Are all numbers rational?

Question

Is it correct to say that some number $A$ is rational because $A$ is $\frac{A}{1}$?

Because the only restriction I had see for rationality is $A=\frac{a}{b}$ with $b\neq0$, but in that way I can say $\pi$ is rational because $\pi=\frac{\pi}{1}$. What's the problem here?

Answer 1

$a$ and $b$ must be integers. $\pi$ is not

Answer 2

The definition of a rational is: $ a \in \mathbb Q \Leftrightarrow \exists (n,m) \in \mathbb Z \times \mathbb N ^ *, a = \frac n m$

And no not all real numbers ($\mathbb R $) are rational. It is easy to show that $ \sqrt 2 $ is not (ref. on Wikipedia)

• assume that $ \sqrt 2 $ is a rational number, meaning that there exists a pair of integers whose ratio is $ \sqrt 2 $

• if the two integers have a common factor, it can be eliminated using the Euclidean algorithm

• then $ \sqrt 2 $ can be written as an irreducible fraction $ \frac a b $ such that $a$ and $b$ are coprime integers (having no common factor).

• it follows that $ \frac {a^2} {b^2}=2 $ and $ a^2 = 2 b^2 $

• therefore, $a$ is even because it is equal to $2b^2$ and $b^2$ is integer. Say $a=2k$

• but then $4k^2 = 2b^2$ so $b^2=2k^2$ - for same reason b is even

• $a$ and $b$ share 2 as common divisor which is excluded since they a coprime

  Conclusion: $\sqrt 2$ cannot be rational

Answer 3

$A=\frac A1$ proves rationality of (left-hand side) $A$ iff (right-hand side) $A$ is integer.

Which implies integers are rational.