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- June 25th 2012, 02:25 PM #1
## Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

I'm reading a differential geometry book named "Elementary geometry of differentiable curves: An undergraduate introduction" written by Gibson.

I'm on page 2. On that page, 4 properties of scalar product(dot product) of vectors are given.

One of them is:

I know the proof of this and the proof shows that the statement is true for

I accept this.

But if isn't it definite that has to be equal to and not greater than ?

Can anyone find a such that when then is greater than ?

Why did the author said that ? Why did he mention greater and equal, and not just equal to ?

- June 25th 2012, 02:32 PM #2

- June 25th 2012, 02:40 PM #3

- June 25th 2012, 02:50 PM #4

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- June 25th 2012, 02:52 PM #5

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## Re: Why is that if and only if z = 0 then z.z >= 0 and not just only = 0?

it's just a concise of saying:

if z ≠ 0, z.z > 0

if z = 0, z.z = 0.

from these 2 statements, we can conclude that if z.z = 0 (that is, z.z is not greater than 0), then z must itself be 0 (for if it were non-zero, then z.z > 0).

in general, all we can say is that z.z ≥ 0, for an arbitrary vector z, because of the fact that z might be 0.

it's like saying: "a square (of a real number) is always positive"....except for 0. Zero-objects are funny like that, they often introduce "exceptions" which make simple rules more complicated to say.

- June 25th 2012, 02:52 PM #6

- June 25th 2012, 03:16 PM #7

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- June 25th 2012, 03:22 PM #8

- June 25th 2012, 03:27 PM #9