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oadrian96

One fishy thing that I see is that square roots are defined to give you two values a positive and a negative one so in the strictest sense it fails the first rule of a norm. It seems like we just disregard the negative square root.

Osoii

|v|>=0: ?x is always equal to or greater than 0

|v|=0 only when v is a zero vector: vi^2>=0, so if ?vi^2 = 0, then all the vi must be zero

|av|=|a||v|: |av| = ??(avi)^2 = ??(a^2vi^2) = ?a^2?vi^2 = ?a^2 * ??vi^2 = |a||v|

|u|+|v|>=|u+v|: |u|+|v| = ??ui^2 + ??vi^2, (|u|+|v|)^2 = ?ui^2+?vi^2 + 2??ui^2vi^2 |u+v| = ??(ui+vi)^2 = ??(ui^2+vi^2+2uivi), |u+v|^2 = ?ui^2+?vi^2 + 2?uivi 2??ui^2vi^2 = 2?|ui||vi| >= 2*?uivi so, |u|+|v|>=|u+v|

Seems like the root and sigma symbol cannot display correctly, they are all replaced by '?', sad :(

Azure

Square roots are actually defined to only give you the positive square root. This is why we write -sqrt to give the negative square root.

The Euclidean norm is the distance between the point and the origin in space as given by the Pythagorean theorem. If we take this interpretation, then we can just use our geometric examples. If attempting to formally prove this, we would need tools of analysis, and likely the Cauchy Schwarz theorem and the inner product to show the triangle inequality property of the Euclidean norm.

jlessioh

To reply to the first comment above, while i agree that -2sqrt(5)^2 = sqrt(4^2+2^2), we obviously cannot have a negative distance, so we don't need to consider the negative case.