proving irrationality by using root 2

Question

Suppose $a$ is a positive integer such that it's square root is between a positive integer $x$ and $x+1$. ($x<\sqrt{a}<x+1$) Prove that $a$ must be irrational.

Okay so I have some idea of how to advance here. By using the irrationality prove of $\sqrt 2$ (if $\sqrt 2$ can be written as a rational fraction in lowest term $\frac{m}{n}$ then $\sqrt 2$ can be also written as $\frac{2n-m}{m-n}$ which would be a contradiction as this can be simplified)maybe we can extend this to every a...

After thinking a bit, I think I figured out this explanation: $x<\sqrt{a}<x+1$ implies that a is not a perfect square. However I have a difficult time trying to explain for every positive integer a such that $a$ is not a perfect square $\sqrt{a}$ is irrational: this is common sense but I just cannot come up with a good and rigorous explanation. I think I need to use the same technique of proving $\sqrt 2$'s irrationality but I just cannot apply this to $a$...

Answer 1

If you want to prove that $\sqrt a$ is irrationnal. Assume that is not and take $m,n$ positive integers such that $gcd(m,n) = 1$, then $a = \frac{m^2}{n^2}$ and $m^2 = an^2$ this implies that $n | m^2$ and, by the Gauss theorem $n | m$. Having $gcd(n,m) = 1$ so $n= 1$. So $\sqrt a = m \in \mathbb N$

Answer 2

For any non-zero integer $A$ and any prime number $p$ let $D(A,p)$ be the largest $n\in \{0\}\cup \Bbb N$ such that $p^n$ divides $A.$ That is, $Ap^{-D(A,p)}\in \Bbb Z$ and $Ap^{-1-D(A,p)}\not \in \Bbb Z.$

For non-zero integers $A,B$ and prime $p$ we have $$(\bullet)\quad D(AB,p)=D(A,p)+D(B,p).$$ In particular $ D(A^2,p)=2D(A,p).$

Suppose $1<A\in \Bbb N.$ Then the set $S$ of prime divisors of $A$ is not empty, and $A=\prod_{p\in S}\;(p^{D(A,p)}).$ If $D(A,p)$ is even for every $p\in S$ then $D(A,p)/2\in \Bbb N$ for each $p\in S,$

so $\sqrt A=\prod_{p\in S}\;(p^{D(A,p)/2})\in \Bbb N,$

so $A$ is the square of the natural number $\sqrt A.$

Therefore if $A\in \Bbb N$ and $A$ is not the square of a natural number then $1<A$ (obviously), so by the preceding paragraph, $A$ has at least one prime divisor $p$ such that $D(A,p)$ is odd.

Let $a,x\in \Bbb N$ with $x<\sqrt a<x+1.$ Then $a$ is not the square of any $y\in \Bbb N$ because (i) if $y\leq x$ then $y^2\leq x^2<a,$ while (ii) if $y>x$ then $y\geq x+1,$ implying $y^2\geq (x+1)^2>x.$ So, by the preceding paragraph, let $p$ be a prime divisor of $a$ such that $D(a,p)$ is odd.

Now by contradiction suppose that $\sqrt a=m/n$ with $m,n\in \Bbb N.$ Then $n^2a=m^2,$ so by $(\bullet )$ we have $$2D(n,p)+D(a,p)=D(n^2,p)+D(a,p)=D(n^2a,p)=D(m^2,p)=2D(m,p).$$

This requires that the odd integer $2D(n,p)+D(a,p)$ is equal to the even integer $2D(m,p),$ which is impossible.