Proof that 1 = 0 using a common logical fallacy | Hacker News Reader

Proof that 1 = 0 using a common logical fallacy

nathanmarz.com

21 pointsnathanmarz27 comments

There are a couple of subtle pieces lurking just under the surface of this post, but diving into them requires a lot of time and words, so I don't mind them not being included. For example, proofs by contradiction require the Law of the Excluded Middle[1], which some reject. You could go into this for pages and pages, and it's very interesting, but it's also a complete tangent.

However, there's one important part that I think deserves some more depth. One of the main points driving the logic on display here is that the mathematical concept of "implies" is different than the normal one, because in logic "A -> B" does not suggest any sort of relationship between A and B. The sentence "My name is Fred, and that implies water is wet" is a true statement in logic, but most non-logicians would consider the statement false because there's no causative relationship. A logic definition of implication is more like "there is no possible universe in which A is true and B is false," which can be satisfied by A being false (or B being true) in all universes, regardless of what B (or A) happens to be and what A and B have to do with each other.

It's possible to see this as the crux of most arithmetic paradoxes. If you can slip in a false hypotheses, you can imply any conclusion, no matter how false it may be. All that's required is some sleight-of-hand so that you don't notice when the false axiom is being invoked. A common trick is dividing by an expression that evaluates to zero, without evaluating the expression.

For related shenanigans, check out the Principle of Explosion: a single logical inconsistency suffices to prove all statements, including their negations.

On an unrelated note, talking about the logical implication operator is difficult because you have to be careful not to use the word "implies." Good fun!

[1] "A or ~A", or equivalently, "~~A -> A" for all logical propositions A.

so is there a similar constructivist proof for x*0 = 0?

Deleted comment

sweet - thanks. oh! it's gone... and it seemed so reasonable!

I'm sorry, but this post is utterly pedantic.

The proof by Daniel Levine is absolutely "logically sound." The idea that what was shown was the conditional, "if y=y then x0=0" and not "x0=0 is true" simply ignores the fact that the first proposition in the proof was NOT an implication or derivation - it was an axiom.

Because it is an axiom, we are guaranteed that the argument is sound (and the conclusion true under whatever interpretation allows proposition #1 as an axiom) as soon as we prove that the argument is valid. Which is exactly what Levine set out to show. Logic 101.

Proof from the beginning of this article is ok but you have to reduce hypothesis to axiom using only operations that transform one line into second line that is true "If and only if" the first line is true.

Following proof of 1 = 0 does not have this property.

I was taught that this kind of proof is named "proof through equivalence" (my own translation) and should be avoided because it's easy to mistakenly assume that two lines are equivalent but there is only one way implication between them.

I apologize for not reading the entire post, but I don't have the patience tonight. Couldn't this problem be solved by saying that 0 is not a normal number with normal properties? Therefore, it might not fit into all of the properties of equality.

1/0, 0/0, (5)x(0) = (88)x(0), 5^0 = 77^0

are a few example of really weird math things. Accepting 0 can not be used as a number but as a placeholder for an ideal would be better because then you won't make logical fallacies.

In general, this is sort of how things are done. In abstract algebra you often define 0 as the additive identity (and then prove there can be only one).

But, you don't want to give it too much special treatment. You can prove that x * 0 = 0. Why is this nice? Because you can apply abstract algebra to systems that are not just arithmetic. It can, for example be applied to set theory where you do this:

Addition --> Set intersection Multiplication --> Set union

If you do that, and all the axioms are met, then you get a bunch of theorems about set theory for free. And its not just set theory. I've heard about this being applied to material science as well.

The thing to remember is that the operators such as +, *, -, / are very specific with arithmetic but in the more abstract fields, they are place holders for operators that share some generic properties.

Is there a valid proof that x * 0 = 0? The definition of multiplication under the Peano axioms makes it axiomatic.

It depends on what your axiom system is, what 0 is, and what your universe is. In Peano arithmetic it's axiomatic. Meanwhile, for fields (and possibly less-structred rings) in abstract algebra, it's a theorem. An important note is that the proof for fields requires negative numbers, while Peano arithmetic only describes the natural numbers.

Here's the field proof I'm pulling out of memeory:

  x*0 = x*(y + -y) = x*y + x*(-y) = x*y - x*y = 0

This relies on the existence of additive inverses for the first step, distribution in the second step, and the fact that x(-y) = -(xy) in the third step. That third property can be derived from sufficiently-specific ring axioms, but I forget how specific they have to be. It might be true in any ring by virtue of distribution but I forget the proof.

So to answer your question, yes there are... but it depends on the context. In some situations it may need to be axiomatic. A recursively-defined system like Peano may need to take it axiomatically as a base case.

x(-y)=x(-y+y-y)=x(-y)+xy+x(-y)

Cancellation yields 0=xy+x(-y)

Therefore -(xy)=x(-y)

Thank you. That's beautiful.

0 is additive identity. 0+y=y for all integers y. 0 is an integer, so 0+0=0. 0+0 is closed under addition, so (0+0) is an integer, so x(0+0) = x0 for all integers x. By distributive law, x0 + x0 = x0. By closure under multiplication, x0 is an integer. By additive inverses, there exists an integer (-x0), such that x0 + (-x0) = 0. Because (-x0) is an integer, x0 + x0 + (-x0) = x0+(-x0). By associative property of addition and the transitive property of equality, x0 + 0 = 0. By the additive identity, x*0 = 0.

Excellent walk-through, thank you. You also removed one of my uses of additive inverses, which gets us closer to not needing the negative integers. I have two ideas for getting rid of the last use of them in your last step, but I'm sure they're kosher.

1) From the equation x0 + x0 = x0, you don't need inverses, just cancellation. I believe that cancellation is a strictly weaker property.

2) From the equation x0 + x0 = x0, note that x0 is the additive identity. Inverses are unique, thus x0 = 0.

Unfortunately, I'm not familiar enough with Peano arithmetic to know if proving either of these statements requires the statement we're trying to prove, that x*0 = 0. I'm more familiar with algebra, where inverses exist axiomatically. But at least we've weakened the hypotheses!

To prove something about multiplication you first need to provide a definition of multiplication. So what is you definition of multiplication? The usual definition is:

    a*0 = 0
    a*S(b) = a*b+a

Cancellation follows from injectivity of S by induction.

No, you can't say x0 is the additive identity because 0 is the only additive identity. That would be equivalent to x0=0. Your way is fine if you just use what archgoon wrote to complete it.

This is exactly the proof in the case of vector spaces.

See:

http://us.metamath.org/mpegif/mul01i.html

http://us.metamath.org/mpegif/mul02i.html

I'm thinking something that has to do with set theory. I don't remember the steps.

This is another way of explaining why you can't divide by zero (which is essentially what he's doing in step 2), since it opens the door for all kinds of "proofs" that are actually fallacies.

He's not dividing by zero. He's just using ...

  a = b -> a - b = 0 -> c * (a - b) = 0 -> c * a = c * b

The whole point of the article is that implication is a binary function that doesn't have an intuitive behavior to many people.

Like the following statement which evaluates to TRUE ...

  I have 6 legs *implies* it will rain tonight

That's because the definition of the function is this ...

  A -> B => NOT(A) or B

And this actually makes sense ...

  I have 6 legs *implies* I am not normal

Which is true. But the reverse is false ...

  I am not normal, therefore I have 6 legs

A -> B doesn't imply that B -> A. The equivalence function would be ...

  (NOT(A) and NOT(B)) or (A and B)

I think this is what he meant: Sometimes you can extract a valid proof from this kind of bad proof by reading it in reverse:

    3. 0=0
    2. 1*0=0*0
    1. 1=0

However, in this case the step from 2 to 1 requires "division by 0".

Jules is right; I wasn't talking about the Levine proof, but rather the 3 step proof the blog author mentions after that.

Step 2 is division by zero.

I have a similar proof that doesn't quite take so long to explain.

1 + 1 = 1

One number plus one number becomes one number.

Is this a joke or are you saying something deep that I'm missing?

Yes. The singularity is near. Resistance is futile. We are the borg.

IMO

    1 = 0 in > time (t)

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