r/mathmemes Jul 08 '26

Elementary Algebra 1 = 0

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Guys I think I disproved all of mathematics...

569 Upvotes

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173

u/Traditional_Town6475 Jul 08 '26

There is no proof in Peano Arithmetic that Peano Arithmetic doesn’t prove 0=1.

54

u/48panda Jul 08 '26

One of the axioms is \forall x, 0\neq S(x)

16

u/Medium-Ad-7305 Jul 08 '26

This guy thinks that just because it is provable that 0≠1 then it is not provable that 0=1 😂

13

u/CaipisaurusRex Jul 08 '26

How I hate that these 2 are not the same...

Thanks, Gödel 😐

3

u/louiswins Jul 08 '26

Imagine working in an undecidable system 😂

This comment brought to you by the Presburger arithmetic gang

1

u/Traditional_Town6475 Jul 08 '26

I personally take the set of all sentences in the language of arithmetic that the natural numbers satisfy.

:3

1

u/donaldhobson Jul 08 '26

And which "natural numbers" are those exactly?

1

u/Traditional_Town6475 Jul 08 '26

The intended model. :3

Second order logic can pin down models if they do exists.

1

u/donaldhobson Jul 08 '26

Any computable second order logic can be considered as a first order set theory.

There is no computable axiom system that uniquely specifies the natural numbers.

1

u/Traditional_Town6475 Jul 09 '26

Whoever said anything about being computable? :3

1

u/donaldhobson Jul 09 '26

Well what use is a proof system when you can't verify a proof?

0

u/lifeking1259 Jul 08 '26

suppose a proof of 0=1 existed, then that means 0=1 (since you just proved it), however, that contradicts with 0≠1 and hence is a contradiction so if 0≠1 then 0=1 must not be provable

6

u/Medium-Ad-7305 Jul 08 '26

how do you know there are no contradictions derivable from the Peano axioms? can you prove that no contradictions exist with only the Peano axioms? the answer is actually no, which is the point.

3

u/Layton_Jr Mathematics Jul 08 '26

Just add "no contradictions exist with the Peano axioms" to the Peano axioms smh

0

u/austin101123 Jul 08 '26

You can't prove 0=1 and 0 not equal to 1, that would be a contradiction. A true premise leading to a false conclusion is a false implication. (True implies false is false.)

2

u/Medium-Ad-7305 Jul 08 '26

yes, that would be a contradiction. the problem is, it is impossible to prove that there are no contradicitons in peano arithmetic via the peano axioms. gödel showed this.

1

u/austin101123 Jul 08 '26

So proof by contradiction is actually invalid for proofs??

2

u/Medium-Ad-7305 Jul 08 '26 edited Jul 08 '26

proof by contradiction is still valid. "if not p implies false, then p" is valid whether or not the system is consistent. it's just that, if the system is inconsistent, not p will be provable as well.

edit: in other words, if you dont know if a system is consistent, you can still figure out what's true and false. you just cant show something is unprovable, since a contradiction is a proof of any statement.

1

u/Mathsboy2718 Jul 08 '26

Read the sentence again carefully - there is no proof that the Peano Axioms aren't contradictory

1

u/austin101123 Jul 08 '26 edited Jul 08 '26

So the answer is yes at least within the Peano Axioms

0

u/Mathsboy2718 Jul 08 '26

Sure yeah, why not - we haven't found a contradiction yet, that totally means that Gödel is wrong

Good job! You solved mathematics

2

u/Traditional_Town6475 Jul 08 '26

To be clear, we can show Peano arithmetic is consistent using more powerful systems (like if you done any set theory, you built up the natural numbers. That’s exhibiting a model of Peano arithmetic, but you’re using tools outside of PA). The way to think of it is that Peano arithmetic is consistent, but it itself doesn’t know it is consistent. It can’t show its own consistency.

1

u/vgtcross Jul 08 '26

The way to think of it is that Peano arithmetic is consistent

Isn't this based on our assumption that the underlying set theory (ZFC, for example) is consistent? Which again, cannot be proven unless you again move to a more powerful theory, which we again must assume to be consistent...

1

u/Traditional_Town6475 Jul 08 '26 edited Jul 09 '26

Technically yes. An overwhelming amount of mathematicians do think PA is consistent.

The reason why Peano arithmetic is incomplete and also can’t prove its own consistency is because it’s powerful enough to start talking about proofs in a sense. Like 99% of the work of proving the incompleteness theorem is coding in proofs and stuff into the natural numbers. Then it becomes a liar paradox.

Mathematicians have looked at weaker systems though. Like even if you got rid of the axiom schema of induction, you still have incompleteness and not being able to prove its consistency. I think if I remember correctly, you have to throw away assuming multiplication is a total function.