There are many elementary proofs for the Pythagorean Theorem, but no matter they use areas, similarities, even algebraic proofs, it is not straightforward to tell why it is true tracing back to the (Euclidean geometry) axioms. Are all these proofs equivalent? Do they all track back to the same axioms?
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1Studious gives a nice answer here, for proofs that rely on the notion of area, http://math.stackexchange.com/questions/675522/whats-the-intuition-behind-pythagoras-theorem/676731#676731 . – Sawarnik Mar 31 '14 at 14:05
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@Sawarnik, thanks. Studious answer fills the gap between the area proofs and the underlying theorem. – ahala Mar 31 '14 at 14:16
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Sure, the Pythagorean theorem is an item in the theory of Euclidean geometry, and it can be derived from the modern axioms of Euclidean geometry.
A full set of Euclidean geometry axioms contains the information about similarity and area that are sufficient to prove the Pythagorean theorem "synthetically," that is, directly from the axioms. The algebraic proofs are a little different, though!
It turns out that after defining the real numbers and basic algebra, you can create a model of Euclidean geometry in $\Bbb R\times \Bbb R$ which obeys all the Euclidean axioms. The algebraic operations in an algebraic proof reflect the synthetic axioms being used, but the direct connections are not obvious. You are still indirectly using the synthetic axioms, but they are all hidden assumptions about $\Bbb R\times\Bbb R$ and coordinate geometry.
rschwieb
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Your Question is much more complicated than it looks.
First some philosophy:
Are all proof equivalent?
What do you mean by equivalent in relation to proofs? (not just in relation to this proof but to any proof at all)
Euclidean geometry:
Euclid was a bit lacks with his Postulates and Common Notions, The axioms of his geometry were only found in late 19 , beginning 20 century so it is reasonably fresh.
see for example: (there are many more and even within these examples there are different options)
- Hilbert's axioms http://en.wikipedia.org/wiki/Hilbert_axioms
- Tarski's axioms http://en.wikipedia.org/wiki/Tarski_axioms
PS: Make sure you use the axioms for Euclidean geometry, you need to add the parallel axiom or an axiom that (together with the other axioms) can proof it.
Theories (and Euclidean geometry is a theory) are defined by their theorems (everything that follows from the axioms and rules of inference) not by their axioms, so many different axiomatisations can give the same Theory. But for being the same theory they have to proof the same theorems.
Then the Pythagorean Theorem.
Yes you can proof the Pythagorean Theorem in any axiomatisation, if anything if you could not prove it it would not be Euclidean geometry.
Do they all track back to the same axioms?
No, there are different axiomatisations possible so they are not even able to track all back to on and the same set.
Hope it all helps
Willemien
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Equivalent = Relying the same set of axioms, usually. Therefore a proof that uses less axioms is "better", and a proof that uses different non-equivalent axioms is simply "different". – yo' Mar 31 '14 at 19:45
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@tohecz but then you get in a spin what are non-equivalent axioms, for example Tarski's axioms don't use angles does that make them 'better'? and you can string all axioms to one big axiom, (not very readable) again better? – Willemien Mar 31 '14 at 20:19
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@William if $A_1\wedge A_2\wedge\dots\wedge A_n \Leftrightarrow B$, then $B$ is equivalent to the system of axioms $A_1,\dots,A_n$, so I'm no quite sure what you speak to in the second part. And if $A\wedge B\Rightarrow T$ and $A\wedge C\Rightarrow T$ and $A\wedge B\not\Rightarrow C$ and $A\wedge C\not\Rightarrow T$, then you got two non-equivalent proofs of your theorem $T$. I would never "guess" which axioms are "better" in any other way than possibility to derive ones from the others. (But maybe it's just too late and I overlook some stupidity in my arguments.) – yo' Mar 31 '14 at 20:31
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If you skipped 10th grade plane geometry because you just loved algebra, your path would take you right to the complex numbers. You don't know any geometry but you like the idea of representing complex numbers
z = x + yi
as points/vectors in a two coordinate Real Number 'Plane', Of course you naturally put the pure imaginary numbers (y axis) 'orthogonal' to the pure real numbers (x axis).
As you play with your new toy, you begin to recognize geometric type things going on. You decide that no matter what, the 'vectors'
1 + 0i, 0 + 1i, -1 + 0i, 0 - 1i
must all be at a distance of 1 from 0, and you want to create an absolute value (length) for any complex number. You also want the length to be an extension of the absolute value function defined for (pure) real numbers.
If $z = 5i$ and $w = 3$, $zw = 15i$, so you think that, tentatively, you want to keep
$\lvert z w\rvert$ = $\lvert z\rvert \lvert w\rvert$
One day while fooling around you start to examine the conjugates of complex numbers. It then hits you! The distance traveled by going x steps horizontally and then y steps vertically is invariant under the sign of y.
So your 'killer' axiom is:
The absolute value of a complex number is equal to the absolute value of its conjugate:
$\lvert\bar z\rvert = \lvert z\rvert$
But now that is all you need, since
$z \bar z = x^2 + y^2 = \lvert z \bar z\rvert = \lvert z\rvert \lvert\bar z\rvert = \lvert z\rvert ^2$
You define the absolute value of z to be the square root $x^2 + y^2$, and wonder where that will take you.
Ans: The Euclidean 2-Dimensional Plane.
Exercise: Show that with this definition, the absolute value of a product is the product of the absolute values.
CopyPasteIt
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Pythagoras is equivalent to the parallel postulate.
Jason Zimba
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To improve your answer I suggest you explain what the parallel postulate is, as not everyone knows it. – Jori Mar 31 '14 at 14:17
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Ok not a bad idea - however I would not want to merely recapitulate the discussion here http://en.m.wikipedia.org/wiki/Parallel_postulate so I will merely link to it. – Jason Zimba Mar 31 '14 at 14:55
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2I think this statement is misleading. To prove the equivalence, it is required to first state the pythagorean theorem in neutral geometry, and this is not obvious (see https://math.stackexchange.com/questions/1292453/how-to-state-pythagorean-theorem-in-a-neutral-synthetic-geometry). – Julien Narboux Jun 28 '17 at 11:51
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All proofs I have seen so far trackback to the parallel axiom, or the axioms of similarity(which I think are equivalent). I'd be interested to see these "algebraic" proof though. Anything proved by pure algebra alone must be universally true, which is not so in the case of the Pythagorean theorem. It must use (euclidean) geometry in some form or the other.
Guy
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Hmm, "universally true": this is a dangerous and misleading concept :) – rschwieb Mar 31 '14 at 13:48
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Actually, I am the one who needs to as you to explain what "universally true" means. Otherwise I don't know how I would respond. – rschwieb Mar 31 '14 at 13:49
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@rschwieb okay. so I meant "universally true" to mean that algebra doesn't use a lot of axioms(apart from things like $a+b=b+a$, which can very well be seen as part of definition of
+) so it doesn't have any implicit "A is true, but only so long as B true". What has been shown to be true, is always true. I might be wrong about algebra's (non)axiomatic status though. – Guy Mar 31 '14 at 13:53 -
OK, that helps me get a better picture of what you're thinking :) It looks like you have a lot of ideas about truth which I probably am not very good at addressing. Here's my best shot: in mathematics, only concern yourself about truth within a set of axioms. Synthetic geometry has its own axioms, and algebra (usually via set theory) has its own axioms. Within the axioms of algebra, you can make a model of the axioms of Euclidean geometry, so you can do geometry in algebra and prove things there that the synthetic axioms prove. – rschwieb Mar 31 '14 at 14:03
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@rschwieb i was talking purely of algebra and not geometry. Could you mention one of these axioms? Also I don't understand you're statement about asking a lot of having ideas about truth. I understand that the question of whether something is "true" can be philosophically approached, but that wasn't my intention. :) In any case your answer seems more complete (+1). – Guy Mar 31 '14 at 14:10
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@rschwieb Algebra(usually via set theory) apparently I am mssing something obvious. – Guy Mar 31 '14 at 15:06
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Oh, OK, that's fine too! I meant that we use ZFC set theory axioms, and define the real numbers and vector spaces as sets with certain properties, etc. That is the axiomatic framework of coordinate algebra, right? – rschwieb Mar 31 '14 at 15:12
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@rschwieb Yes it is. But then again all science relies on this basic framework of axioms. In this case, I just define the truth value of something by whether it is experimentally verified/refuted. By universally true I meant that $(a+b)^2=a^2+2ab+b^2$ will always hold, without any axioms. (But I guess since the axioms are there, I am just willing to say they are known to be true and "basic") – Guy Mar 31 '14 at 15:17
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But $(a+b)^2=a^2+2ab+b^2$ does not hold in the quaternions for example! Nothing against science and experimentation, but defining truth via experimental verification is not consistent with modern mathematical thinking. The lion's share of what is done is deriving truth from the axioms at hand, not making many observations. Of course, experimentation as it is used in the physical sciences is perfectly fine: nobody has told us the axioms of the universe, so experimentation is the right way to go, there :) – rschwieb Mar 31 '14 at 15:27
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@rschwieb well my example was just for $\Bbb R$( I know it holds for $\mathbb C$ as well, but I was just talking about the reals.) And yes, experimenting and observing goes against the usual outlook of math, but I mostly view math as a tool for the physical sciences(just my opinion) but I see your point of course, and I think it mostly comes down to what I view mathematics as. – Guy Mar 31 '14 at 15:30
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I suppose that is the case, but I imagine you will change your mind over time :) I don't agree with "$(a+b)^2=a^2+2ab+b^2$ will always hold, without any axioms." That expression doesn't even have any meaning without the axioms of the real numbers. Distributivity and commutativity, not to mention what "equality" means are all established by axioms. At the end of the day you can point to the axioms and say "that expression holds for any field," but bringing philosophy, science and "universal truth" into it seems distracting. – rschwieb Mar 31 '14 at 15:42
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But this is all part of your intellectual growth: I think your inquisitiveness and willingness to think about these points will eventually lead to good understanding. Good luck! – rschwieb Mar 31 '14 at 15:44
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@rschwieb thanks. Just one final comment, "... "equality" means are all established by axioms" I look at that more on the levels of definitions. I guess maybe because it's simpler that way. "What does it mean for two things to be equal" seems like an unnecessary philosophical question, when the 'intention' behind
=is obvious, but then again, it could be that I see it that way because it is just do deeply ingrained into the way we do math that its near impossible to think anything violating those boundaries. But I think the discussion at this point is becoming.... – Guy Mar 31 '14 at 15:51 -
(continued comment)..... too philosophical, even though both of us understand each other. Good day cheers. – Guy Mar 31 '14 at 15:52
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too philosophical that's exactly what I thought when you brought up "universal truth" :) seems like an unnecessary philosophical question This is a fair observation (at least for now but maybe not later), but I hope you don't conclude that and then stop! Have you ever heard of equivalence relations? They study exactly what you described: when two things are equal. I don't think there is a single branch of mathematics where using multiple equalities doesn't happen all the time :) – rschwieb Mar 31 '14 at 16:20
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About "algebraic universally truth", see Field (mathematics) : commutativity of addition and multiplication is an axiom in the abstract algebraic definition of field. Thus, a field has this property "by definition". Of course, there are structures that do not satisfy it (and they are not field), because not all math binary operations are : subtraction and division between numbers are not commutative, matrix multiplication is not commutative. – Mauro ALLEGRANZA Apr 01 '14 at 08:54