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Trying to work with the Pythagorean Theorem when the hypotenuse is equal to 1 in some weird attempt to make some unit-circle equivalent for a right triangle.
Let c = 1
a2 + b2 = c2
c2 = 1
a2 + b2 = 1
a2 = 1 - b2
b2 = 1 - a2
- Here's where it gets fucked up, because I'm not sure how to proceed without some kind of complex conjugate, but the following is what my friend claims is mathematically illegal, but it still winds up a true statement:
a = 1 + bi
b = 1 + ai
1 = bi - a
1 = ai - b
bi - a = ai - b
bi - ai = a - b
i (b - a) = a - b
i = (a - b) / (b - a)
ai - bi = b - a
i(a - b) = b - a
i = (b - a) / (a - b)
(b - a) / (a - b) = (a - b) / (b - a)
(b - a) (b - a) = (a - b) (a - b)
b2 - 2ab + a2 = a 2 - 2ab + b2
So, it winds up being a true statement, but I don't know if it proves anything or has any significance whatsoever.
I'm just putting it out here in case anyone has anything to say about it other than that I'm a complete lackwit.
What the hell would a and b be if c were 1, anyway?
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>>8505629
It's nonsense.
> a = 1+ bi
You start with a triangle with set sides, and then you set the values to some complex value, whatever that means. Then you do a bunch of manipulations that go back and forth but end up in a triviality. But they have nothing to do with triangle.
If c=1 then a and b are sine and cosine of acute angles.
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>>8505653
So, you're saying that there's no way to represent what the two sides would be in real numbers? They'd have to be the same length, wouldn't they?
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>>8505653
And you can't reduce a^2 = 1 - b^2 to anything? There's no way to get a square root of (1 - b^2)?
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>>8505629
try [math]a=\frac{2t^2}{t^2+1}, b=\frac{t^2-1}{t^2+1}[/math]
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>>8505670
this. paging Mr. Euclid.
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>>8505670
but if a^2+b^2=c^2=1 then plugging in you get:
4t^4+t^4-2t^2+1=t^4+2t^2+1
t^2=1
a=1
b=0
and it isn't a triangle.
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>>8505780
should be a = 2t/ (t^2+1) and b = (t^2-1)/(t^2+1), giving a^2 + b^2 = (4t^2 + t^4-2t^2+1)/(t^2+1)^2 = 1, for any t.
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