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March 27th, 2019, 07:54 PM   #1
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Pythagoras theorem

https://en.m.wikipedia.org/wiki/Pythagorean_theorem

Can we apply Pythagoras theorem to decimal points
where a,b & c are decimal points sides in a right angled triangle?

a^2 + b^2 = c^2

a & b are sides
c is Hypothenuse

Examples of a,b & c : 3.1,2.3,4.3,5.2,7.3,11.4,13.3 etc

Thanks & Regards,
Prashant S Akerkar

Last edited by prashantak; March 27th, 2019 at 07:58 PM. Reason: Content updates
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March 27th, 2019, 09:09 PM   #2
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Thanks.

Can we have Pythagorean triplets which
are decimal point numbers?

Thanks & Regards,
Prashant S Akerkar
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March 27th, 2019, 09:12 PM   #3
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scale up a 3,4,5 right triangle until the sides are > 100

(147,196,245) is one example

Now divide by 100

(1.47, 1.96, 2.45) is a right triangle with sides that have decimal length.
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March 27th, 2019, 09:22 PM   #4
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Thanks.

So we can have Pythagorean triplets,
and the Pythagorean theorem is modified & tested
to work with decimal points triplets.

Similar to the above decimal points Pythagorean triplets, Can
we list all ?

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Prashant S Akerkar

Last edited by prashantak; March 27th, 2019 at 09:39 PM.
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March 28th, 2019, 01:42 AM   #5
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Not quite. By definition, a Pythagorean triplet is a triplet of natural numbers. Corresponding triplets of rationals can be obtained by applying a rational scale factor.
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March 28th, 2019, 04:14 AM   #6
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Quote:
Originally Posted by skipjack View Post
Not quite. By definition, a Pythagorean triplet is a triplet of natural numbers. Corresponding triplets of rationals can be obtained by applying a rational scale factor.
Thanks.

We are enhancing Pythagorean theorem.

Can we create a list of Pythagorean triplets of rationals?

Thanks & Regards,
Prashant S Akerkar
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March 28th, 2019, 05:39 AM   #7
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Quote:
Originally Posted by prashantak View Post
Can we create a list of Pythagorean triplets of rationals?
Yes we can, and the previous post said how to do it.
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April 13th, 2019, 09:48 AM   #8
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Thanks.

Can we have a Dictionary of Pythagorean triplets
of rationals?

Thanks & Regards,
Prashant S Akerkar
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April 13th, 2019, 04:02 PM   #9
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Math Focus: Wibbly wobbly timey-wimey stuff.
I don't know about a library, but there is a way to generate them (all of them I think.) Given two positive integers m, n such that m > n we can construct:
$\displaystyle a^2 + b^2 = c^2$
with
$\displaystyle a^2 = (m - n)^2$
$\displaystyle b^2 = 4mn$
$\displaystyle c^2 = (m + n)^2$

The only trick is the value for $\displaystyle b^2$. We have to put in "by hand" m, n such that $\displaystyle b^2$ is actually the square of a number. For example, if we choose m = 4, n = 1 we get
$\displaystyle a^2 = (m - n)^2 = (4 - 1)^2 = 3^2$ So a = 3.
$\displaystyle b^2 = 4mn = 4(4)(1) = 4^2$ so b = 4,
$\displaystyle c^2 = (m + n)^2 = (4 + 1)^2 = 5^2$ so c = 5.

Thus we know that 3, 4, 5 is a Pythagorean triple.

And, of course, for rational numbers you can use skipjack's comment. For otherwise real numbers you can simply pick a value of a and b can just calculate c. No troubles there.

-Dan
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