I was reading 'Mathematical circles' and it used parity to solve a lot of questions. I was curious that, is there a standard definition of parity over rational numbers or maybe real numbers?

By 'standard definition' do you mean accepted "universally accepted by mathematicians", or "most common extension"? – R. Burton Nov 25 '18 at 16:24

A widely accepted definition – Number_Fanatic_SoyBoy Nov 25 '18 at 16:25
2 Answers
The notion of parity can be extended to many rings  but certainly not all rings, since if $\,2\,$ has an inverse $\,u\,$ then $\,2u = 1\ \Rightarrow 2\mid 1\mid x\,$ so everything is "even". So we can't extend it to all rationals, but if we restrict to the subring of rationals with odd denominator then integer parity has a unique extension: the parity of $\,m/(2n+1)\,$ is the parity of $\,m,\,$ by $\bmod 2\!:\ (2n+1)^{1}\equiv 1^{1}\equiv 1.\,$
More generally we can always apply parity arguments in any ring which has $\,\Bbb Z/2 = $ integers $\!\bmod 2\,$ as a homomorphic image, e.g. the Gaussian integers $\,\mathbb Z[i],\,$ where the image $\ \mathbb Z[i]/(2,i\!\!1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\ a+b\,i\ $ is even $\iff a\equiv b\pmod{\! 2}\ $, i.e. if $\ a+b\,i\ $ maps to $\,0\,$ via the above isomorphism, which maps $\ 2\to 0,\ i\to 1\,$.
Similar ideas work for many rings of algebraic integers (though parity need not exist nor be unique in general). See this answer for further discussion. See also this answer which uses parity in $\,\mathbb Z[\sqrt{5}]\,$ to show the integer $\,(9+4\sqrt{5})^n + (94\sqrt{5})^n\,$ is even, and see this answer for parity in an ordered ring with infinite elements.
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There is no standard definition of parity for rational numbers. You can look at the power of $2$ that appears when you write a nonzero rational number in lowest terms. That function is usually written as $\nu_2$, so for example $$ \begin{align} \nu_2(8) &= 3 \\ \nu_2(24) &= 3 \\ \nu_2(5/8) &= 3 \\ \nu_2(5/3) &= 0 \\ \end{align} $$ Then a nonzero integer $n$ is even just when $\nu_2(n) > 0$.
You can generalize by defining $\nu_p$ for any prime $p$.
There's no good way to do any of this for real numbers.
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I think it does have a name, but I don't know it. See https://en.wikipedia.org/wiki/Padic_number – Ethan Bolker Nov 25 '18 at 16:27