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March 23rd, 2018, 10:06 AM   #1
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triangle inequality

My proof to the triangle inequality:
(1) I draw two point A & B.
(2) By the first axiom of Eculid: "between 2 points you can draw only one straight line'.
(3) I draw a line between A & B.
(4) => This line is the shortest distance between A & B. (Euclid axiom)
(5) If i draw a broken line between A & B (Zigzag line), this line is longer becuase of (4).
(6) So the line that pass A & B & C (C is the point that the vertix of the line).
The triangle A & B & C - is inequality triagnle.
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March 23rd, 2018, 10:10 AM   #2
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All triangles are half a rectangle: isn't that enough?
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March 23rd, 2018, 10:25 AM   #3
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I don't understand what you mean...
Why? Is My proof not simple?
You can explain what you mean... please!!!!
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March 23rd, 2018, 10:37 AM   #4
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Originally Posted by Denis View Post
All triangles are half a rectangle: isn't that enough?
All triangles are half a parallelogram, not a rectangle.

-Dan
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March 23rd, 2018, 11:25 AM   #5
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Quote:
Originally Posted by topsquark View Post
All triangles are half a parallelogram, not a rectangle.
Agree
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March 23rd, 2018, 11:42 AM   #6
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After I ask a colleague I have some notes:
1. the triangle ineqaulity in the metric space is axiom.
2. The Euclid space is example to metric space. There are another examples of other spaces. To prove that a space is Euclid space I need to prove the 3 axioms of the Euclid space that are true and especially triangle inequality.
3. My proof is not execptable proof. The axiom of between to points pass only one line don't tell nothing one their lenghts (even if is seem intuitive). The proof that the trangle ineqaulity is been in the Euclid space need to be a much more work. Actually, I need to use the inequality of Cauchy-Swartz.
So can one's show me the way to prove it by inequality of Cauchy-Swartz?
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