Talk:Pythagorean theorem/Archive 2

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Proof using Ptolemy's theorem

Using the Ptolemy's theorem, the Pythagoras theorem can be proved in a much easier method.

Image:http://web7.twitpic.com/img/22498214-dabf85fe36ef0ea9c1e5274caa24c9f4.4a80338f-full.png | The Figure

As seen in the above figure, there is a right angled triangle with sides a, b and hypotenuse c. It is extended to for rectangle of side a and b.

Since rectangles are always cyclic quadrilaterals, thus we can apply the Ptolemy's theorem. According to Ptolemy's theorem, The product of the diagonals is equal to the sum of the products of the opposite sides.

In this case,

c*c = a*a + b*b (By Ptolemy's theorem)

Thus a² + b² = c²

Hence proved.

Midhul (talk) 14:58, 10 August 2009 (UTC)

Yes, the Pythagorean theorem follows from Ptolemy's theorem, because the latter is a generalization of the former. But of course Ptolemy's theorem also requires a proof. Ishboyfay (talk) 15:45, 10 August 2009 (UTC)

But perhaps it would be good to mention Ptolemy's theorem in this article's section on generalizations. I have no strong feeling about this, but am just throwing it out as a suggestion. Ishboyfay (talk) 15:51, 10 August 2009 (UTC)

Formulas and MathML

Hi, It seems that putting characters like "\," at the end of math equations prevents these from being rendered as MathML equations (this is useful for coherence of fonts, and for the possibility of copying and pasting an equation, for example), even if the user has selected the MathML rendering option. Most of the equations in this page have "\," at the end, but I do not see the reason for this. If it is not necessary, would you consider removing these (I'm OK to do this myself if the moderators allow me to) ? And if it is necessary, could someone explain me why please, as I'm new to Wikipedia editing ? Thank you very much for your work. Yanske (talk) 08:59, 17 September 2009 (UTC)

i still don't get this at all —Preceding unsigned comment added by 68.255.254.30 (talk) 20:48, 15 October 2009 (UTC)

Proof using the subtraction formulas

{{editsemiprotected}} Hello there,

A counterintuitive proof of the Pythagorean theorem has just appeared in Forum Geometricorum (Forum Geom. 9 (2009), pp. 275-278). See http://forumgeom.fau.edu/FG2009volume9/FG200925index.html.

The proof is counterintuitive because it uses trigonometry...and many people naturally assume that this can't be done without a vicious circle. So perhaps it would be worth adding a section on the Pythagorean theorem page, under Proofs, something like: "Proof using the subtraction formulas."

Best, Jason Zimba (who published the proof) Jzimba (talk) 23:29, 1 November 2009 (UTC)

Wouldn't this be better off here?   Set Sail For The Seven Seas  324° 43' 45" NET   21:38, 2 November 2009 (UTC)

Hullo Jason, and congratulations on the breakthrough. I wonder if we should wait until secondary sources have covered your proof, to avoid misintrepretation? Best,  Skomorokh, barbarian  14:48, 3 November 2009 (UTC)

Both observations make sense to me! Just wanted to put it on your radar screen. Best, Jason Jzimba (talk) 19:38, 9 November 2009 (UTC)

Proof by differential equations

In “Proof by differential equations”: Why is da/dc = c/a? —Preceding unsigned comment added by 85.225.242.42 (talk) 05:04, 10 December 2009 (UTC)

Hyperbolic case

I can't make sense of the following:

In hyperbolic geometry, for a right triangle one can also write,

$\sin {\bar {a}}\sin {\bar {b}}=\sin {\bar {c}}$

where $\scriptstyle {\bar {a}}$ is the angle of parallelism of the line segment AB that $\scriptstyle \mu (AB)\,=\,a$ where μ is the multiplicative distance function (see Hilbert's arithmetic of ends).

• The sentence doesn't quite parse.
• Is $\scriptstyle {\bar {a}}$ really related to the segment AB, rather than to BC? That is weird notation.
• Why are we told about $\mu (AB)\,=\,a$ even though a doesn't even show up in the formula?
• In the sequel, the letters a, b and c are used. Are they still the ordinary side lengths of the right triangle, as in the rest of the article?

AxelBoldt (talk) 01:07, 16 December 2009 (UTC)

I just cut out the following material from the article:

In hyperbolic geometry, for a right triangle one can also write,

$\sin {\bar {a}}\sin {\bar {b}}=\sin {\bar {c}}$

where $\scriptstyle {\bar {a}}$ is the angle of parallelism of the line segment AB that $\scriptstyle \mu (AB)\,=\,a$ where μ is the multiplicative distance function (see Hilbert's arithmetic of ends).

In hyperbolic trigonometry, the sine of the angle of parallelism satisfies

$\sin {\bar {a}}={\frac {2a}{1+a^{2}}}.$

Thus, the equation takes the form

${\frac {2a}{1+a^{2}}}{\frac {2b}{1+b^{2}}}={\frac {2c}{1+c^{2}}}$

where a, b, and c are multiplicative distances of the sides of the right triangle (Hartshorne, 2000).

My concerns are: the material is not clearly presented (see above), it likely uses the letters a, b and c in a different sense from the rest of the article, and in any event it is not an equation about the side lenghts of a right triangle, so it isn't really relevant to the Pythagorean theorem. AxelBoldt (talk) 00:58, 17 December 2009 (UTC)

Lp space

I added the following:

In a hypothetical L^P space the Pythagorean theorem would have the form

$a^{P}+b^{P}=c^{P}\,$

P=2 is the only case in which the length of an object does not depend on the orientation of your coordinate system.

It was immediately deleted by oli filth with this comment: (Reverted 4 edits by Em3ryguy; Please provide a source for the idea that the Pythagorean theorem is a meaningful concept in an Lp space)

AFAIK, its the very definition of an lp space. just-emery (talk) 21:21, 18 June 2009 (UTC)

You're right, c is the length of a vector defined by components a and b in L^P space. But show me where this relationship is described as the Pythagorean theorem! Oli Filth^{(talk|contribs)} 21:40, 18 June 2009 (UTC)

It doesnt have to be called the Pythagorean theorem. It is a GENERALIZATION of the Pythagorean theorem. just-emery (talk) 00:07, 19 June 2009 (UTC)

Equally, show me a source where someone describes it as such... Oli Filth^{(talk|contribs)} 11:22, 19 June 2009 (UTC)

Agreed. I would think we would need a reliable source to avoid original research. Plastikspork (talk) 15:41, 19 June 2009 (UTC)

The definition is just that the sum (or integral) of the pth powers of the absolute value is finite. It's not defined by such an identity as the one above. Michael Hardy (talk) 00:29, 19 June 2009 (UTC)

this is the equation from the lp space article itself:

$\ \|x\|_{p}=\left(|x_{1}|^{p}+|x_{2}|^{p}+\cdots +|x_{n}|^{p}\right)^{1/p}$

just-emery (talk) 08:51, 20 June 2009 (UTC)

This is not an equation, this is not a theorem, this is a definition of the Lp norm. (Igny (talk) 18:35, 20 June 2009 (UTC))

The definition of ℓ^p says that

$|x_{1}|^{p}+|x_{2}|^{p}+|x_{3}|^{p}+\cdots <\infty .\,$

You can't leave out the part that says "< ∞". And if it only goes up to n terms rather than going on forever, then the part about "< ∞" is ALWAYS true, so there's no point in having such a definition. But what was asked about was L^p, not ℓ^p. So we're looking at

$\int _{\mathbb {R} }\left|f(x)\right|^{p}\,dx<\infty ,$

and again, the part that says "< ∞" is the essence. Michael Hardy (talk) 20:17, 20 June 2009 (UTC)

Verbatim quote from the L^p space article:

The space ℓ^p is then defined as the set of all infinite sequences of real (or complex) numbers such that the p-norm is finite.

Michael Hardy (talk) 20:24, 20 June 2009 (UTC)

Thank you so much for patiently explaining that to me. Wikipedia is so lucky to have such knowledgable and unbiased editors such as yourself and oli filth who are voluntarily working so energetically to improve its articles. just-emery (talk) 22:24, 20 June 2009 (UTC)

Moreover I didnt say anything about ℓ^p space. My comment was about L^P space. ℓ^p space has an infinite number of dimensions. L^P space does not. just-emery (talk) 21:42, 24 June 2009 (UTC)

L^p is also infinite-dimensional, except when you're dealing with a measure space with only finitely many points. Michael Hardy (talk) 22:16, 24 June 2009 (UTC)

Enough! Everyone is making this sooooo complicated "a" squared + "b" squared = "c" squared. Thats it! Thats all there is to it! If you are missing a side and eqations would look like this (lets say you are missing side "a"):

EX: "a" squared + "b" squared = "c" squared

   "a" squared + 24 squared = 25 squared
   "a" squared + 567 = 625
               - 567 - 567
   "a" squared = 49
   "a" = square root of 49

"a"= 7 Bold text —Preceding unsigned comment added by 67.251.76.91 (talk) 23:05, 7 January 2010 (UTC)

Scalene triangle image

Generalization for scalene triangles,
green area = blue area

How does this image, copied from the page with its caption, illustrate Pythagoras's theorem? and the shapes aren't similar. The green and red areas may be equal but there's no indication why, or how it relates to the article.--JohnBlackburne^words_deeds 12:20, 25 April 2010 (UTC)

John: Pythagoras' theorem states that for a right triangle the square on the hypotenuse is the sum of the squares on the other two sides.

The construction in the figure shows that for a scalene triangle, the area of the parallelogram on the longest side is the sum of the areas of the parallelograms on the other two sides, if the parallelogram on the long side is constructed as indicated. This replacement of squares with parallelograms bears a clear resemblance to the original Pythagoras' theorem, and was considered a generalization by Pappus of Alexandria in 4 A.D.

The diagram indicates how the construction is done by indicating with the double arrowheads the dimensions that are equal and similarly oriented. The proof that the construction is valid can be found in the linked source, and depends upon parallelograms of the same base and height having the same area.

The pre-existing diagram above using triangles is neither explained nor referenced, so maybe further discussion is needed there as well? Brews ohare (talk) 13:35, 25 April 2010 (UTC)

Proof by differential equations

Construction for proof using infinitesimal increments da and dc in sides a and c. Notice that although c > a , da > dc .

The section Proof by differential equations was unclear to me, so I revised it using a new figure. The second from last paragraph in this section is not clear to me. Brews ohare (talk) 00:11, 30 April 2010 (UTC)

I rewrote this paragraph and dropped the reference to a similarity with line integrals, which was too obscure to convey anything to me. Brews ohare (talk) 14:36, 30 April 2010 (UTC)

I haven't read it carefully yet, but I certainly don't like the inclusion of the cosine of the angle; this is supposed to be very elementary geometry not needing trigonometric functions. I also have some issues with the new picture. More later.... Michael Hardy (talk) 18:40, 30 April 2010 (UTC)

The cosine is not actually used; it's purpose here is to provide reinforcement that the two ratios are the same thing, namely the cosine. Of course, similarity of the triangles is sufficient, but addition of this fact might make it more transparent to some readers.

I'd be interested to know what your comments about the figure might be. Brews ohare (talk) 03:46, 4 May 2010 (UTC)

Just my opinion, but I don't think the cosine is worth mentioning, and its purpose isn't explained except for the phrase "for those readers who might prefer trigonometry to a geometric argument" and the trigonometric argument isn't explained. --Bob K31416 (talk) 05:53, 4 May 2010 (UTC)

Also, there seems to be a sticky part that was passed over. The similar triangle that supposedly has sides da and dc isn't clear. The similar triangle has side dc and a side that is constructed by drawing a line tangent to the circle at the intersection of dc and the circle, to the line segment da. Such a triangle would be similar to the triangle of sides a+da, b, c+dc, but one of its sides wouldn't be da but something smaller. --Bob K31416 (talk) 06:24, 4 May 2010 (UTC)

P.S. Is there a source for any of this? --Bob K31416 (talk) 06:43, 4 May 2010 (UTC)

Bob K31416: Regarding clarity of similar triangles: did you read the footnote that explains there is no difference between the arc length and the chord when differentials are involved? What it means is that the infinitesimal triangle has the same side lengths and same angles as the figure with an arc as one side. What do you think about that? §6 here provides a more elaborate discussion.

The only source I have come up with is an on-line reference referred to Michael Hardy in Mathematical Intelligence in 1988. This source is mentioned elsewhere as well. However, I have been unable to retrieve that source, as this magazine does not archive that far back. Another reference provided is Mike Staring in Mathematics Magazine Feb. 1996. I didn't chase this down. This source discusses them both briefly: "Proof #40". Brews ohare (talk) 14:19, 4 May 2010 (UTC) I was able to find Staring's article using U of A library search, and added it as a source. Brews ohare (talk) 14:50, 4 May 2010 (UTC)

I like the change you made re cosine. More concise and more informative.

But unfortunately, there remains the problem with the proof that I identified above. Looking at the above link that you provided, "Proof #40"(btw thanks), the type of problem mentioned with Hardy's proof is similar (but different) to the type of problem that I have with the proof in Pythagorean_theorem#Proof_using_differentials. From your response, I apparently wasn't able to communicate to you the problem I had with that proof regarding the side of the triangle being less than da. Maybe you might see what I was trying to get at by looking at the proof of Michael Staring which mentions some line segments being slightly less than others. Staring's proof seems fine, except possibly for the part "Passing to the limit as Δx tends to 0+" which may need more explanation since it is referring to an inequality in the previous line instead of an equality. --Bob K31416 (talk) 15:46, 4 May 2010 (UTC)

Bob K31416 Hi Bob: Perhaps the construction of the figure is not clear? Describing the triangle made up of infinitesimals, the side da is exactly of length da by construction. Also the side dc is exactly the length dc by construction. The contained angle is exactly θ, again by construction. The right angle is exactly a right angle, created by construction.

The only possible issue with this "triangle" is that the side opposite θ is not a straight line, but is an arc of a circle described with radius c centered on the lower left corner of the original right triangle. Because this side is an arc, and not a straight line, the figure made up of differentials is not exactly a triangle when the sides are not differentials. However, as da becomes a differential, the length of this curved side actually is the same as the length of the chord joining its ends, so there is no problem: the curve can be replaced by its chord, and the differential figure becomes a triangle. Brews ohare (talk) 16:02, 4 May 2010 (UTC)

I still don't think you understood my point that the side of the triangle is less than da, since you haven't yet specifically mentioned that point AFAICT.

Bob K31416: This is a semantic issue, it seems. The side da in my construction must be of length da as it is part of a rectangle with top and bottom sides both exactly da. You are insisting upon a different construction based upon erecting a perpendicular to the line c + dc and letting it intercept the line da. I am not taking that route, but leaving the opposite side of the differential figure as the arc of a circle of radius c. On the other hand, Staring uses your approach and supplements it with the second choice of choosing the corner of the rectangle and dropping a perpendicular to c + dc. Thus, the curved side I am referring to is pinned between two right triangles that become the same triangle as the change Δa → da. Brews ohare (talk) 16:38, 4 May 2010 (UTC)

May I ask you what you thought about the comments made at "Proof #40" about the problems with Hardy's proof, e.g. "It is easy to take an issue with this proof. What does it mean for a triangle to be approximately right?" I felt that what you are claiming is in the same style as what Hardy was claiming, viz. everything works out somehow because they are taken in the limit as things go to zero. It reminds me of Suspension of disbelief, but not that extreme. --Bob K31416 (talk) 16:10, 4 May 2010 (UTC)

Bob K31416: There is actually no difference in concept between any of these proofs. The second-hand version of Hardy's proof (who knows what the original was like) is a bit elliptic, and you have to connect some dots. Staring's version begins with differences Δa and Δc and explicitly describes the limiting procedure. The version I've supplied starts with differentials, and suggests to those disturbed by the equating of an arc to a chord that sin (Δθ) → Δθ as Δθ becomes small. There is no lack of rigor involved: it is just a different way to reach the differential limit. Brews ohare (talk) 16:38, 4 May 2010 (UTC)

Well, I don't see any progress towards a meeting of the minds here so I'll just pull out. Thanks anyhow for the discussion. --Bob K31416 (talk) 16:45, 4 May 2010 (UTC)

Bob: Sorry to see you leave. I changed the note in the article in the hope of reassuring you. Brews ohare (talk) 17:24, 4 May 2010 (UTC)

I'll look at this again soon. Michael Hardy (talk) 17:18, 4 May 2010 (UTC)

Recent addition including figure

Hi Brews. I can see you put in some good effort. When I see a proof like that which uses words in some places instead of math expressions, from my view it is OK as long as I can see that it can be done with math expressions instead of words if need be. For a mathematically rigorous proof, I think that everything has to be written with Δ's, then get a semi-final expression in terms of the Δ's, and then take the limit as the Δ's → 0.

For example, the angles φ in the isosceles triangle that go to π/2 (φ → π/2) should be expressed in terms of Δa and used to get the semi-final expression, for the proof to be rigorous. Sometimes it's not enough that a quantity goes to a certain value, but how it goes to that value in terms of the primary quantity that is going to its value (i.e. Δa → 0) is important. I think this was the kind of thing that the commentator "Proof #40" had in mind when he criticized Hardy's "approximately right angle" part of the proof, although he didn't mention it. So, instead of saying that φ → π/2 as Δa → 0, I think the semi-final expression should include φ (or sin φ, etc) in terms of Δa for the proof to be rigorous. --Bob K31416 (talk) 22:06, 5 May 2010 (UTC)

Of course, you are right. But will anyone stand for that kind of detail? I think it is clear that it will work, so the details are simply that. For example, it is perfectly clear that as the base of the isosceles triangle goes to zero the base angle must go to 90 degrees, eh? Where these things can hang up is with ratios of Δ quantities that go to forms like 0/0 and need care to evaluate. None of that happens here.Brews ohare (talk) 05:12, 6 May 2010 (UTC)

Actually, I would like you to try to do it in private and let me know here how you do. Perhaps you might then see the problems with the proof.

Here's something to consider that might get you started. If the Δ's were used, instead of the first equation in the section there would be the relation,

${\frac {a+\Delta a}{c+\Delta c}}>{\frac {\Delta c}{\Delta a}}\ .$

The inequality arises from the Δa in the denominator on the right being too large.

I haven't looked at the full article by Staring, but you might get some help with the above by looking it: Mike Staring (Mathematics Magazine, V. 69, n. 1 (Feb., 1996), 45-46). There seems to be similarities with the above inequality in the article. (I tried to find it online but it wasn't at their website. Regards, --Bob K31416 (talk) 05:29, 6 May 2010 (UTC)

Hi Bob: The objective is to show that in the limit as Δa→0, the figure with sides Δa, Δc, and the curved side cΔθ becomes a right triangle. First, Δθ→0 as Δa→0. Inasmuch as Δa is the originating cause of Δθ, that seems incontrovertible to me. It remains to show the obtuse angle becomes a right angle. It is the supplement of the angle φ, so it is sufficient to show the angle φ becomes a right angle as Δa→0. Applying the theorem on triangles that the sum of interior angles is π to half of the isosceles triangle, φ + Δθ/2 = π/2. Hence, as Δθ→0, φ→π/2. Already it has been show that the chord length becomes the same as the arc length. Thus, in the limit as Δθ→0, the curved side can be replaced by the chord, and the angle made with the side dc is π/2. In other words, the figure made up by Δa, Δc, and the curved side cΔθ becomes a right triangle as Δθ→0. QED How's that? Brews ohare (talk) 13:32, 6 May 2010 (UTC)

BTW, as noted in the footnote, one can bracket the curve with two right triangles and proceed to show that the two triangles become one as Δa→0, forcing the curve to become coincident with the approaching straight sides of the bracketing triangles. That is what you are attempting with the inequalities you mention. That is a different construction of the argument than the one based upon the isosceles triangle. Maybe you can focus on the suggested argument for the moment? Brews ohare (talk) 13:55, 6 May 2010 (UTC)

It doesn't work for me. I need mathematical expressions and operations on those expressions, instead of words, to make a proof. Anyhow, thanks for your explanations. --Bob K31416 (talk) 16:59, 6 May 2010 (UTC)

Continued discussion of limits

This discussion continues that above regarding the aside here.

Hi Bob: Is it that Δθ is used instead of Δa that is bothersome? Because the limits in terms of Δθ are expressed mathematically. Brews ohare (talk) 17:58, 6 May 2010 (UTC)

My view of the advantage of the isosceles triangle approach is that the limits in terms of Δθ are easier to understand than when using Δa. So long as we agree that Δθ = f(Δa) and that f(Δa) → 0 as Δa → 0, there is no need to know exactly what function f  is. Brews ohare (talk) 18:05, 6 May 2010 (UTC)

I guess the basic problem I have with it is that an expression should be obtained before any limits are taken and then the limit should be taken for that expression. What this means is that the expression should first be obtained without first taking a limit where supposedly both base angles of the isosceles triangle become right angles. --Bob K31416 (talk) 15:50, 7 May 2010 (UTC)

Hi Bob: I've reworded things to fit better your criteria. Before the limit is taken, the chord is of length $2c\sin(\Delta \theta /2)\,$. The arc is of length $c\ \Delta \theta \,$. In the limit, both become $c\ d\theta \,$. Before the limit is taken, the corner angle is $\pi /2+\Delta \theta /2=\pi /2\left(1+{\frac {\Delta \theta }{\pi }}\right)\,$. In the limit this corner angle is $\pi /2\,$. Does that meet your requirements? If not, perhaps you could define what the "expressions" needed actually must express? Brews ohare (talk) 16:11, 7 May 2010 (UTC)

BTW, it has just occurred to me that you are discussing the overall construction of the section, and not the aside establishing the right triangle. Could that be the case? Brews ohare (talk)

Yes. : ) --Bob K31416 (talk) 17:26, 7 May 2010 (UTC)

OK; that's a bigger issue. I thought the article was OK if one accepted that the differential triangle was in fact a right triangle. The present structure of the article is then simple geometry. Assuming the right triangle has been established by the aside, do you see more than geometry as needed? Brews ohare (talk) 19:49, 7 May 2010 (UTC)

BTW, both the sourced proofs are based upon similarity of triangles. Brews ohare (talk) 19:51, 7 May 2010 (UTC)

Image simplifications in section Proof using similar triangles

Proof using similar triangles

Re: the image "Proof using similar triangles" at right:

Suggest removing the arrows for a, b, d, and e. And removing the partition line between arrows for d and e. --Bob K31416 (talk) 21:12, 4 May 2010 (UTC)

Introduction of variables d and e avoids the use of labels AH and HB in the equations of the proof; making it all algebraic; personally I like that. The arrows for a b and c can be dropped at the expense of some statements like "Let the length of AB = c and that of CA = b and that of BC = a. However, the reader then must refer back to both this lexicon and to the figure to understand things, which seems to me more complicated. It's all a matter of taste, I'd guess. Brews ohare (talk) 21:55, 4 May 2010 (UTC)

No problem. I appreciate the effort you made in making the image. --Bob K31416 (talk) 22:07, 4 May 2010 (UTC)

Here's a revised version: is it better? Brews ohare (talk) 22:21, 4 May 2010 (UTC)

Yes. Thanks. I would prefer the d and e above the line, but like you said, it may be a matter taste. --Bob K31416 (talk) 22:24, 4 May 2010 (UTC)

Here we go. Thanks for the suggestions. Brews ohare (talk) 22:36, 4 May 2010 (UTC)

You've been so nice that I fear being a pill, but I would suggest not having any whitespace in the triangles and not having the line dashed and having it black and of the same weight as the others. --Bob K31416 (talk) 22:42, 4 May 2010 (UTC)

Well, here it is again; pretty much as requested. Brews ohare (talk) 23:03, 4 May 2010 (UTC)

Whoops. I didn't realize that getting rid of the whitespace would also get rid of the arc that indicates an angle. I thought that a line in the shape of an arc would still be at each of the two vertices with θ. Would it be possible to have those arcs put in without the whitespace? --Bob K31416 (talk) 00:37, 5 May 2010 (UTC)

Hi Bob: Got it. Brews ohare (talk) 05:00, 5 May 2010 (UTC)

It looks really nice. I think the usual convention is to have the θ outside the arc, rather than inside, but I'm not sure. And with it outside, a smaller arc would do. See for example Angle. In any case it's a fine image. --Bob K31416 (talk) 13:29, 5 May 2010 (UTC)

Center one-line figure captions?

I centered the one-line figure caption in the section Pythagorean_theorem#Proof_using_similar_triangles and then I realized that there were others in the article that this could be done with. So I thought I should check here first. Note that the suggested centering would be only for one-line captions. Feel free to revert the one that I have done. --Bob K31416 (talk) 14:16, 5 May 2010 (UTC)

Hi Bob: The centered caption looks better than the left adjusted one. Brews ohare (talk) 16:22, 7 May 2010 (UTC) A possible problem is the centering of figures that are not default size (which I seem to recall is about 220px). If a reader uses their own style settings, or if the figure is re-sized along the way, maybe the single-line caption will become a two-line caption. Brews ohare (talk) 16:34, 7 May 2010 (UTC)

Merge discussion

I have suggested the the article Pythagorean addition be merged into Pythagorean theorem. Please discuss the merger here. —Preceding unsigned comment added by Deathsythe (talk • contribs) 19:06, 7 May 2010 (UTC)

Converse

Can we put the converse statement on the exact words from Euclid's Elements, Proposition 48 of Book 1? I think it's a good only-words statement: If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right. 201.21.11.54 (talk) 20:47, 7 May 2010 (UTC)

I added this statement. Brews ohare (talk) 18:57, 12 May 2010 (UTC)

Even easier

I thought of this this morning 3a + 4b = 5c We can call it the Handyside Theory.

Handyside (talk) 21:07, 7 May 2010 (UTC)

Staring's proof

Staring's proof

Bob: Here is Staring's differential proof of Pythagoras theorem:

Triangle PBC is the original right triangle and triangle ABC is the modification of PBC when side PB is extended by increasing a to a + Δa. The arcs show curves of radius c and c + Δc , where Δc is the change in c that occurs as a result of the change Δa.

Construct right triangle ADP (upper panel). Triangles ADP and ABC are similar right triangles. Then

$\cos \theta ={\frac {a+\Delta a}{c+\Delta c}}={\frac {AD}{AP}}>{\frac {\Delta c}{\Delta a}}\,\ .$

The inequality results from inspecting the figure, which shows Δc < AD.

Next construct right triangle APQ (lower panel). Triangles APQ and PBC are similar right triangles. Then:

$\cos \varphi ={\frac {a}{c}}={\frac {PQ}{PA}}<{\frac {\Delta c}{\Delta a}}\,\ .$

The inequality results from inspecting the figure, which shows Δc > PQ. Hence,

${\frac {a}{c}}<{\frac {\Delta c}{\Delta a}}<{\frac {a+\Delta a}{c+\Delta c}}=\left({\frac {a}{c}}\right){\frac {1+\Delta a/a}{1+\Delta c/c}}\,$

We now have the ratio Δc/Δa pinned between two values. As Δa → 0, and Δc → 0, the ratio becomes the derivative dc/da, the left bound doesn't change, and the right bound becomes the same as the left bound. Consequently,

${\frac {dc}{da}}={\frac {a}{c}}\ .$

The rest of the proof is the same.

Bob, I think this is more to your taste. It has a drawback as presented, namely, it appeals to the figure to establish the inequalities instead of arguing for them geometrically.

Although Staring's proof is neat, it is has the drawback already mentioned of appealing to the figure, and IMO it is less transparent than the proof presented in the article because the added triangles are a clever, but unexpected, invention. Would you recommend switching this for the proof in the article? Brews ohare (talk) 15:10, 8 May 2010 (UTC)

This proof is fine for me. (But see P.S. below.)

Re "It has a drawback as presented, namely, it appeals to the figure to establish the inequalities instead of arguing for them geometrically." - Not sure what you mean by this. Does your remark have something to do with, for example, the construction of the triangle ADP? --Bob K31416 (talk) 15:39, 8 May 2010 (UTC)

P.S. I just noticed you modified your original message. My above comments were referring to your original message. I'll look at your changes to see if I have to change anything I wrote above. Cheers, --Bob K31416 (talk) 15:49, 8 May 2010 (UTC)

From the figure, it is evident that point D lies inside the circle of radius c. However, a rigorous proof would establish that fact by arguing from theorems that (for example) for point D to lie on or outside the circle angle ADP would have to be obtuse. Likewise, for point Q to lie on or outside the circle of radius c + Δc. I'm not clear just how this would be accomplished. Brews ohare (talk) 16:20, 8 May 2010 (UTC)

Actually it's simpler than that. CD < c since CD is a leg of the right triangle whose hypotenuse is c . --Bob K31416 (talk) 16:44, 8 May 2010 (UTC)

(edit conflict)After reading your message with its changes:

Re " Would you recommend switching this for the proof in the article?" - Yes. We'll tweak it but it is basically sound.

To start the tweaking, suggest changing ${\frac {AD}{AP}}\ \ to\ \ {\frac {AD}{\Delta a}}\,\ \ \ and\ \ {\frac {PQ}{PA}}\ \ to\ \ {\frac {PQ}{\Delta a}}\,\ .$

--Bob K31416 (talk) 16:32, 8 May 2010 (UTC)

I've made the switch, with a note covering my reservations. Brews ohare (talk) 16:35, 8 May 2010 (UTC) I don't agree with your suggestion to change the ratio of sides, as that explains how the similar triangles work. An additional step like yours could be inserted to show the length of the side. Brews ohare (talk) 16:37, 8 May 2010 (UTC)

Not sure I understand your comment since AP, PA, and $\Delta a$ are different names for the same line segment. --Bob K31416 (talk) 16:52, 8 May 2010 (UTC)

Hi Bob: I added this step. The point for me is that using AP identifies a side in the diagram enabling the reader to see what triangle is in use, while $\Delta a$ is length of this side, but also is the dimension of a number of other elements in the diagram, with various locations. Brews ohare (talk) 16:56, 8 May 2010 (UTC)

In the diagrams, only one line segment is identified as $\Delta a$ so there's no ambiguity. I would prefer to use $\Delta a$ in the inequality expression because it is more clear that the inequality is due solely to the numerator. --Bob K31416 (talk) 17:03, 8 May 2010 (UTC)

Bob: I got that. I put both versions into the article, if you take a look. Brews ohare (talk) 17:08, 8 May 2010 (UTC)

I just looked, and what's there is clear, as far as this $\Delta a$ aspect. Thanks. --Bob K31416 (talk) 17:14, 8 May 2010 (UTC)

Hi Bob: OK. Thanks for pushing me on this one. I had to cross a threshold to make a diagram and sort through Staring's proof, because I already had set up the alternative. However, Staring's proof leads to a shorter article, and it is sourced. Brews ohare (talk) 17:20, 8 May 2010 (UTC)

Is Staring's proof circular?

BTW, the caveat regarding proof that the points D and Q lie within their respective circles probably cannot be established based upon the right triangles becoming obtuse, as this argument is based upon Pythagoras' theorem itself. Thus, this hole in the proof may be substantial. Brews ohare (talk) 17:48, 8 May 2010 (UTC)

In all the back and forth, you might have missed might little comment above of 16:44, 8 May 2010? Regards, --Bob K31416 (talk) 21:06, 8 May 2010 (UTC)

Hi Bob: The comment you refer to says:

Actually it's simpler than that. CD < c since CD is a leg of the right triangle whose hypotenuse is c . --Bob K31416 (talk) 16:44, 8 May 2010 (UTC)

Is this comment germane? What we need to know is that Δc < AD, which is the same thing as knowing that point D lies inside the circle of radius c, and not outside. Brews ohare (talk) 21:13, 8 May 2010 (UTC)

It seems germane to the above comment of yours, "...the caveat regarding proof that the points D and Q lie within their respective circles..." . Doesn't it show that D lies within its circle? --Bob K31416 (talk) 21:23, 8 May 2010 (UTC)

Of course it does. Sorry. Brews ohare (talk) 21:42, 8 May 2010 (UTC)

No problem. I'm happy to be making good progress together. --Bob K31416 (talk) 21:48, 8 May 2010 (UTC)

Hi Bob: I've rewritten the footnote 11 to reflect your suggestion. At this point I'm persuaded this proof is OK. Do you have other suggestions? Brews ohare (talk) 21:53, 8 May 2010 (UTC)

Suggest deleting the line, "The inequality results from inspecting the figure, which shows Δc < AD."

Regards, --Bob K31416 (talk) 22:03, 8 May 2010 (UTC)

What's the reasoning, Bob? To me, for the reader wondering where this inequality comes from, this sentence guides the reader to the diagram and suggests they compare Δc with length AD. Do you feel that is redundant? Brews ohare (talk) 22:12, 8 May 2010 (UTC)

Seems like nearly all the expressions come from inspecting the figure, so I don't think it needs to be mentioned. There may be a lot of fine points that come up, but I think we are trying to summarize the Staring article. The reference is given for a reader to look it up if they want to, but I think readers will just go along with the proof since D being inside the circle just seems right. If one tries to picture it in one's mind outside the circle, things look askew, so I doubt if the vast majority of readers will consider that possibility. --Bob K31416 (talk) 22:25, 8 May 2010 (UTC)

No doubt I'm in a statistical minority, but due to my own confusion on the point, I'd choose to leave it as is. Brews ohare (talk) 22:27, 8 May 2010 (UTC)

Sometimes it helps if something like this is put aside for awhile and looked at again in a week or so. For now, I wouldn't contest you on this. --Bob K31416 (talk) 22:34, 8 May 2010 (UTC)

Regarding the comment:

The reference is given for a reader to look it up if they want to

Which reference is in mind? Everyone doesn't have quick access to articles on JSTOR. - Ac44ck (talk) 17:39, 9 May 2010 (UTC)

That's a good point. I'm leaning towards Brews Ohare's position. Not to argue this point, but FYI in case you didn't notice, here's the second part of the citation:

An abbreviated version of the proof in this article is in the second half of Proof #40 at Bogomolny, Alexander. "Pythagorean Theorem". Interactive Mathematics Miscellany and Puzzles. Alexander Bogomolny. Retrieved 2010-05-09. {{cite web}}: External link in |work= (help) Archived version May 8, 2010.

Regards, --Bob K31416 (talk) 18:59, 9 May 2010 (UTC)

Bob, nice job on reorganizing the intro to this section. Clear and concise. Brews ohare (talk) 19:02, 9 May 2010 (UTC)

Thanks. Just following up on what you started. --Bob K31416 (talk) 19:10, 9 May 2010 (UTC)

Status

Most of the subsections have been modified to improve clarity. Some odd statements have been removed. Some proofs have been fixed up. Some figures have been added or improved. However, judging by the mess that existed, a careful combing through all the sections still is needed. Brews ohare (talk) 15:58, 9 May 2010 (UTC)

Simple Euclidean proofs involving area

In the this subsection an effort is made to introduce some paradoxes of set theory. Aside from questions about the pertinence of this diversion, the statement is made that "simple Euclidean proofs involving area; for instance, deriving the area of a right triangle [based upon] the assumption that it is half the area of a rectangle with the same height and base" are fundamentally suspect. Which appears to suggest opening the topic of validity of the Euclidean formulation of geometry (and presumably of non-Euclidean geometries as well). If that is to be a topic, it should be properly discussed and not simply thrown up as dust in the reader's eyes. The further statement is made "For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles". As Euclidean geometry is normally considered an "axiomatic formulation", this sentence seems to suggest some (unspecified) axiomatic formulations exist that are better. I'd suggest that if “mathematization” of this article is to be attempted, it should be done elsewhere and linked as a "See also" or, at least, done with some kind of clarity. Dropping in a link to the Banach–Tarski paradox is insufficient. Brews ohare (talk) 17:22, 9 May 2010 (UTC)

I have removed this paragraph in anticipation of a more useful successor. Brews ohare (talk) 17:22, 9 May 2010 (UTC)

This topic seems to fit more naturally in Euclidean geometry, and I put a form of this assertion as a note in that article. Brews ohare (talk) 14:36, 10 May 2010 (UTC)

I added a note on Lebesgue measure here. Brews ohare (talk) 16:21, 10 May 2010 (UTC)

Last paragraph of Proof using differentials

Regarding the last paragraph of Proof using differentials:

"As can be seen, the squares are due to the particular proportions between the changes and the sides: the change in a side decreases inversely with the length of the side so c dc = a da . The same approach made by letting b increase while a is held fixed would show b db = c dc . More generally, if both a and b change by da and db respectively, the total change in c is c dc = a da + b db . The sum of squares therefore is an expression of the independent contributions of the changes in the sides a and b, a point not evident from the geometric proofs."

This paragraph is unclear to me. For instance, the conclusion at the end needs more explaining, "The sum of squares therefore is an expression of the independent contributions of the changes in the sides a and b...". Is there a source for it? Thanks. --Bob K31416 (talk) 00:28, 11 May 2010 (UTC)

Bob, this is my attempt to make sense of a paragraph I couldn't follow. The basic idea is easier to follow if we start with the total differential:

$d(c^{2})=d(a^{2})+d(b^{2})\ .$

This result shows that the increase in the square of the hypotenuse is the sum of the independent contributions from the squares of the sides.

Next, one asks: Why the squares? Why not:

$dc=da+db\ ?$

The answer is that the increases are inverse to the lengths of the sides, so looking at the case when only a changes, it is c dc = a da, not dc = da.

See what you can do to make this clearer. Brews ohare (talk) 17:05, 11 May 2010 (UTC)

I noticed you made a change, perhaps because you thought I was aquiescing. I'm not. I'm still thinking about the above since you posted it about an hour ago. No deadlines here. : ) Regards, --Bob K31416 (talk) 18:20, 11 May 2010 (UTC)

Bob: I rewrote it because I thought you felt it was not clear. That may still be the case. However, IMO there is nothing debatable about its content. Brews ohare (talk) 20:15, 11 May 2010 (UTC)

Re:

"Next, one asks: Why the squares? Why not:

$dc=da+db\ ?$"

Isn't the simple answer:     because   $c\neq a+b\quad for\ a,b>0$ ?     --Bob K31416 (talk) 03:47, 13 May 2010 (UTC)

We don't need a simple answer here to the wrong situation. We're looking for a correct answer for why a large c leads to a small dc while a small a leads to a large da so that c dc = a da. Maybe there is no intuitive way to arrive at this result qualitatively - you just have to depend entirely upon the math. Maybe the last sentence should be dumped. What do you think? Brews ohare (talk) 04:55, 13 May 2010 (UTC)

I just looked at what you deleted in the article. That's better.

Re "This result shows that the increase in the square of the hypotenuse is the sum of the independent contributions from the squares of the sides, a point not evident in the geometrical proofs." - I suspect that the last part, "a point..." , is your own idea? I wouldn't make that statement without a source. --Bob K31416 (talk) 05:28, 13 May 2010 (UTC)

Hi Bob: No, both this "point" and the one I removed just now were inherited observations. Actually, I find this "point" pretty obvious because no other proof considers such "changes" or "contributions", so it is pretty obvious that their effect is not evident in the other proofs. Brews ohare (talk) 15:14, 13 May 2010 (UTC)

I removed it. Brews ohare (talk) 16:06, 13 May 2010 (UTC)

Pentagon

You have removed this section as a digression and as OR.

On the second point, I'd take it that you think the cited link to Brodie is an inadequate source. Is that so? The actual construction is a direct use of Pythagoras' theorem, and hardly a controversial matter. However, there are a number of ways to construct an inscribed pentagon, and this one may be original with Brodie.

On the first point, we have the examples of the equilateral triangle, the square, and so I added the regular pentagon. The actual construction is a simple use of Pythagoras' theorem. The more elaborate follow-on is the connection to the golden ratio. That subject is interesting, with a long history in the arts, in "magical" thinking, and in geometry itself. It seems to me worthwhile to use Pythagoras' theorem two more times to make the connection.

More comments? Brews ohare (talk) 17:14, 11 May 2010 (UTC)

I didn't notice the citation! But I do think it is too much of a digression. The part of the Brodie article that I think might be mentioned somewhere in the article would be the idea regarding "no right angles" in the comment, "The construction and its verification give another illustration of how the Pythagorean Theorem is woven into the structure of geometry, even though there are no right angles in a pentagon." And the reader could be referred to the construction in Brodie's paper. --Bob K31416 (talk) 17:32, 11 May 2010 (UTC)

Bob: I trimmed it down to the bare establishment of the length of the side, which is Pythagoras applied twice. The final result is sourced to an old geometry text, and is correct. The construction is very similar to the classical one by Richmond; see also Peter R. Cromwell (1999). Polyhedra. Cambridge University Press. p. 63. ISBN 0521664055.; but a little different. Brews ohare (talk) 20:20, 11 May 2010 (UTC)

But it makes no sense as trimmed. You show how to apply Pythagoras to compute the length of the constructed segment s, but nothing says or suggests that that segment s is the same length as the side of the pentagon. We know it is, because we read the cited page that proves it, but your section leaves it as a complete non-sequitur. The part about the sines is similarly disconnected; what is it there for? What is it sourced to? How does it relate to the figures or to the math before it? I can figure it out, but as a WP section it's very broken. Dicklyon (talk) 23:54, 13 May 2010 (UTC)

Hi Dick: I'll reply to your comments one by one.

1. “You show how to apply Pythagoras to compute the length of the constructed segment s”; Yes, that's right.

2. “but nothing says or suggests that that segment s is the same length as the side of the pentagon”; A source is given that says it is the right length.

3.“ We know it is, because we read the cited page that proves it”; Yes, that is correct. You also know that the objective from the outset was to find the side of the regular pentagon, as stated in the first paragraph.

4. “but your section leaves it as a complete non-sequitur”; not exactly - it says it is the side of the regular pentagon. It also is obviously irrational, containing two square roots, one being √5. So the pentagon fits in as the third example of a regular polygon related to construction of irrational numbers.

5. “The part about the sines is similarly disconnected; what is it there for?” It is explained that trig can find the side directly, which is (I thought) a point some readers would notice. Or, turning it around, they could use the analysis to determine the sine. This info is an aside, but once you've gone to Milan, you might as well visit La Scala.

6. “What is it sourced to?” Does "it" refer to the sine, or to the use of Pythagoras' theorem? If you are concerned that the sine of π/5 is the number given, anyone can calculate it using their pocket calculator. Are you suggesting a table of sines be cited?

7. “How does it relate to the figures or to the math before it?” The value of s was calculated using the figures. The diagram makes clear that the trigonometric formula is correct. Do you wish another figure to show the bisection of the isosceles triangle?

Dick, please take the time to make clear just what you would recommend to fix this section. Or, is it your position that it is all a waste and should be removed? Brews ohare (talk) 01:54, 14 May 2010 (UTC)

I have rewritten the sine portion of the article to meet what I interpret as your reservations. Brews ohare (talk) 02:07, 14 May 2010 (UTC)

It is all a waste and should be removed. There is no logical flow in it. You've shown a segment whose length you can compute, and you've given a ref the says this segment is the same length as the side of the regular pentagon. There's no logical connection given and no sense that can be taken from this. And the sources are not very good at supporting the statements they're attached to, or clarifying the intended relationship. The section purports to be about an application of the Pythagorean theorem; yet it is unclear what the application is, or what the relationship of the constructed segment is to the pentagon, except indirectly via a ref that gives the same length as the side of the pentagon. It makes no sense. Dicklyon (talk) 05:05, 14 May 2010 (UTC)

Dick: What is wrong here? The construction creates the required side of the pentagon, and that is first derived and then compared with sources. It is the correct length. Why is there no sense to be taken from this? The goal is to construct the side and side has been constructed. Pythagoras' theorem is used several times in doing this. What is your point? What exactly is unclear about this? Why is it "indirect"? And if it is, what is so bad about being indirect? All constructions of the inscribed pentagon are a bit complicated, if that is what you are worried about. Brews ohare (talk) 05:48, 14 May 2010 (UTC)

It's not clear what you mean by "is first derived". There's nothing in the construction that suggests that the length s is the side of the regular pentagon. You rely on the fact that the length you come up with matches what someone came up with via an actual derivation by a different method. It doesn't make any logical sense, being presented as merely a cryptic coinciendent. Yes, it uses Pythagoras's theorem, but it remains unclear why you call it an application of the theorem. Logically, what you've done is to show how to construct a length a length given from another source; the application of Pythagoras here is simply that to construct such a length involving square roots, you set up triangles to get there; that process and construction has nothing to do with pentagons; it's just that the construction you did was designed to match the number that a source had provided as the length of the side of regular pentagon.

If this makes sense to anyone else, please say so. Evidently Brews and I are unable to communicate. Dicklyon (talk) 15:19, 14 May 2010 (UTC)

Maybe for now, it might be best to lay this discussion aside for a day or two. Wikipedia has no deadlines. : ) --Bob K31416 (talk) 15:48, 14 May 2010 (UTC)

Good idea. As I said, I'm on wikibreak, so I don't intend to get drawn more into it, now that I've said what I had to say. Dicklyon (talk) 18:52, 14 May 2010 (UTC)

Hi Dick: Your description is accurate: the construction comes up with the correct side length, but it is a bit of a deus ex machina. It seems to me a short but satisfactory approach to simply point out that the result is correct. The alternative, to demonstrate that the result is correct, rather than pointing out merely that sources agree that it is correct, is actually a digression, not a necessity. You may wish to see that proved, but citing a source that it has been proved is a lot simpler and shorter, and serves the purpose.

To put things more in perspective, I also looked into the construction at Pentagon. A very much liked method is due to Richardson. I contributed a calculation of the side it produces, just as I did here for Brodie's method. I did not demonstrate that five such sides fit inside the circle. An animation presented there illustrates that fact, but isn't a proof, of course.

The problem with my proof that Richmond's method produces the correct length is that it requires use of the double-angle formula, which is more complicated mathematically than Brodie's method I've used here. Brodie's method needs only Pythagoras.

If you look at Brodie's presentation, which is much more complete, you may be more satisfied with it. I've abridged it, maybe you can suggest some additions that would improve matters? Brews ohare (talk) 18:18, 14 May 2010 (UTC)

Dick Lyon continues

Yes, Brodie's is much more complete, and has a logical point. He uses the pythagorean theorem to prove that the constructed length is the side of the regular pentagon. This is quite different from what you do, which is to compute the length of the segment and then refer to another source that says that computed length is the side of regular pentagon. It leaves your section with no point, and the references attached to pointless statements. Like "Finally, the side is determined taking the square root, a well-known result.[]" Yes, it's well known that you can get a length from a length squared by taking a square root; but the cited source says nothing about any well-known result; the point, if there was one, is lost. And your statement "Pythagoras' theorem can be used to find the side and the diagonal of such a pentagon, despite the figure having no right angles,[]" attributed to Brodie, is a trivialization of what Brodie shows. What he used to find the length of the side of regular pentagon was some mathematical logic involving a right triangle that's not even in your drawing. Again, if there was a point you were trying to make about application of the Pythagorean theorem, it was lost in the trivialization. Can I suggest some additions that would improve matters? Generally, it is easier to improve such things by subtractions, which is what I have recommended for this case. If a point became clear, that might be a different matter. Dicklyon (talk) 06:22, 19 May 2010 (UTC)

Dick: It would be correct to say that Brodie has a different objective, but not to say that what I wrote "has no point". You simply have chosen to see what I wrote as a (botched) summary of Brodie, which it is not. What it is, is an extraction from Brodie of his method for establishing the length of the side, which is an application of Pythagoras' theorem and therefore exactly the topic of this article. So indeed, the material I wrote does "use the Pythagorean theorem to prove that the constructed length is the side of the regular pentagon". It is not untoward to derive the length and then refer to established sources to show that it is indeed the correct length. It is not a "trivialization of Brodie". I am not attempting to replicate what Brodie said or did, except for the application of Pythagoras' theorem. It is not a critique of what I wrote to say it is not what Brodie wrote in its entirety. My objective is different, and is well stated: to show (as Brodie also showed as part of his exposition) that Pythagoras' theorem can be used to derive the side of the regular pentagon. I know of no other approach that does this using only Pythagoras' theorem. That is the point.

In sum, I believe that you have unfortunately assumed my purpose was the same as Brodie's, and seen what I wrote as deficient because it does not afford a summary of what he has done. However, the objective here was different, and relied only upon a small portion of what Brodie did, namely the portion that relies upon Pythagoras' theorem, which is what the article here is talking about.

I've rewritten the section to make these points obvious. Brews ohare (talk) 18:54, 20 May 2010 (UTC)

Former Featured Article

This article has no nomination history, so it is hard to know what was found wrong with it. I'd say it is now in pretty good shape. Maybe it could be renominated? Brews ohare (talk) 16:58, 12 May 2010 (UTC)

About the intro

"... understood the principle, if not the mathematical significance ..."

I didn't get it :X 187.107.13.253 (talk) 21:54, 12 May 2010 (UTC)

I've rewritten this sentence and provided a source where the matter is further discussed. Brews ohare (talk) 22:48, 12 May 2010 (UTC)

The existence of irrational numbers

I've made a major reorganization of the section The existence of irrational numbers, except for the part re pentagon. If anyone doesn't like it, feel free to revert. Thanks. --Bob K31416 (talk) 11:04, 15 May 2010 (UTC)

Irrational numbers.

The citations in this section do not support the text. The role of the right triangle is not historically as indicated. I also find the flow of the section is not as well served by dropping headers. Below is my suggestion for the beginning of this section:

The existence of irrational numbers

One of the consequences of the Pythagorean theorem is that lengths with a value that is an irrational number can be constructed. The notion of irrational numbers was contrary to the long-held belief that everything was rational. According to one legend, Hippasus was drowned at sea for making known the existence of the irrational or incommensurable.

Three examples follow based upon the dimensions of regular polygons.

Equilateral triangle

First is an example that shows how to construct the irrational number $\scriptstyle {\sqrt {3}}$. Pythagoras' theorem determines the height of an equilateral triangle in terms of its sides. The height of the equilateral triangle divides it into two congruent right triangles. With $l\,\!$ the hypotenuse, $h\,\!$ the height, $l/2\,\!$ the base, from Pythagoras' theorem it follows:

$l^{2}=h^{2}+\left({\frac {l}{2}}\right)^{2}=h^{2}+{\frac {l^{2}}{4}}\,\!$

$h^{2}=3\ {\frac {l^{2}}{4}}\,\!\ .$

Taking the square root of both sides of the equation and rearranging,

${\sqrt {3}}={\frac {h}{(l/2)}}\,\!\ .$

The diagonal of the square

See also: Square root of 2 § Proofs of irrationality

The second example is the square, used to construct $\scriptstyle {\sqrt {2}}\,.$ Pythagoras' theorem determines the length of the diagonal of a square in terms of the sides. The diagonal of the square divides it into two congruent right triangles. Because $l\,\!$ is the side and $d\,\!$ the diagonal, from Pythagoras' theorem it follows:

$d^{2}=l^{2}+l^{2}=2\,l^{2}\,\!$

Finally, $\scriptstyle {\sqrt {2}}\,$ is found as:

${\sqrt {2}}={\frac {d}{l}}\,\!.$

This result appears to be the first example of an irrational number to be discovered.

Brews ohare (talk) 15:22, 15 May 2010 (UTC)

Re "The citations in this section do not support the text." - I don't have access to the first source but I see that you are right with regard to the pentagon sources and expect that if I saw the other source it would confirm what you said. Then isn't the title of the section "The existence of irrational numbers" inappropriate? BTW, is the first source at Google books? --Bob K31416 (talk) 15:45, 15 May 2010 (UTC)

Hi Bob: the sources are Stillwell and Heath. I implemented this version. I wonder if you know of a construction for √5 or maybe some other simple irrational using Pythagoras to supplant the pentagon? Brews ohare (talk) 15:52, 15 May 2010 (UTC).

Thanks for the links! Re √5, nope. I'll be looking at what's been presented. It may take me awhile. : ) --Bob K31416 (talk) 15:58, 15 May 2010 (UTC)

See root rectangle for a simple construction of square roots of integers. Dicklyon (talk) 16:42, 15 May 2010 (UTC)

Thanks Dick. I added this link to a ‘See also’ template. Brews ohare (talk) 16:52, 15 May 2010 (UTC)

Seems like a bad idea. That linked article doesn't have anything to add to the section topic. Dicklyon (talk) 21:41, 15 May 2010 (UTC)

Bob: Take a look at Square root of 5#Relation to the golden ratio and Fibonacci numbers. Brews ohare (talk) 16:11, 15 May 2010 (UTC)

Bad image

Hi, I noticed that the image in the equilateral triangle section isn't an equilateral triangle. 187.107.19.43 (talk) 18:13, 15 May 2010 (UTC)

I changed the figure to be equilateral. Brews ohare (talk) 21:15, 15 May 2010 (UTC)

Pythagoras's theorem in 7 dimensions

In another article, Lagrange identity, it is correctly reported that,

For a and b as vectors in ℝ⁷, Lagrange's identity:

${\biggl (}\sum _{k=1}^{n}a_{k}^{2}{\biggr )}{\biggl (}\sum _{k=1}^{n}b_{k}^{2}{\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2},$

takes on the particular form involving the cross product sometimes referred to as the Pythagorean theorem:

$|\mathbf {a} |^{2}|\mathbf {b} |^{2}-|\mathbf {a} \cdot \mathbf {b} |^{2}=|\mathbf {a} \times \mathbf {b} |^{2}\ ,$

that is, the same form found in ℝ³.

This tends to suggest to me that if Pythagoras's theorem holds in any other dimensions apart from 3, that it could only be 7. In this article, we have a section on Pythagoras's theorem in 'n' dimensions. While there has been quite alot of debate on how to best present a proof of Pythagoras's theorem in 3 dimensions, has anybody ever seen a proof of Pythagoras's theorem in 4 dimensions? I may be overlooking something as regards to the argument above, and so I welcome any comments. David Tombe (talk) 12:28, 16 May 2010 (UTC)

Hi David: The pythagorean theorem is a very specific statement about sides of a triangle in 3D. In n-D it has to be extended, and there may be more than one way to do this. So, for example, one may focus upon the areal interpretation or the interpretation in terms of lengths. For example, Lounesto refers to it using the cross-product, an area interpretation clearly restricted to 3-D and 7-D. A much more common way is to take a length interpretation as the square of the norm of a vector in n-space as given by the sum of the squares of its orthogonal constituents. That version applies in a space of any dimension, and has no area connotation. Such is my take of the matter, at any rate. Brews ohare (talk) 13:35, 16 May 2010 (UTC)

References

I'm just adding a "references" so the footnotes will appear here. TomS TDotO (talk) 12:53, 16 May 2010 (UTC)

Brews and TomS, Isn't it the old story of presuming to extrapolate something that we fully understand in 3D into 'n'D, without perhaps having fully thought through all the implications? I've just been looking at a proof of the sine relationship in the vector cross product. It hinges on the Jacobi identity. The Jacobi identity as we all know only holds with the 3D cross product. The Lagrange identity is certainly a general identity for 'n'D. But many threads of circumstantial evidence are tending to point to Pythagoras's theorem as being purely a 3D geometrical affair. Like I said above, let's see somebody proving Pythagoras's theorem in 4D. David Tombe (talk) 11:01, 17 May 2010 (UTC)