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

**x3bnm** · June 25th 2012, 01:25 PM

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:

$\mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z = \mathbf 0$

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

I accept this.

But if $\mathbf z = \mathbf 0$ isn't it definite that $\mathbf z \cdot \mathbf z$ has to be equal to $0$ and not greater than $0$ ?

Can anyone find a $\mathbf z$ such that when $\mathbf z = \mathbf 0$ then $\mathbf z \cdot \mathbf z$ is greater than $0$ ?

Why did the author said that $\mathbf z \cdot \mathbf z \geq 0$ ? Why did he mention greater and equal, and not just equal to $0$ ?

**Plato** · June 25th 2012, 01:32 PM

Originally Posted by x3bnm

One of them is:
$\mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z = \mathbf 0$

That is simply saying $z\cdot z=0\text{ if and only if }z=0$

**x3bnm** · June 25th 2012, 01:40 PM

Plato said:
That is simply saying $z\cdot z=0\text{ if and only if }z=0$

So the $>$ sign has no role here. I've to overlook it.

Roger that. Thanks Plato.

**xxp9** · June 25th 2012, 01:50 PM

z . z >= 0 for any z, and the z . z = 0 when z=0. Is this clearer?

**Deveno** · June 25th 2012, 01:52 PM

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.

**Plato** · June 25th 2012, 01:52 PM

Originally Posted by x3bnm

So the $>$ sign has no role here. I've to overlook it.

It has role. That is a fairly standard notation.
Here is another example:
$a^2+b^2\ge 2ab,~\text{ equality holds if and only if }a=b=1~.$

**HallsofIvy** · June 25th 2012, 02:16 PM

Originally Posted by x3bnm

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:

$\mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z = \mathbf 0$

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

I accept this.

But if $\mathbf z = \mathbf 0$ isn't it definite that $\mathbf z \cdot \mathbf z$ has to be equal to $0$ and not greater than $0$ ?

Can anyone find a $\mathbf z$ such that when $\mathbf z = \mathbf 0$ then $\mathbf z \cdot \mathbf z$ is greater than $0$ ?

Why did the author said that $\mathbf z \cdot \mathbf z \geq 0$ ? Why did he mention greater and equal, and not just equal to $0$ ?

You are misreading the statement. He is NOT talking about only z= 0. The statement says that $z\cdot z\ge 0$ for all z and, additionally, $z\cdot z= 0$ if and only if z= 0.

**x3bnm** · June 25th 2012, 02:22 PM

Thanks all for answering my question. So I understand that zero is an exceptional case that creates exceptional situations.

**x3bnm** · June 25th 2012, 02:27 PM

Originally Posted by HallsofIvy

You are misreading the statement. He is NOT talking about only z= 0. The statement says that $z\cdot z\ge 0$ for all z and, additionally, $z\cdot z= 0$ if and only if z= 0.

You are right. In that sense the statement has to use $\geq 0$ and not just $= 0$

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