# logic question, understanding contrapositive

I'm tring to understand supposedly simple logic question.

I'm given the definition of vector space (listed below)

Then I want to prove the following

Suppose a in R, v in V and av = 0. Prove that a =0 or v = 0 (S1)

Proof goes like this,

First I prove: L1) a0 = 0 for every a in R (a number times the vector 0 is 0)

a0 = a(0+0) = a0 + a0
a0 + (-a0) = a0 + a0 + (-a0)
0 = a0 (done)


Then I prove a != 0 => v = 0

Here's the proof (pf1)

suppose a != 0
since av = 0, 1/a * av = 0 (by L1)
1/a * av = (1/a*a)v (by associativity) = (1)v = v (by multiplicative identity)
= 0 (done proving a!=0 => v=0)

since we have shown a!= 0 => v=0
and if we use contrapositive
we can also say v != 0 => a = 0


Therefore we have shown

a != 0 => v = 0
v != 0 => a = 0


Which seems to suggest that (a =0 or v = 0) (S1) that I was trying to prove in the first place.

or (pf2) I could argue somewhat differently,

a =0 or a!=0.
when a!=0, v = 0 as shown above.
therefore a = 0 or v = 0


or (pf3) I could argue,

when a = 0, then S1 is trivially true,
if a != 0  then v = 0 as shown above.
therefore a = 0 or v = 0


The first proof is somewhat weird in a sense that,

In other words, I thought that the meaning of "the contrapositive statement is logically equivalent to the original statement", was roughly similar to the idea that "contrapositive statement doesn't say anything new beyond the original statement".

Yet I don't feel confident my understanding is correct in this case.
Because, if someone proved only one way (p=>q, in pf1) or prove only half of the case in pf2 or pf3, I would have thought proof is not thorough

Q2. Since I proved a!=0 => v=0, i know v!=0 => a =0 is true (if I accept contrapositive argument is valid)
But then, it is not trivial to prove v!=0 => a=0 by tracing back the argument (a!=0 => v=0)

I can do something like,
v != 0
v = 1v (by multiplicative identity)
1v = (1/b*b)v = (1/b)(bv) != 0 for any b != 0

and we know bv != 0 by L1's contrapositive
ie, we have shown for any b!= 0 => bv != 0.
Then, again by contrapositive,
we can say bv = 0 => b = 0 (av=0 => a=0)


So it doesn't seem to be trivial to prove the other way.
More specifically, for someone hesitant to accept contrapositive argument, there doesn't seem to be another way to prove the contrapositive argument ($$v!=0 => a =0$$) without using contrapositive reasoning yet again.

Which I find a bit annoying, although I don't think it affects the legitimacy of the contrapositive reasoning, I hoped to find a way to confirm the legitimacy of the contrapositive reasoning by alternative proof.
My reasoning is that, there might be an alternative system with axioms (alternative way of thinking) but not including the contrapositive reasoning as an axiom to derive the same statements which is derived by the system with contrapositive reasoning as an axiom.

If I try to understand the legitimacy of contrapositive reasoning, I find it assumes that the statement in question is either $$true$$ or $$false$$ or something similar.
I think it's called the law of excluded middle. And I guess I feel a little bit uneasy about the assumptions that it requires me to make.
I feel the law of excluded middle can be safely applied to this problem, but I feel uneasy if the law is universally applicable..

If I dig deeper, think harder on the proof of legitimacy of contrapositive (https://en.wikipedia.org/wiki/Contraposition), should I be able to answer my question1 and 2?

Is there something I can study to understand my uneasiness?

definition of vector space i use

• commutativity
• associativity
• multiplicative identity
• distributive property
• Too long and messy... Having said that, you want to prove that "if $av=0$, then either $a=0$ or $v=0$". But $p \lor q$ is equiv to $\lnot p \to q$. Thus, you have to assume $av=0$ and prove that "if $a \ne 0$, then $v=0$". Commented Apr 20, 2023 at 14:15

The hypothesis is that $$a v = 0_V$$ ($$a \in \mathbb{R}$$ and $$v \in V$$, say), and the claim is that $$a = 0$$ or $$v = 0_V$$. You proved that if $$a \ne 0$$, then $$\frac{1}{a} (a v) = \frac{1}{a} (0_V) \implies 1 v = v = 0_V$$. For the purpose of showing the initial claim, this is enough, since given any $$a \in \mathbb{R}$$, either $$a = 0$$ or $$a \ne 0$$ (in which case $$v = 0_V$$). QED.
The contrapositive $$v \ne 0_V \implies a = 0$$ does not add anything new to the proof (notice we did not use it at all in the above).
• I have a related question: If I accept contrapositive argument, I can derive $av=0 => a=0$ But if I don't accept, I don't seem to be able to derive it. Which I find a bit annoying, although I don't think it affects the legitimacy of the contrapositive reasoning, I hoped to find a way to confirm the legitimacy of the contrapositive reasoning by alternative proof. ... I did edited op about this. What do you think about my thought? I find it interesting to find "Many modern logic systems replace the law of excluded middle with the concept of negation as failure" in wikipedia.. Commented Apr 21, 2023 at 3:03
• @eugene That's an interesting idea, the fact that some statements can not be easily proven without the contrapositive. The easiest way to show, with hypothesis $a v = 0_V$, that $a \ne 0 \implies v = 0_V$ is indeed by showing the contrapositive, $v \ne 0_V \implies a = 0$. It hinges on the fact that a statement is either true, or it is false: $P \lor \neg P$ is a tautology. I agree with you, that lack of a straightforward alternate direct proof may be unsatisfying, but the contrapositive proof must be accepted. Commented Apr 23, 2023 at 19:03