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.


closed as off-topic by Saad, A. Goodier, Inactive - avoiding CoC, Dietrich Burde, Xander Henderson Apr 8 '18 at 15:12

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    $\begingroup$ Are the operations the usual addition and multiplication of real numbers? $\endgroup$ – carmichael561 Apr 8 '18 at 3:20
  • $\begingroup$ Yeah, they are the binary operators addition and multiplication. $\endgroup$ – Star Platinum ZA WARUDO Apr 8 '18 at 3:21
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    $\begingroup$ "I think it should be a field as it satisfies the multiplicative and additive identities" Seriously? What do you think the multiplicative and additive identities are? $\endgroup$ – fleablood Apr 8 '18 at 3:24

As the other answers point out, there are many reason why $\mathbb{I}\subseteq\mathbb{R}$ is neither a field nor a ring. However, it might be insightful to consider the smallest ring $\mathcal{I}$ that $\mathbb{I}$ is contained in. Our strategy will be to add elements to $\mathbb{I}$ whenever $\mathbb{I}$ fails to be a ring, correcting this failure.

First, we clearly need $0,1\in\mathcal{I}$ in order to have identities in both operations. Next, we should check closure in $(\mathcal{I}, +)$ and $(\mathcal{I}, \cdot)$.

Question: Is it required that $\mathbb{Z}\subseteq\mathcal{I}$?

Yes! The user ml0105 pointed out in their answer that $\langle 1\rangle = \mathbb{Z}$. Since $1\in\mathcal{I}$, we must have $\mathbb{Z}\subseteq\mathcal{I}$.

Question: Is it required that $\mathbb{Q}\subseteq\mathcal{I}$?

As it turns out, yes! Let $a,b\in\mathbb{Z}$ be coprime integers, and consider the element $\frac{a}{b}\in\mathbb{Q}$. Now, take your favorite irrational number $x\in\mathbb{I}$ and consider the elements $ax, \frac{1}{bx}\in\mathbb{I}$. We want $(\mathcal{I},\cdot)$ to be closed, and so $\left(ax\cdot\frac{1}{bx}\right) = \frac{ax}{bx} = \frac{a}{b}\in\mathcal{I}$. Thus $\mathbb{Q}\subseteq\mathcal{I}$.

Therefore, we have

$$\mathbb{I}\subseteq\mathcal{I}; \quad \mathbb{Q}\subseteq\mathcal{I}. $$

This implies that $\mathbb{R}\subseteq\mathcal{I}$, and so $\mathcal{I} = \mathbb{R}$. Hence, any subring of $\mathbb{R}$ containing every irrational must be $\mathbb{R}$ itself.


It neither a ring nor a field for the law induced by the real since it does not contains neutral elements for the addition and multiplications.


Irrationals are not closed under addition or multiplication.

for example $$\sqrt 2 \sqrt 8 =4 $$

Thus they do not form a field or a ring.


$\sqrt 2$ and $10-\sqrt 2$ are two irrational numbers whose sum is not an irrational number. So, with the usual addition, it's not a ring or a field because it's not closed under addition. It also contains no additive identity.


Counterexamples are useful in illustrating concepts.

Consider $\pi$. Observe that the multiplicative inverse of $\pi$ (under the inherent operation of multiplication in $\mathbb{R}$) is $\frac{1}{\pi}$. Now $\pi \cdot \frac{1}{\pi} = 1$, which is clearly in $\mathbb{Q}$. So the irrationals are not even closed under multiplication, inherent from $\mathbb{R}$. Thus, they cannot form a ring.

Similarly, using the standard addition operation from $\mathbb{R}$, we have that $\pi + -\pi = 0$. So the irrationals do not form an additive group.

Edit: If you throw in $0, 1$ into your set, then $\langle 1 \rangle \cong \mathbb{Z}$ is contained in your structure. So consider $\pi \cdot \frac{1}{2\pi} = \frac{1}{2}$, which is clearly not an irrational or integer.

  • $\begingroup$ Sorry, my bad. Is it okay to edit a question after someone has answered? I'd like to include 0 and 1 in the set as well. $\endgroup$ – Star Platinum ZA WARUDO Apr 8 '18 at 3:28
  • $\begingroup$ I don't personally have an objection, but I think you are trying to force structure where there is none. There is a good discussion here on the (lack of) algebraic structure of the irrationals: math.stackexchange.com/questions/649217/irrationals-a-group $\endgroup$ – ml0105 Apr 8 '18 at 3:34
  • $\begingroup$ If you include 0 and 1 and you still want it to be a ring you are going to also have to include all the integers (so it is closed under addition). To be a field it would require all real numbers. $\endgroup$ – fleablood Apr 8 '18 at 5:03
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    $\begingroup$ "which is clearly not in Q" -> you might want to correct this typo $\endgroup$ – user370967 Apr 8 '18 at 8:26
  • $\begingroup$ Fixed. Thanks for catching that! $\endgroup$ – ml0105 Apr 9 '18 at 1:10

"I think it should be a field as it satisfies the multiplicative and additive identities"

Why on earth do you say that? It does not contain $0$ which is the additive identity and it does not contain $1$ which is the multiplicative identity, so it absolutely does NOT satisfies the multiplicative and additive identities.

So it is neither a ring nor a field.


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.


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