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Background

When I was learning Math on Khanacademy, the teacher gave a brief demonstaration about the difference between 'theorem' and 'postulate/axiom'. He indicated that 'postulate' and 'axiom' means the same. I assume so in this post.

My understanding is following:

  • 'theorem' is a true statement established by means of assumptions.
  • 'postulate/axiom' is an assumption cosidered to be true (generally with high probability).

Question

Postulates/axioms used in a proof might turn out to be false. Does this mean it's possible that a theorem which has been proved can be false if one or more postulates/axioms are in fact false?

By the way, the following question is related, but I guess it doesn't answer my question directly. Once a mathematical theorem is proven true like the Halting problem can it ever be disproven?

catwith
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  • In math, axioms are not assumed to be "true" in any absolute sense. The axioms of a vector space are not necessarily true in a group, for example. They are just statements that one uses as a basis from which to prove other statements. If you try to use the axioms of Linear Algebra to prove statements in Group Theory, say, you may well wind up proving statements that do not hold for groups. – Gerry Myerson Jan 11 '21 at 03:28
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    Axioms are used to define the system in which you are working. The axioms of vector spaces can't be false – they define what a vector space is. – Gerry Myerson Jan 11 '21 at 03:30
  • Axioms need not be considered true, in fact maths with axioms that we consider false is perfectly fine, so long as it's not inconsistent. The extra condition that axioms should be considered true is relevant to applying maths in the real world, but in applied maths they don't care about rigor in the first place so why bother with axioms? – SenZen Jan 11 '21 at 03:30
  • In arithmetic, one starting place is Peano's axioms which include that every number has a successor which is $1$ more than the previous number. If the axiom said only that each number has a successor that is $5$ more than the previous number, theorems based on them may still be true if no contradictions can be found. Of course there are consistent systems that contradict real world observations but the theorems can still be true in that they "prove" the relationships among the given "parts" including axioms. – poetasis Jan 11 '21 at 03:31
  • One major advance in math came from assuming that parallel lines meet. There were no contradictions and the author [I forget his name] once said something like, "I have created an entire universe out of nothing." From this we got spherical geometry/trigonometry like the kind use with latitude and longitude on the surface of the Earth. – poetasis Jan 11 '21 at 03:35
  • @GerryMyerson how do you distinguish a definition from an axiom? – SenZen Jan 11 '21 at 03:36
  • @Sen I'd say it should be possible to word a definition in the form "A means B", whereas that form should not make sense for an axiom. You wouldn't write "$a+b$ means $b+a$", but you would write "commutativity means $\forall a\forall b\ a+b=b+a$. – Gerry Myerson Jan 11 '21 at 03:53
  • In that case, the axioms of vector spaces aren't axioms at all? They are parts of the definition of vector spaces. – SenZen Jan 11 '21 at 03:59
  • @poet the quote "Out of nothing I have created a strange new universe" is attributed to János Bolyai. It had nothing to do with parallel lines meeting, nor with spherical geometry. It had to do with a plane geometry in which there are many lines parallel to a given line through a given point not on the line. – Gerry Myerson Jan 11 '21 at 04:00
  • @Sen the definition of a vector space consists of statements which we take to be the axioms in the theory of vector spaces. – Gerry Myerson Jan 11 '21 at 04:02
  • @GerryMyerson What's the difference between "statements which we take to be the axioms in the theory of vector spaces" and "requirements which we take to be the definition in the theory of vector spaces"? – SenZen Jan 11 '21 at 04:31
  • theorems are proved from axioms and previously proven theorems. We assume that the axioms of our theory are true of some piece of (mathematical) world and we work with them accordingly (see e.g. Euclidean geometry or Peano arithmetic). Thus, if the theorems are correctly derived by means of deductive inferences, they will express "true facts" regarding the piece of world that satisfies the axioms. – Mauro ALLEGRANZA Jan 11 '21 at 07:37
  • What happens if the axioms are "false"? Two cases: (i) axioms are contradictory; in this case there is no "world" where the axioms hold and thus the ensuing theorems are useless. (ii) axioms does not hold for a specific world (e.g. the parallel postulate (axiom) with regards to non-euclidean geometry). In this case the theory is still correct: we have only to be careful about its "applicability" (silly example: the Peano axiom stating that there is no number less than zero does not hold for integers or rationals; this does not mean that it is not true of the naturals). – Mauro ALLEGRANZA Jan 11 '21 at 07:40
  • @Sen there's a difference between "a Schmidlap is defined to be a set in which $x+x=x$ for all $x$" and "$x+x=x$ for all $x$." But maybe I don't understand your question. – Gerry Myerson Jan 11 '21 at 08:11
  • @GerryMyerson that's right, and that's exactly my point. You would say $x + x =x $ for all $x$ is a definition requirement for schmidlaps, you wouldn't say it is an axiom of schmidlaps. Same thing with vector spaces? – SenZen Jan 11 '21 at 10:16
  • A schmidlap is defined to be a set in which $x+x=x$ is the sole axiom. Of course it's an axiom. It's the defining axiom of Schmidlap Theory. – Gerry Myerson Jan 11 '21 at 11:54

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