

Proof that 1 = 0 using a common logical fallacy - nathanmarz
http://nathanmarz.com/blog/common-logical-fallacy/

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lmkg
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.

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

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najirama
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 x _0=0" and not "x_ 0=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.

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scotty79
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.

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dbz
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.

~~~
roundsquare
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.

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

~~~
lmkg
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.

~~~
python123
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) = x_ 0 for
all integers x. By distributive law, x _0 + x_ 0 = x _0\. By closure under
multiplication, x_ 0 is an integer. By additive inverses, there exists an
integer (-x _0), such that x_ 0 + (-x _0) = 0. Because (-x_ 0) is an integer,
x _0 + x_ 0 + (-x _0) = x_ 0+(-x _0). By associative property of addition and
the transitive property of equality, x_ 0 + 0 = 0. By the additive identity,
x*0 = 0.

~~~
lmkg
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!

~~~
jules
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.

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dpapathanasiou
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.

~~~
bad_user
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)

~~~
jules
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".

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hkuo
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.

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

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

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known
IMO

    
    
        1 = 0 in > time (t)

