Friday, October 5, 2018
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.
Labels:
infinite descent,
irrational,
modulo,
Number Theory,
square root
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