Prove square root is irrational

Prove that for any positive integer $n$, then $$\sqrt{\frac{n}{n+1}}$$ is irrational.

Solution

If $n+1$ is a square then we only need to show that \sqrt{n} is irrational, which amounts to showing that it's not a square. This follows the fact that if $n+1$ and $n$ are both squares then $n=0$, contradicting problem condition.

If $n+1$ is not a square, then it has prime factors with odd powers in $n$. Let $k$ be the largest of such factor and $a$ is its power. Meaning, $k^a | n+1$ but $k^{a+1}$ does not divide $n+1$.

Now suppose there exist $p,q$ relatively prime such that $$\sqrt{\frac{n}{n+1}}=\frac{p}{q}$$ $$(n+1)p^2 = nq^2$$ Because $k^a | LHS$ then $k^a | RHS$. If $k^a | n$ then $k | (n+1)-n = 1$, a contradiction (because $k$ is a prime).

And because $k$ is prime then $k^a | q^2$ which means $k^{a+1} | q^2$ (because $a$ is odd). This means that $k$ divides $p$, contradicting our assumption that $p,q$ are relatively prime.

Sum over hypercube

Let $n$ be a positive integer, and suppose $a = a_1, \dots, a_{2019}$ is a sequence of integers each ranging from $1,\dots,n$. Determine the sum of $$\sum_a \frac{a_1 - a_2 + a_3 + \dots -a_{2018} + a_{2019}}{a_1 + \dots + a_{2019}}$$ Where the sum is taken over all possible sequences in that range.