(N.B. : The answer to the question was already given by both fleablood and gnasher729, so I'm just going to summarise a few points from this discussion for those who encounter problems like this in proof-writing.)
The problem I have encountered here is the confusion whether there exists a proof that there can be no number greater than the GCD of two numbers that divides them. I tried to prove it, but I did it the worng way, and hence ended up with a stupid write-up(LOL). So here are a few points for people to get away from puddles while proving.
- One can't prove a definition using its properties, as a definition is self-explanatory and is also an abbreviation of the properties. As from gnasher729's answer, if $x$ is defined as the largest number with a property $X$, then it is obvious that no number $x' > x$ can have that property.
- Trying to use a property of a number (here, $\gcd(a,b)$ and the property that $\gcd({{a}\over{\gcd(a,b)}},{{b}\over{\gcd(a,b)}}) = 1$) to prove its definition is useless as you are actually trying to use the property to prove itself and thus prove the definition, which is quite senseless and has no meaning. Therefore, try assuming the converse of that property (in this case, the GCD of the quotients is $1$) to try proving it (and still I won't get it in my case as GCD is defined to be so, which means negating the property and trying to prove is equivalent to saying that the GCD of $a$ and $b$ is another number greater than the one we take).
I didn't go for any formal training in number theory nor did I have time to learn fully from an online book that a forum member here offered me to take as a course, but I attempted the proof for "There exists no number greater than the GCD of $a$ and $b$ that divides both of them" by myself with the small amount of my knowledge in set theory and a bit on prime numbers. Here's how I did it, and I'd like anybody to tell me if I did it right or if I made a mistake.
The proof:
Let $\gcd(a,b) = d \implies a = dm, b = dn \space$ & $\space\gcd(m,n) = 1$
$ \exists k: k > d, k \mid a , k \mid b $
Let $pf(x) = \text{prime factors of a number} \space x \space$ $\text{considering each occurrence of a prime in the factorisation as a distinct one}$
$\because \gcd(a,b) = d, k \mid a \space \text{and} \space k \mid b, pf(d) \subseteq pf(k)$
$\implies k = ds , s >1 (\because k > d)$
$ \implies ds \mid a, ds \mid b$
$\implies s \mid m, s \mid n$
$\because \gcd(m,n) = 1 (i.e., pf(m) \cap pf(n) = \phi), s = 1$
Thus, $k \mid a \space\text{and}\space k \mid b$ only if $k = d$ (since $s = 1$ contradicts our assumption that $s > 1$).
$\therefore, \nexists k : k > \gcd(a,b), k \mid a, k \mid b$
The Result
As a result of posting this question, it seems to me that this will be a question for novices in proof-writing may view for reference on how to write proofs.
Best Answer
Let $x$ be the largest number having a property $X$. Then no $x' > x$ has the property $X$, because if $x'$ had the property $X$, and $x' > x$, then $x$ wouldn't be the largest number with that property.
In your case the property $X$ is "is a common divisor of $a$ and $b$". The GCD of $a$ and $b$ is defined as the largest x with that property. Therefore no $x' > x$ can have that property. There's nothing to prove. We don't even have to look at anything about the GCD except that it is "the largest number with some property".
Go into a room with $50$ people. Find the tallest person in the room. Then I can guarantee that nobody in the room is taller. I don't have to measure anything, it's just obvious that nobody in the room can be taller than the tallest person in the room.