Is the set of all Irrational Numbers a ring or a field? I would really appreciate a proof of either one.
I think it should be a field as it satisfies the multiplicative and additive identities and is commutative.

A ring requires a valid operation (addition) with an inverse (subtraction) and an identity or zero satisfying $x+0=x$ for all $x$.
Likewise a field requires a further operation (multiplication) which is distributive over addition (i.e. $a(b+c)=ab+ac$) and which also has an identity or unit satisfying $1\times x=x$ for all $x$, also having a unique inverse ($\frac{1}{x}$) for every $x$ satisfying $\frac{1}{x}\times x=1$
The other answers here all appear to assume standard binary functions of addition, multiplication etc.
But the irrational numbers $\Bbb{I}$ in themselves are only really become well-defined as an algebraic object once the operations of addition and multiplication are defined.
The irrational numbers are uncountable set and the complex numbers endowed with with standard addition and multiplication are a field.
Therefore a bijection $f:\Bbb{I}\to\Bbb{C}$ can be defined such that $f^{-1}(f(c_1)\cdot f(c_2))$ is a valid addition and $f^{-1}(f(c_1)\circ f(c_2))$ is a valid multiplication in which case $f^{-1}(0)$ is your additive identity and $f^{-1}(1)$ your multiplicative identity.
Once you have done so, $(\Bbb{I},\cdot,\circ)$ is a field and a ring.  So in actual fact you are right on both counts, although perhaps not in the sense you expected.
