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Questions about Pythagorean triples (lemmy.world)

submitted 2 weeks ago* (last edited 1 week ago) by [email protected] to c/[email protected]

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I’m gathering data for a hobby project and notating where the triangles in the data correspond to Pythagorean triples, but sometimes it doesn’t seem clear to me with certain data.

Can there exist Pythagorean triples in which the leg lengths are not coprime with each other but both are coprime with the hypotenuse? i.e., a right triangle in which (leg~1~, leg~2~, hypotenuse) = (a * n, b * n, c), in which a, b, c, and n are whole numbers and n is not a factor of c?

How can I determine if a right triangle with given lengths can scale to be a Pythagorean triple? If any of the values in (leg~1~: leg~2~: hypotenuse) are irrational, that does indeed mean the values cannot scale to be whole numbers?

Once it is determined that the triangle can scale to a Pythagorean triple, what is the best method of scaling the values to three whole numbers?

Thanks for any help

Edit: I’ve found an effective way to determine primitive Pythagorean triples from given leg lengths. Using a calculator that can output in fractional form, such as wolfram alpha, input leg~1~ / leg~2~ and the output will be a fraction with the numerator and denominator denoting the leg lengths of a primitive Pythagorean triple. Determining the hypotenuse is then simply using the Pythagorean Theorem.

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[–] [email protected] 6 points 2 weeks ago (3 children)

Can there exist Pythagorean triples in which the leg lengths are not coprime with each other but both are coprime with the hypotenuse?

I don't think so. Using the theorem, (an)² + (bn)² = c² ... Algebra... c = n(root(a² + b²))

If c is a whole number, then the root must give a whole number, and therefore n divides c

If any of the values in (leg~1~: leg~2~: hypotenuse) are irrational, that does indeed mean the values cannot scale to be whole numbers?

Hypothetically I can see it working from the algebra. We can construct some trivial cases like taking 3:4:5 and dividing all those lengths by root(2). All lengths are now irrational and could be made whole by multiplying them all by root(2). We can also do it with different roots in a way that is hard to type on mobile but essentially involves breaking up, say, root(30) into component roots and having different pairs of components between the top and bottom of the fraction line but that is essentially the same thing as dividing them all by the same root. I can't see a general way of doing it if they are not all irrational in the same way, though I am just a maths teacher not a proper researcher.

[–] [email protected] 5 points 2 weeks ago (2 children)

If only some side lengths are irrational, then, by definition, there is no ratio of whole numbers that describes those lengths relative to each other - therefore, you can't divide them out to be whole. Unless all side lengths are irrational and also share the same irrational multiplier (like root(2) in your example).

[–] [email protected] 3 points 2 weeks ago (1 children)

Yeah I think I was overcomplicating it in my head, I kept getting caught in the hows and whys of squaring everything in Pythagoras' theorem, that maybe it could get around the problems of some being irrational and some not.

Ultimately I think you're right that the scaling itself is the problem: scaling irrational numbers to be whole numbers generally won't work outside of those specific cases.

[–] [email protected] 3 points 2 weeks ago

Also, not all irrational numbers can even be squared into rationals