Prove that $\forall a, b \in \mathbb{Z}^+. a \neq b. \exists \text{ infinite } n \in \mathbb{Z}^+, \gcd(a+n, b+n) = 1$
This question was asking to prove that for every two distinct positive numbers $a$ and $b$ there exist infinite positive numbers such that $\gcd(a +n, b+n) = 1$.
I tried proving it. This is my proof.
Proof:
Lemma 1: There exist infinitely many relative primes to a number which isn't 0. Let the number be $m$ and $m \neq 0$. Let $n$ range over all integers and $r$ be such that $0 < r < m$. So the set $S = \{mn + r\}$ consists of all such numbers.
Let $a$ and $b$ be the two numbers. Since they are different let us assume that a is the lesser one. So, $b = a + k$ and $k> 0$.
We know for every number, except 0, there exist infinitely many relatively prime numbers but none of $a, b, k$ are 0 over here by Lemma 1.
Let $s$ be any of those numbers and $s = a + n$.
So,
$$\gcd(a + n, k) = 1 \Longleftrightarrow \gcd(a+n, a+n +k) = 1 \Longleftrightarrow \gcd(a+n , b+n ) = 1$$
Thus, the proposition is proved.
Is my proof correct? Could you suggest some better way to prove this?
Your proof seems to work, but here is (in my opinion) a simpler one:
Let $p > \max(a,b)$ be any prime number strictly bigger than both $a$ and $b$, and then set $$n = p- \max(a,b).$$
Suppose that, $a > b$, then $n = p-a$ and $$\gcd(a+n, b+n) = \gcd(p, b+n) = 1$$ because $b+n < a+n = p$ and $p$ is a prime number.
Of course, there are infinitely many prime numbers bigger than $\max(a,b)$.
I hope this helps $\ddot\smile$