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

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:

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

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

I accept this.

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

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

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

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

Originally Posted by x3bnm
One of them is:
$\displaystyle \mathbf z \cdot \mathbf z \geq 0 \text{ with equality if and only if } \mathbf z = \mathbf 0$
That is simply saying $\displaystyle z\cdot z=0\text{ if and only if }z=0$

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

Plato said:
That is simply saying $\displaystyle z\cdot z=0\text{ if and only if }z=0$
So the $\displaystyle >$ sign has no role here. I've to overlook it.

Roger that. Thanks Plato.

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

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

5. ## 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.

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

Originally Posted by x3bnm
So the $\displaystyle >$ sign has no role here. I've to overlook it.
It has role. That is a fairly standard notation.
Here is another example:
$\displaystyle a^2+b^2\ge 2ab,~\text{ equality holds if and only if }a=b=1~.$

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

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:

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

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

I accept this.

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

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

Why did the author said that $\displaystyle \mathbf z \cdot \mathbf z \geq 0$? Why did he mention greater and equal, and not just equal to $\displaystyle 0$?
You are misreading the statement. He is NOT talking about only z= 0. The statement says that $\displaystyle z\cdot z\ge 0$ for all z and, additionally, $\displaystyle z\cdot z= 0$ if and only if z= 0.

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

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

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

Originally Posted by HallsofIvy
You are misreading the statement. He is NOT talking about only z= 0. The statement says that $\displaystyle z\cdot z\ge 0$ for all z and, additionally, $\displaystyle z\cdot z= 0$ if and only if z= 0.
You are right. In that sense the statement has to use $\displaystyle \geq 0$ and not just $\displaystyle = 0$