# Prove the impossible?

• Jun 11th 2010, 04:30 AM
CropDuster
Prove the impossible?
Hi all,

So I've just started a linear algebra course, and this is my first exposure to formal proofs.

Here is a question from my book:

Prove the following by contrapositive: Let X be a vector in R^n. If XdotY=0 for every vector Y in R^n, then X=0.

So the contrapositive is: If X=/=0, then XdotY=/=0 for every vector Y in R^n. Right?

But if Y can be every vector in R^n, then couldn't Y be the zero vector, hence XdotY=0.

I can't figure out anyway around this fact. I have been specifically told not to alter any restrictions (ie "for every vector Y in R^n") when using the contrapositive.

• Jun 11th 2010, 04:55 AM
Ackbeet
Your statement of the contrapositive is incorrect. The only way to state the contrapositive is to have the "not" affect the entire "if" part. The correct contrapositive statement would be the following:

Let $\displaystyle \vec{x}\in\mathbb{R}^{n}$. If $\displaystyle \vec{x}\not=0$, then it is not the case that for every $\displaystyle \vec{y}\in\mathbb{R}^{n}$, we have $\displaystyle \vec{x}\cdot\vec{y}=0$.

The way you worded the contrapositive, the "not" applied only to the inequality of the dot product, and it did not apply to the quantifier "for all". Do you see what I mean?
• Jun 11th 2010, 04:56 AM
Prove It
Quote:

Originally Posted by CropDuster
Hi all,

So I've just started a linear algebra course, and this is my first exposure to formal proofs.

Here is a question from my book:

Prove the following by contrapositive: Let X be a vector in R^n. If XdotY=0 for every vector Y in R^n, then X=0.

So the contrapositive is: If X=/=0, then XdotY=/=0 for every vector Y in R^n. Right?

But if Y can be every vector in R^n, then couldn't Y be the zero vector, hence XdotY=0.

I can't figure out anyway around this fact. I have been specifically told not to alter any restrictions (ie "for every vector Y in R^n") when using the contrapositive.

I assume that it would have been stated in the beginning that $\displaystyle \mathbf{Y}\neq \mathbf{0}$
• Jun 11th 2010, 05:18 AM
CropDuster
Quote:

Originally Posted by Ackbeet
Your statement of the contrapositive is incorrect. The only way to state the contrapositive is to have the "not" affect the entire "if" part. The correct contrapositive statement would be the following:

Let $\displaystyle \vec{x}\in\mathbb{R}^{n}$. If $\displaystyle \vec{x}\not=0$, then it is not the case that for every $\displaystyle \vec{y}\in\mathbb{R}^{n}$, we have $\displaystyle \vec{x}\cdot\vec{y}=0$.

The way you worded the contrapositive, the "not" applied only to the inequality of the dot product, and it did not apply to the quantifier "for all". Do you see what I mean?

Ugh! Of course! (smacks palm against head) that makes perfect sense. Thank you so much!

And no, it was not stated that $\displaystyle \vec{y}\not=0$
• Jun 11th 2010, 05:20 AM
Ackbeet
You're very welcome. Have fun!
• Jun 11th 2010, 08:53 AM
HallsofIvy
Quote:

Originally Posted by Prove It
I assume that it would have been stated in the beginning that $\displaystyle \mathbf{Y}\neq \mathbf{0}$

No, it said "for all y" which includes y=0 but also includes y not equal to 0.