A very pedestrian question about proving $a \implies b$?

Question

I have a very silly doubt about proofs, I believe I know the correct way to do it but I want to confirm if it is indeed correct. Take as an example the following proposition:

$x$ is a linear combination of $\{x_i \}$ if $\{x_i\}\cup \{x\}$ is linearly dependent.

$\{x_i \}$ is a set of linearly independent vectors.

My guess Is that I have to perform the following steps:

• Assume "$x$ is a linear combination of $\{x_i \}$" is true;
• Check if "$\{x_i\}\cup \{x\}$ is linearly dependent" is true.

My question is: Do I have to check what happens if "$x$ is a linear combination of $\{x_i \}$" is false or that is irrelevant?

Answer 1

Your strategy is correct: assume the premise ("if $\ldots$") and try to prove the consequent ("then $\ldots$"). You don't have to check anything regarding when the premise is false.

To pick a simpler example: if you want to prove that "Every lion has claws" (i.e. "if x is a lion, then x has claws"), you don't need to worry about animals that aren't lions.

Answer 2

It's irrelevant, and I think the problem is you've phrased this backwards - it's really if $\{x_i\} \cup \{x\}$ is linearly dependent, then $x$ will be a linear combination of $\{x_i\}$. Try and think of theorems in terms of conditions and the things which are true when those conditions hold: for this example, it might to help to break down all the constituent parts.

Let $\{x_i\} \subset V$, where $V$ is a vector space, be a set of linearly independent vectors. $\forall x \in V$, if $\{x_i\} \cup \{x\}$ is linearly dependent, then $x$ is a linear combination of $\{x_i\}$.

Moreover, we know a little something about what happens if the conclusion is false. If we have $p \implies q$, then $\lnot q \implies \lnot p$. So if $x$ is not a linear combination of $\{x_i\}$, $\{x_i\} \cup \{x\}$ is not linearly dependent.

When might it be relevant what's going on with what you asked? If your theorem as stated was an "if and only if". Then you have to check forwards implication and backwards.