# Contrapositive of a (if p and q, then r) statement?

1. Nov 8, 2012

### SithsNGiggles

1. The problem statement, all variables and given/known data
I hope this is the right place to post this.

For my linear algebra homework, I have to prove that
"If $\vec{u} \neq \vec{0}$ and $a\vec{u} = b\vec{u}$, then $a = b$."

2. Relevant equations

3. The attempt at a solution
I'm trying to prove the contrapositive, but I'm not sure how to alter the statement.

I think it's something like
"If $a \neq b$, then $\vec{u}=\vec{0}$ or $a\vec{u} \neq b\vec{u}$."

But then I don't know where to begin, so I'm guessing a proof by contrapositive isn't the right method to use here. It seems to me that a direct proof might me simpler, but I guess I wanted to challenge myself. Any ideas? Thanks.

2. Nov 8, 2012

### Zondrina

A direct proof would be simpler here. Basically what you're trying to prove is the cancellation law.

I'm sure you're in a field here, particularly a vector space. What do you know about the properties of a vector space? Think about axioms.

Last edited: Nov 8, 2012
3. Nov 8, 2012

### SithsNGiggles

Well we know quite a few axioms, and we also know other properties as a result of earlier proofs.

My "direct proof" goes like this:
Let $U$ be a vector space and $\vec{u} \in U$. Let $a,b$ be arbitrary scalars.

Given that $a\vec{u} = b\vec{u}$, by substitution we have
$\frac{1}{a} (a\vec{u}) = \frac{1}{a} (b\vec{u})$.

By axiom 9 (an associative property),
$(\frac{1}{a} \cdot a) \vec{u} = (\frac{1}{a} \cdot b) \vec{u}$
$(1) \vec{u} = \frac{b}{a} \vec{u}$

So for equality to hold, we must have
$1 = \frac{b}{a}$, and from this it follows that
$a = b$. Q.E.D.

I wrote this up pretty quickly, so I agree that the direct proof is easier. Is there a way to work with the contrapositive, though? It doesn't look like it to me.

4. Nov 8, 2012

### Zondrina

Wait what substitution? You multiplied by the left inverse of a, that's not a substitution. Also, your vector u doesn't automatically cancel out of thin air. The point is, you have to argue as to why u is going to cancel.

You have that u is non-zero and au = bu. You want to show that a = b.

Since u is a non-zero vector... hmm ;). Another hint is to abandon the fraction notation and use inverse notation.

5. Nov 8, 2012

### SithsNGiggles

My professor considers "multiplying both sides by the same thing" to be substitution. I think what he means by that is since, for example, for some vectors u and v, if u=v then au = av. Does that clear it up?

As for getting rid of the vector u, I just matched up the scalar coefficients. I don't how else to describe what it is I did there... I guess I'm saying I don't actually get rid of u. How is this normally shown in a proof knowing that u is non-zero?

Edit: I think I can answer my own question.
$(1)\vec{u} = \frac{b}{a}\vec{u}$
$(1)\vec{u} + (-\frac{b}{a}\vec{u}) = \frac{b}{a}\vec{u} + (-\frac{b}{a}\vec{u})$
$(1)\vec{u} - \frac{b}{a}\vec{u} = \vec{0}$
$(1 - \frac{b}{a})\vec{u} = \vec{0}$

Since u is non-zero, it follows that (1 - b/a) = 0, and then I solve to get a=b. Is this right?

Last edited: Nov 8, 2012
6. Nov 8, 2012

### Zondrina

If u is non-zero, then it has an inverse? What happens if you apply the inverse to what you're trying to show.

7. Nov 8, 2012

### SithsNGiggles

We haven't learned anything about inverse vectors, but I think I get what you're saying here. I figured another way which I added to my last comment.

Edit: Oops, I misinterpreted "inverse." I was thinking multiplicative for some reason...

8. Nov 8, 2012

### Zondrina

Ah you haven't been taught about it. The problem with what you're doing is you're assuming a has an inverse. What if a=0? The whole thing goes down the drain.

9. Nov 8, 2012

### SithsNGiggles

Sorry, I was mistaken. We actually did learn about the inverse of a vector, namely the negative of a vector. I don't know... I knew of the additive inverse but it didn't quite register.

And it looks like I have to provide a proof for when either a or b are zero. Thanks for reminding me.

10. Nov 8, 2012

### Dick

Why are you dividing by anything? You've got (a-b)u=0 where u is nonzero. Isn't that enough?

11. Nov 8, 2012

### 5hassay

I believe the contrapositive here would be what you guessed it to be. As others have said, I think its easier to prove directly. For fun, let's prove the contrapositive, worded better for proving:

"For any vector $u$, in some vector space $V$, and any scalars $a,b$, in some field $F$, consider the equation $au=bu$, with truth of equality undetermined. If $a \neq b$, then either $u=0$ (so that $au=bu$), or $au \neq bu$."

Let's begin the proof. For the first case, suppose $u\neq 0$. Because by assumption $a\neq b$, we thus have $a-b\neq 0$. Therefore, we can conclude that

$(a-b)u\neq 0$

From this we can easily conclude by moving terms around that it must be that $au\neq bu$. (How?) Next, suppose instead (working on the second and final case) that $u=0$. To be clear, we want to draw from this that $au=bu$. Of course, that is obvious.

Note that, because this is a mathematical statement, the "or" component means we could have both parts of the conclusion of the statement, i.e. both that $u=0$ and $au \neq bu$. However, this cannot be. (Why?) So, I guess for this theorem (problem), it must be that we can only have one or the other condition, i.e. it is an "exclusive-or" conclusion, rather "inclusive-or."

This concludes the proof.

My bad if I did anything wrong. I'm a bit tired, but I feel like there's enough good info there to give reason for me to submit this reply, anyway

12. Nov 9, 2012

### SithsNGiggles

I looked over the proof last night and I realized that now (that I have (a - b)u = 0). Thanks!

And to 5hassay, I think I follow. Since it's an exclusive-or statement, would that mean I would only have to show one for the statement to hold?

13. Nov 9, 2012

### 5hassay

If I understand correctly, since our conclusion is a "result 1 OR result 2" conclusion, and it appears we can only have one or the other, but not both, i.e. the "OR" as an "exclusive-or," then we must show both that result 1 can occur while result 2 does not occur, and, likewise, that result 2 can occur while result 1 does not occur