation without multiplying the other like you did absolutely does not yield an equivalent proposition (x=y is certainly not the same as 2*x=y), so I fail to see your point.
    – armand
    Oct 20 '20 at 3:49
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    This is not correct mathematically. If it were, then one could do the reverse and say that 666*0 = 0, so 0/0 = 666. But then one could do the same with 12 * 0 = 0, so 0/0 = 12. Or 7928, or x or n. The reason this is incorrect is that when we say a/b = c is equivalent to a = b * c it is an algebraic shorthand. In everyday life it works as a shorthand as long as one uses real numbers and does not divide by 0. But if one does each step to get that equivalency, it would fail at divide each side by 0. Or in this case step 1 multiply each side by 0, step 2 cancel 0/0 on the left is not allowed.
    – Damila
    Oct 20 '20 at 5:32
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    Yes, you can't do the reverse operation, because dividing by zero is wrong. Yet multiplying is fine. I shall correct "is equivalent to" by "implies", since the operation is not reversible. The result of 1/0 might be undefined, but each of its parts 1, 0 and / are very well defined. 1/0 = x is unambiguously false for any x, the rest is hair splitting. Curiousdanii also nailed the fact that "defined = undefined" would also be unambiguously false, anyway.
    – armand
    Oct 20 '20 at 7:03

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