§ GCD is at most difference of numbers

• assume WLOG $l< r$. Then, Let $g \equiv gcd(l, r)$. Claim: $g \leq r - l$.
• Proof: we have $g \div r$ an $g \div l$ by definition, hence we must have $g \div (r - l)$, and $g$, $(r-l)$ are nonnegative. So $g \leq (r - l)$.
• Intuition: the gcd represents the common roots of $l, r$ in Zariski land. That is, if $l, r$ are zero at a prime then so is $r - l$.
• So, the GCD equally well represents the common roots of $l$ and $(r - l)$.
• Now, if a number $x$ vanishes at a subset of the places where $y$ vanishes, we have $x < y$ (the prime factorization of $y$ contains all the prime factors of $x$).
• Since the GCD vanishes at the subset of the roots of $l$, a subset of the roots of $r$, and a subset of the roots of $(r-l)$, it must be smaller than all of these.
• Thus, the GCD is at most $r - l$.
• Why does GCD not vanish at exactly the roots of $r-l$? If $l$ and $r$ both take the same non-zero value at some prime then $(r - l)$ does too. But this is not a loacation where $l$ and $r$ vanish.