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Let $R=\{m+n\sqrt{2} \mid m,n\in \mathbb{Z}\}$. Prove that $R$ is a subring of the real numbers.

I just want to know how to get started really. My professor has used the same example for the past two months on everything we've done and hasn't really showed us any other examples on proving whether something is a subring or not.

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  • $\begingroup$ What have you tried? What you need to do is prove that $R$ is a subgroup of $\mathbb{R}$ that is closed under multiplication. (If you are going by the definition of ring that says it must contain a multiplicative identity, then you also need to show that $R$ contains 1). $\endgroup$ – Tom Oldfield Feb 18 '13 at 15:36
  • $\begingroup$ I know you have to prove the R is not the empty set, that any real number multiplied by any value in the subgroup belongs to the subgroup, and the any value in the subgroup subtracted from any real number is also in the subgroup. So those three things make it a subring. $\endgroup$ – USC Feb 18 '13 at 15:38
  • $\begingroup$ You seem a little confused. You need to show that $R$ is closed under addition and multiplication. What this means is that for any two elements $x,y \in R$, $x+y$ and $x\times y$ are in $R$. You don't need to show that any real number $a$ multiplied by, or added to an element of $R$ is also in $R$ since that is not true for subrings in general. This alone is also not enough to show that $R$ is a subring. You also have to show that for each element $x \in R$, it's additive inverse is in R, and that $0$ is in $R$, making it a subgroup. $\endgroup$ – Tom Oldfield Feb 18 '13 at 15:43
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Have you been given the definition of a ring? If not then there is a list of properties that must hold ont he wiki page: https://en.wikipedia.org/wiki/Ring_%28mathematics%29#Definition_and_illustration/wiki/Ring_%28mathematics%29

For example, is there an additive identity. Well, 0 would seem like an obvious choice. Is it true that $0 +(a+b\sqrt 2 ) = a+b\sqrt 2$? It is, since $0 +(a+b\sqrt 2 ) = (0+a) + (0+b\sqrt 2) = a+b\sqrt 2$. You need to check all the parts of the definition in similar manner. Feel free to ask in the comments if any parts of the definition don't make sense.

Edit: As pointed out in the comments, you do not need to prove that 0 is the identity since you are proving it is a subring, rather than just proving it is a ring. It should still serve as an example for proving the other parts of the definition are satisfied though.

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  • $\begingroup$ Actually, since the questions is to show that $R$ is a subring, it is not necessary to check things like whether or not $0$ is the identity. It is the identity in $\mathbb{R}$ so will be in $R$. All you need to do is check that $0 \in R$. Similarly, checking all the ring axioms would be tedious and unnecessary, all that needs to be shown is that $R$ is a subgroup and $R$ is closed under addition (and also possibly that $1 \in R$. $\endgroup$ – Tom Oldfield Feb 18 '13 at 15:48
  • $\begingroup$ So being closed under addition, you mean some like (m1+n1sqrt(2))+(m2+n2sqrt(2)) $\endgroup$ – USC Feb 18 '13 at 16:02
  • $\begingroup$ @USC: Exactly. You also have to show closure under multiplication, which takes two lines instead of one. You get distributivity for free. $\endgroup$ – Ross Millikan Feb 18 '13 at 16:03
  • $\begingroup$ Since you can rearrange that to be (m1+m2)+(n1sqrt(2)+n2sqrt(2)) and we know than that m1+m2 belongs to R and n1+n2 belongs to R so it is closed under addition? Sorry if I'm not getting it my professor just said this is the definition of a subring but never really show us how to prove it $\endgroup$ – USC Feb 18 '13 at 16:04
  • $\begingroup$ @TomOldfield Indeed, thank you for pointing that out - I will edit the answer accordingly. $\endgroup$ – Joe Tait Feb 18 '13 at 16:10
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You want to prove that $R$ is a subring of the real numbers. First note that this just means that you want to show that $R$ is subset and that $R$ itself is a ring. That $R$ is a subset (containing $0$ and $1$) is clear, so you "just" want to prove that that the set $R$ satisfies the axioms of a ring. Now you get most of these axioms for free. For example: $R$ being a subset of the real numbers means that it already satisfies the other properties of a ring. So you for example will have $a(b + c) = ab + ac$ simply because this is true in the real numbers. So the only things left for you to prove are:

(1) First you would need to prove that the sum of two elements in $R$ is again in $R$. Say $m_1 + n_1\sqrt{2}, m_2 + n_2\sqrt{2}\in R$. Then $$ m_1 + n_1\sqrt{2} + m_2 + n_2\sqrt{2} = (m_1 + m_2) + (n_1 + n_2)\sqrt{2} $$ and this is also in $R$. (2) Second need to prove that the product of two elements is again in $R$. So $$ (m_1 + n_1\sqrt{2})(m_2 + n_2\sqrt{2}) = \dots $$ (you can probably figure this out.)

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  • $\begingroup$ Actually, all that needs to be proved to show that $R$ is a subring is that it is a subgroup, closed under multiplication (and possibly that it contains $1$). Anything more is unnecessary and probably not helpful. $\endgroup$ – Tom Oldfield Feb 18 '13 at 15:50
  • $\begingroup$ @TomOldfield: True. $\endgroup$ – Thomas Feb 18 '13 at 15:58
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Hint $\ $ By the subring test it suffices to show that $\,R\,$ is closed under subtraction and multiplication, which are straightforward verifications.

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How to get started: Prove that it is closed under subtraction and multiplication (maybe and contains the $1$ of $\mathbb{R}$), e.g. why is $(m+n\sqrt{2})+(k+\ell\sqrt{2})$ again in $R$?

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This are the important conditions for a set to be a subring of a main ring. Let s be a subring. Then show that zero is in s. Then show that for all a,b in s a-b is in s. Finally for all a,b in s then ab is in s.

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  • $\begingroup$ In my answer above the last condition is for all a,b in s then an is in s. and not an in s. $\endgroup$ – Gekonga Chacha Nov 20 '16 at 21:19
  • $\begingroup$ You can write your equations with MathJax (see the instructions at meta.math.stackexchange.com/questions/5020/…) to make it easier to read. But I don't think answering an almost 4 years old question is particularly useful if it already has 4 answers, including one equivalent to yours. $\endgroup$ – Arnaud D. Nov 20 '16 at 21:48

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