Is even divided by even a rational or irrational number?

Question

For any rational number, $\frac{p}{q}$ , $p$ and $q$ should be integers, $q\neq0$ and $p,q$ should not have any common factors. Now, if we have two even numbers, say $2m$ and $2n$ where $m$ and $n$ are integers. $$\frac{\text{even}}{\text{even}}=\frac{2m}{2n}=\frac{m}{n}$$ where($\frac{m}{n}$) nature is still unknown. So, what nature does $\frac{\text{even}}{\text{even}}$ have, rational or irrational?

additional reference: https://www.quora.com/Is-2-4-rational

Answer 1

The fact is that if you have $p$ and $q\neq 0$ integers then $|p|$ and $|q|$ are positive integers, or $p=0$ when $\frac pq=0$ is rational. If you cancel a common factor $2$ to obtain $|m|\lt p$ and $|n|\lt q$ you have smaller positive integers.

You can keep dividing common factors and obtain a decreasing sequence of positive integers for numerator, and another for denominator. Since a decreasing sequence of positive integers must eventually be constant this process will terminate in a (form of the) fraction in which numerator and denominator have no common positive integer factor apart from $1$. This is what you have defined as a rational number.

Answer 2

If $m$ and $n$ are integers and $n \ne 0$, then $\frac{m}{n}$ is rational.

Answer 3

https://artofproblemsolving.com/wiki/index.php/Proof_by_contradiction

Proof (by contradiction) that $\sqrt{2}$ is irrational.

Assume for a minute, that $\sqrt{2}$ is rational, that is: it can be written as $\sqrt{2}=p/q$ for two specific integer numbers $p$ and $q$, which don't share any common divisors greater than $1$.

If now $\sqrt{2}=p/q$, then we can square to get $2=p^2/q^2$ or multiply both sides by $q^2$ and get $2q^2 = p^2$. Now the number $q^2$ is an integer, because $q$ is an integer, and the number $2q^2$ is an even number, because we are multiplying the integer $q^2$ by $2$ and the result of multiplication of any integer number by $2$ gives an even number.

If now $2q^2$ is an even number, then $p^2$ is an even number, because $2q^2=p^2$. Thus the number $p$ is an even number, because $p^2$ is an even number.

If now $p$ is an even number, then we can express it as $p=2k$ for some integer $k$ and if $p=2k$, then $p^2 = 4k^2$.

So, we get $q^2=2k^2$, because $2q^2=p^2$ and $p^2=4k^2$. If we have $q^2=2k^2$, then $q^2$ is an even number. If $q^2$ is an even number, then $q$ is an even number.

If $p$ is an even number and $q$ is an even number, then $p/q$ can be further simplified. This is a contradiction to the assumption, that $p$ and $q$ should not share any common divisors greater than $1$. Thus, the assumption is false, it's opposite is true. This means $\sqrt{2}$ must be irrational.