TSOTSROATSOAITIETTSROTRS
Created on: October 31st, 2008
a2 + b2 = c2. (correction... *Not* a2 + b2 = c2.)
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Let a, b, c >= 0 be side lengths of a triangle satisfying the claim. We have that sqrt(a) + sqrt(b) = sqrt(c) <=> (sqrt(a) + sqrt(b))^2 = c <=> a + b + 2*sqrt(a*b) = c, but by the triangle inequality, a+b >= c. Hence, the original equality holds if and only if 2*sqrt(a*b) = 0 <=> at least one of a, b are zero. WLOG we may assume that a=0. Now apply the equality again, but with a, b, c cyclically permuted: sqrt(b) + sqrt(c) = sqrt(a) = 0, since a=0. Consequently a=b=c=0. (Proof is for ANY triangle!)
AIDIR?
YOU ARE DOING IT WRONG!
The triangle inequality does not imply that 2sqrt(ab)=0. It's even easier than that. The claim is that the sum of the square roots of any to sides of an isoceles triangle is equal to the square root of the remaining side. Consider an isoceles tranigle, two of whose sides have length a and a third of length b. The claim says that sqrt(a)+sqrt(b)=sqrt(a). Thus, sqrt(b)=0 => b=0. The claim further states that sqrt(a)+sqrt(a)=sqrt(b)=0=2*sqrt(a). Thus, sqrt(a)=0 and a=0.
The Pythagorean Theorem.
of course it only applys to right triangls due to the massive amount of different triangles you can get from two lines of the same length and another line connecting them.
The simple formula : A^2 + B^2 = C^2 also only applies to certain sides., so if you have the value of the Hypotenuse (C) and want to find the value of one of the other sides (We'll say B) you need to treat it just like an algebra problem.
B^2 = C^2 - A^2
so to say 'the remaining side' even is wrong.
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