Requesting Clarification Regarding A Proof Drafted By A Novice fellow StackExchange users.
I am a mathematics novice with a learning disorder requesting verification of a proof that I have devised. The majority of my "mathematical writing" has consisted entirely of unconventional and otherwise useless theorizing.
The proof is devised to show that the product of variables a and b is equivalent to variable c.
This is the proof:
$(a ∙ b) = c ⇔ Sigma(1,a,b) = c$
I am attempting to prove that the product of a and b is equivalent to c if and only if the series of a sequence beginning from the 1(th) term to the (b)th with 'a' consecutively added to itself is equivalent to 'c.'
I apologize if I am using the website's format incorrectly. I am rather inexperienced with the subject and would like to know if my mediocre proof is somewhat accurate.
The only condition that I was able to devise in which this would not apply was the premise of negative quantities. Although, I am somewhat confused on how to adjust this accordingly.
Also, if possible, could any of the experienced users on this StackExchange subsidiary refer me to an objectively-detailed resource that is suited for novices?
Thank you for reading.
 A: The easiest way to prove that
$$ \sum_{i = 1}^n a = an$$
is to note that adding $a$ to itself $n$ times gives $an$. If you're looking for a formal proof, then you'll want to induct on $n$. The induction principle states that if $P(1)$ true, and $P(n)$ implies that $P(n + 1)$ is true for all integers $n$, then $P(n)$ holds for all integers $n$. Here, $P(n)$ is the statement that the sum from $1$ to $n$ of $a$ is $an$. Since $P(1)$ is obvious, suppose that $P(n)$ holds for some integer $n$. Then
$$ \sum_{i = 1}^{n + 1} a = \sum_{i = 1}^n a + a = an + a = a(n + 1) $$
where the third step follows from the hypothesis that $P(n)$ is true. If you're concerned about these kinds of proofs, I would suggest picking any textbook on discrete mathematics or proofs (sometimes called introductions to advanced mathematics by textbook publishers).
A: That is true if
$b$ is an integer:
adding up $b$ copies of $a$
gives the same result as
$b$ multiplied by $a$.
I am not sure why 
this is important to you,
but you are correct.
A: Your proof is correct also for negative numbers, for example, a=-3.
