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A proof of Pythagoras' theorem using dimensional analysis

A friend of mine suggested that such a proof was possible the other day, and so I resolved to go and think out the details. It turned out to be much simpler than I thought, so I thought I'd post the details here in case anyone is interested. If you don't know what dimensions are, you won't lose anything if you replace every instance of the word 'dimensions' with 'units'.

Pythagoras' Theorem, as you know, says that for any right-angled triangle with hypoteneuse length c and shorter sides of lengths a and b, we have

a2+b2=c2a^2+b^2=c^2

It's obvious that for a right-angled triangle, only the hypoteneuse length c and the size of one of the angles are needed to completely specify the triangle. Once you're given a hypoteneuse length and an angle, then there's only one way to complete the picture to make a right-angled triangle:



Therefore the area of the triangle (let's call the area C) can only depend on c and θ\theta, that is

C=f(c,θ)C = f(c,\theta)

Now we use dimensional analysis. The dimensions of area are length squared, and the dimensions of the length of a side are length (obviously). Therefore if we want the dimensions to match on , the function ff has to have dimensions of length squared. But c has dimensions of length and the angle θ\theta is dimensionless, and so the function must take the form

f(c,θ)=c2g(θ)f(c,\theta) = c^2g(\theta)

where gg is some function of θ\theta. Now we divide the triangle into two in such a way that we end up with two more right-angled triangles, as shown:



By the law of similar triangles, both angles that I've marked in must be equal to θ\theta. Using the same arguments above, the areas of these smaller triangles (lets call them A and B) must be equal to

A=a2g(θ)A = a^2 g(\theta)
B=b2g(θ)B = b^2 g(\theta)

However it's obvious that we have A + B = C, and so we can write

a2g(θ)+b2g(θ)=c2g(θ)a^2g(\theta) + b^2g(\theta) = c^2 g(\theta)

and g(θ)g(\theta) is not zero (so long as θ0\theta\neq 0 because then we would have a zero area, which is just silly) and so we can divide through by it, giving

a2+b2=c2a^2+b^2=c^2
Reply 1
Cexy
But c has dimensions of length and the angle θ\theta is dimensionless, and so the function must take the form

f(c,θ)=c2g(θ)f(c,\theta) = c^2g(\theta)

where gg is some function of θ\theta.


I'm not sure I buy this argument. That's a bit like saying all functions from R to R must be linear.
Reply 2
Also, why do we get the same g for A and B?
Reply 3
dvs
Also, why do we get the same g for A and B?
Because A and B are both right-angled triangles with hypoteneuse length a (resp. b) and angle theta. We argued that the area of a right-angled triangle can depend only on its hypoteneuse and an angle, so we're assuming that there is a general formula for the area of a triangle without knowing its exact form. In fact I think you could strengthen the condition on g and demand that, by symmetry

g(θ)=g(π/2θ)g(\theta) = g(\pi/2-\theta)

Mathmoid
I'm not sure I buy this argument. That's a bit like saying all functions from R to R must be linear.
I don't think that it is, although I definitely see where you're coming from. Here we're demanding that our 'theory of area' if you like makes physical sense, which it certainly should do if we want to use it to measure areas.

General functions from R to R are allowed to have arbitrary constants, and if we were to put ourselves in a dimensional setting we would demand that, for example, a function

f:RRf: R\to R
f(x)=a0+a1x+a2x2+f(x) = a_0 + a_1x + a_2x^2 + \cdots

would have coefficients with the appropriate dimensions to give f the overall dimension of length (assuming that x has dimensions of length). With the triangle example we've argued that the only dimensional variable allowed is the length of the hypoteneuse, so we don't have the same freedom to introduce arbitrary dimensional constants to make the equation dimensionally valid.

Edit: Just to make my motives completely transparent, the reason that I'm posting this here is because I'm thinking of using it as an example for some students that I'm teaching dimensional analysis to, and I want you guys to iron out any errors for me before I do :wink: So please do point out anything you think is wrong with the proof, no matter how trivial, because it will be helpful!
Reply 4
Cexy

General functions from R to R are allowed to have arbitrary constants, and if we were to put ourselves in a dimensional setting we would demand that, for example, a function

f:RRf: R\to R
f(x)=a0+a1x+a2x2+f(x) = a_0 + a_1x + a_2x^2 + \cdots

would have coefficients with the appropriate dimensions to give f the overall dimension of length (assuming that x has dimensions of length). With the triangle example we've argued that the only dimensional variable allowed is the length of the hypoteneuse, so we don't have the same freedom to introduce arbitrary dimensional constants to make the equation dimensionally valid.

So your argument is, that if we disallow any magic constants which aren't dimensionless then f(c)c2f(c) \propto c^2. From a pure maths standpoint this is a big restriction even if, intuitively, it's reasonable, so it's definitely worth stating it explicitly. (Obviously, it doesn't hold in non-Euclidean geometries which often have intrinsic "magic" dimensional constants but neither does pythagoras' theorem so no great surprises really.)
Reply 5
Thanks Mathmoid, that's helpful. It's been a while since I've studied any geometry - can you give me some examples of intrinsic dimensional constants that crop up in non-Euclidean geometry? Are you referring to things like curvature, torsion etc?
Reply 6
Cexy
Thanks Mathmoid, that's helpful. It's been a while since I've studied any geometry - can you give me some examples of intrinsic dimensional constants that crop up in non-Euclidean geometry? Are you referring to things like curvature, torsion etc?


Well, the simplest example is spherical geometry, where the radius of the sphere (R) is a dimensional constant intrinsic to the surface. Small triangles have angles which add up to not much more than pi, but as the sides of the triangle become bigger compared to the radius of the sphere they start adding up to significantly more than pi.

In fact the area of a triangle depends only on its angles and R^2.
http://mathworld.wolfram.com/SphericalTriangle.html
Reply 7
Ah, of course. The way I was taught spherical geometry was 'modulo' the radius of the sphere, and the Euclidean limit was taken by sending the side lengths of a triangle to zero, rather than sending the radius to infinity. I suppose that both viewpoints have their advantages.
Reply 8
Yes, often R is taken to be 1, but along with the practice of taking c to be 1 in special relativity, while convenient it robs you of a useful dimensional crutch so it's something I'm ambivalent about. (Somewhat ironic given the context of the current thread!)
Reply 9
I haven’t done maths for 40 years but I’d like to suggest an alternative proof ; for a right angle triangle with base a and night b and Hypotenuse, as a approaches ; 0 it is obvious that b=c therefore sqr b=Sqr c. But as a increasess to aa to becom cc, which mean c is increasing by aa, it yields sqr b +sqr aa=sqr cc

To all numerical analyst; do you think this has legs?
(edited 6 years ago)
Original post by FirasAli
I’d like to suggest an alternative proof ; for a right angle triangle with base a and night b and hypotonie c, as a —-> 0 it is obvious that b=c tgererofre b2=c2. But as a increasess to aa to becom cc, which mean c is increasing by aa, it yields b2 aa2=cc2


What you wrote is incoherent and most of the words aren't even spelled correctly.
Rushed (sorry) - what I am trying to say concisely is as the base of right angle triangle approches 0, the theorem is obviously true but as it increases, the hypotenuse increases by an equal ration so by adding the square of this ration to both sides of the equation may give a proof of the theorem
Original post by FirasAli
Rushed (sorry) - what I am trying to say concisely is as the base of right angle triangle approches 0, the theorem is obviously true but as it increases, the hypotenuse increases by an equal ration so by adding the square of this ration to both sides of the equation may give a proof of the theorem


"As it increases, the hypotenuse increases by an equal ratio"; the word is "ratio", not "ration", and also, what are you saying it's equal to? Further, this claim as it stands is entirely unjustified.
Original post by Prasiortle
"As it increases, the hypotenuse increases by an equal ratio"; the word is "ratio", not "ration", and also, what are you saying it's equal to? Further, this claim as it stands is entirely unjustified.


I don’t have the answer to the ratio thing, but assuming that if the base increases by the same quantity as the hyponteuse then it should be possible to demonstrate that the theorem holds for any value of the base -

I am trying to clarify my thoughts on this and as such it remains a work in progress.

Thx
Original post by FirasAli
I don’t have the answer to the ratio thing, but assuming that if the base increases by the same quantity as the hyponteuse then it should be possible to demonstrate that the theorem holds for any value of the base -

I am trying to clarify my thoughts on this and as such it remains a work in progress.

Thx


What exactly do you mean by "the base increases by the same quantity as the hypotenuse"? You seem to be misusing all sorts of terms like "ratio" and "quantity" which have specific technical meanings.
A friend of mine suggested that such a proof was possible the other day, and so I resolved to go and think out the details. It turned out to be much simpler than I thought, so I thought I'd post the details here in case anyone is interested. If you don't know what dimensions are, you won't lose anything if you replace every instance of the word 'dimensions' with 'units'.

Pythagoras' Theorem, as you know, says that for any right-angled triangle with hypoteneuse length c and shorter sides of lengths a and b, we have

a2+b2=c2a^2+b^2=c^2

It's obvious that for a right-angled triangle, only the hypoteneuse length c and the size of one of the angles are needed to completely specify the triangle. Once you're given a hypoteneuse length and an angle, then there's only one way to complete the picture to make a right-angled triangle:



Therefore the area of the triangle (let's call the area C) can only depend on c and θ\theta, that is

C=f(c,θ)C = f(c,\theta)

Now we use dimensional analysis. The dimensions of area are length squared, and the dimensions of the length of a side are length (obviously). Therefore if we want the dimensions to match on , the function ff has to have dimensions of length squared. But c has dimensions of length and the angle θ\theta is dimensionless, and so the function must take the form

f(c,θ)=c2g(θ)f(c,\theta) = c^2g(\theta)

where gg is some function of θ\theta. Now we divide the triangle into two in such a way that we end up with two more right-angled triangles, as shown:



By the law of similar triangles, both angles that I've marked in must be equal to θ\theta. Using the same arguments above, the areas of these smaller triangles (lets call them A and B) must be equal to

A=a2g(θ)A = a^2 g(\theta)
B=b2g(θ)B = b^2 g(\theta)

However it's obvious that we have A + B = C, and so we can write

a2g(θ)+b2g(θ)=c2g(θ)a^2g(\theta) + b^2g(\theta) = c^2 g(\theta)

and g(θ)g(\theta) is not zero (so long as θ0\theta\neq 0 because then we would have a zero area, which is just silly) and so we can divide through by it, giving

a2+b2=c2a^2+b^2=c^2


Yep this is a valid proof and is given here https://strathmaths.wordpress.com/2011/10/15/pythagoras-two-favourite-proofs/

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