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Memfis

empty set (math logic)

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Is it correct to say that all elements of an empty set are equal to 2? I mean, there is no way to disprove that statement by picking an element that isn't equal to 2. So is it correct then?

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It's not possible to pick any elements, for that matter.

It's like dividing 2 pies by 0 people: each one gets....wait, each one who?

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The way I see it, the statement 'all elements of such and such set are equal to 2' can be broken down into three parts.
- This set contains elements
- The elements have the property of being equal to a number
- The number they are equal to is 2
So to prove that not all elements of a specific set are equal to 2, you just have to disprove one of these statements. And an empty set does not contain elements, unless I'm more ignorant than I thought on set theory. Even if it did, I don't think the element would have the property of being able to be 'equal' to a number.

Or to put it another way:
Let's assume that all elements of an empty set are equal to 2. Therefore, this set contains at least one element that is equal to 2. Therefore, this set is not empty.
Proof by contradiction that 'all elements of an empty set are equal to 2' is false.

Or to put it another way:
'Equal' assumes existence, by definition. Therefore anything that is equal to 2, must exist. Anything that does not exist cannot be equal to something that does.

Your question is like a scientist producing a sealed box with a vacuum inside. He presents it to you and says 'Everything inside this box is blue!' When you say no, that's ridiculous, he says 'Well can you find something in this box which ISN'T blue?'
It looks like you can't disprove him! Therefore, everything inside a vacuum is blue. Everything inside a vacuum is also red, green, fifty, haggard, conceptual, dynamic, happy, dead, and simultaneously both non-existent and existent. I mean, can you find something inside it that exists? Can you find something inside it that doesn't exist? I think we just broke the logical absolutes.

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Isn't the empty set, by definition, a set with no elements?

Wait, fuck. I think it would be vacuously true to say that "if the empty set contains elements, then all of those elements are equal to 2"...

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bool found = false;    // find a value that's not equal

for(int i = 0; i < length; ++i)
    if(set[i] != 2)
    {
        found = true;
        break;
    }

// If set is empty (length is 0), then found will remain false, i.e. no element found that's not 2
If you ask a computer to solve the OP's logic, it will agree with him.

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I remember once I tried to be a smartass and prove that all numbers were equal, by using raising any number to the zeroth power as "proof":

since any number raised to the zero power equals one, then pow(x,0) == pow(y,0) for any pair of numbers (x,y). Therefore, x == y and all numbers are equal, making the solution of mathematical problems etc. trivial ;-)

My algebra professor didn't quite like it though.

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Memfis said:

Is it correct to say that all elements of an empty set are equal to 2? I mean, there is no way to disprove that statement by picking an element that isn't equal to 2. So is it correct then?

Yes, but vacuously so. http://en.m.wikipedia.org/wiki/Vacuous_truth

The following two propositions are both true, by set theoretic axioms:

1) For every element of the empty set, the property of being equal to 2 holds.
2) There is no element of the empty set for which the property of being equal to 2 holds.

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This we learned on the very first university lesson of Mathematical Logic:

'From any impossible proposition any other proposition follows'.

(not A) => (A => B)

In other words, you can "legitly" deduce anything and everything from an assumption which is false.

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Thanks durian, this answers my question.

Maes said:

I remember once I tried to be a smartass and prove that all numbers were equal, by using raising any number to the zeroth power as "proof":

since any number raised to the zero power equals one, then pow(x,0) == pow(y,0) for any pair of numbers (x,y). Therefore, x == y and all numbers are equal, making the solution of mathematical problems etc. trivial ;-)

My algebra professor didn't quite like it though.

Here is a more interesting "proof" that all powers of 2 are equal to 1, at first it confused me :) -> http://www.cut-the-knot.org/proofs/AllPowersOf2Equal1.shtml

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scifista42 said:

This we learned on the very first university lesson of Mathematical Logic:

'From any impossible proposition any other proposition follows'.

(not A) => (A => B)

In other words, you can "legitly" deduce anything and everything from an assumption which is false.


Yup, though the antecedent needn't be impossible for the proposition to be true - as you indicate, just false will suffice. To illustrate, here's the truth table for material conditional propositions:



A material conditional is false if and only if its antecedent is true and its consequent is false. If its antecedent is false then, regardless of whether its consequent is true or false, the proposition as a whole is true.

So, Memfis's proposition works in the way that it does, in part because, insofar as it involves quantification - by referring to all elements of the empty set - it has the logical form of a conditional. That is, the logical form of the statement, "All elements of the empty set are equal to 2", is equivalent to that of: "For any x, if x is a member of the empty set, then x is equal to 2", and since the antecedent, "x is a member of the empty set", is false for any value of x, given that the empty set has no members, the proposition as a whole comes out as true.

edit - fixed a little sloppiness

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Memfis said:

Thanks durian, this answers my question.

Here is a more interesting "proof" that all powers of 2 are equal to 1, at first it confused me :) -> http://www.cut-the-knot.org/proofs/AllPowersOf2Equal1.shtml


With the difference that that one is actually immediately disproved, while it's harder proving that if two numbers a and b raised to the same power c (a^c, b^c) have equal values, then a and b aren't necessarily equal.

Actually, that's more of a trick with how the "zero power" is defined: it's defined as a "continuity extension", so its value is actually purely speculative ;-)

Even more fun if you note that 0^0 =1....plenty of fun by equating all numbers with zero.

E.g. Since 1^0 = 1 and 0^0 =1 then it follows that 1==0, amirite? ;-)

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@Maes: Amplifying is not an equivalent equation transformation, that's why assumptions like yours don't work.

@Durian: What I was talking about wasn't a simple implication A=>B, but rather (not A)=>(A=>B), but the principle is there.

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Oops! Despite quoting it, I missed the detail in the notation - quite right, that's a different kind of case.

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This is one of the differences between the for all operator (universal quantification) and the there exists (existential quantification) operator. Universal quantification does not require the existence of an element within the set. Therefore if I say that for all elements x in a set R, f(x) is true, this does not necessarily mean that there exists an element x in that set such that f(x) is true.

Logically it occurs as anything is true for nothing. f(x) is always true if no x exists that satisfies the domain requirement.

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printz said:

bool found = false;    // find a value that's not equal

for(int i = 0; i < length; ++i)
    if(set[i] != 2)
    {
        found = true;
        break;
    }

// If set is empty (length is 0), then found will remain false, i.e. no element found that's not 2
If you ask a computer to solve the OP's logic, it will agree with him.


Oh yeah?

bool none_equals_two = true;    // find a value that's equal

for(int i = 0; i < length; ++i)
    if(set[i] == 2)
    {
        none_equals_two = false;
        break;
    }

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kb1 said:

Oh yeah?


Obviously, you should use 3-state logic, and start with "undefined" as a default value. Only a definite find (or at least a failure from a set with at least one element) should be allowed to set a clear true/false value.

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printz said:

If you ask a computer to solve the OP's logic, it will agree with him.

kb1 said:

Oh yeah?

bool found_unequal = false;    // find a value that's not equal
bool found_equal = false;    // find a value that's equal

for(int i = 0; i < length; ++i)
{
    if(set[i] != 2)
    {
        found_unequal = true;
        break;
    }
    else if(set[i] == 2)
    {
        found_equal = true;
        break;
    }
}
Printf("In the %d elements, %s are equal to 2, and %s are unequal to 2\n",
    length, found_unequal ? "some" : "none", found_equal ? "some" : "none");

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