1. ## Set, Ring

I was given this question:
The Set S={2x+1∶x∈I} ,where I denotes the set of all integers, with addition and multiplication defined on I.

I chose false, because in order for this set to have a zero, the x would equal -1/2. It says that x has to be an integer.

The answer given was true, because the set is a ring. I'm still confused because I thought a ring had to have a zero, especially since this says the set is defined for addition and multiplication.

2. ## Re: Set, Ring

Is it because the inverse is -2x-1 and there doesn't need to be a zero?

3. ## Re: Set, Ring

I am not sure I am understanding the addition and multiplication. Let $2x_1+1,2x_2+1 \in S$. Then $(2x_1+1)\bigoplus (2x_2+1) := 2(x_1+x_2)+1$? And $(2x_1+1)\bigotimes (2x_2+1) := 2x_1x_2+1$? To me, that sounds like addition and multiplication defined on $I$, and certainly defines a ring. In fact, $\{S,\bigoplus , \bigotimes \}$ and $\{I, +, \times \}$ are isomorphic rings.

Alternately, if the defined addition and multiplication were $(2x_1+1)+(2x_2+1) = 2(x_1+x_2)+2 \notin S$, then the set is not even closed under addition, and definitely not a ring.

4. ## Re: Set, Ring

Originally Posted by brucebunch
I was given this question:
The Set S={2x+1∶x∈I} ,where I denotes the set of all integers, with addition and multiplication defined on I.
the answer is true, because the set is a ring. I'm still confused because I thought a ring had to have a zero, especially since this says the set is defined for addition and multiplication.
It is clear that $S$ is the set of odd integers. You said that the addition & multiplication are the ordinary ones for integers.
Being the set of odd integers, $0\notin S$. So as written you are correct. Please check the wording of the given question.

5. ## Re: Set, Ring

Thanks so much for responding. I really appreciate it. This is very new to me. I'm trying to understand your math. I don't even know how to type equations into this blog, so I'll try my best to explain my question.

I don't understand how you got (2(x1) + 1) + (2(x2) + 1) = 2((x1) + (x2)) + 1, It makes a lot more sense to me when you did it the second time and the last term was 2. If you factor out the 2(x1) and 2(x2), you are left with 1 + 1. Can you explain how you did it the first time.

I hope this makes sense. I really don't know how to write equations on here or how you inserted the math symbols. My first post, I copied and pasted the question as written so I didn't have to try and type it out.

6. ## Re: Set, Ring

Originally Posted by brucebunch
Thanks so much for responding. I really appreciate it. This is very new to me. I'm trying to understand your math. I don't even know how to type equations into this blog, so I'll try my best to explain my question.

I don't understand how you got (2(x1) + 1) + (2(x2) + 1) = 2((x1) + (x2)) + 1, It makes a lot more sense to me when you did it the second time and the last term was 2. If you factor out the 2(x1) and 2(x2), you are left with 1 + 1. Can you explain how you did it the first time.

I hope this makes sense. I really don't know how to write equations on here or how you inserted the math symbols. My first post, I copied and pasted the question as written so I didn't have to try and type it out.
You can define a different addition and multiplication that do not work like regular addition and multiplication. We turn addition and multiplication into abstractions called binary operators. Binary operators are functions on two variables that return one variable. The phrase a set is "closed" under a binary operator means that if we apply the binary operator to any two elements of a set, we get another element of that set. So, we define a binary operator that works like I showed in the post above. That is what $:=$ means. It means it is defined to be equal. So, rather than defining addition as the normal integer addition, we can define a "variant" addition. That turns the element $2(0)+1$ into the "zero" element and $2(1)+1$ into the "unit" of the ring. Notice that I used a different symbol for the addition and multiplication operators. I used $\bigoplus$ and $\bigotimes$ rather than $+$ and $\times$. That was to differentiate between "normal" addition and multiplication from the addition and multiplication that I defined for this problem.

To see how I format math symbols, you can Reply With Quote to one of my replies and it will show you how I formatted it.

7. ## Re: Set, Ring

Originally Posted by brucebunch
Thanks so much for responding. I really appreciate it. This is very new to me. I'm trying to understand your math. I don't even know how to type equations into this blog, so I'll try my best to explain my question.
I don't understand how you got (2(x1) + 1) + (2(x2) + 1) = 2((x1) + (x2)) + 1, It makes a lot more sense to me when you did it the second time and the last term was 2. If you factor out the 2(x1) and 2(x2), you are left with 1 + 1. Can you explain how yassumeou did it the first time.
I think someone is missing the point of the original post.
It clearly says " with addition and multiplication defined on $I$."
So we must assume that is the ordinary operations on $I$.
With those operations the additive part of set $S$ does not form an Abelian group.
So you are correct to say False, $S$ is not a ring.

8. ## Re: Set, Ring

Originally Posted by SlipEternal
You can define a different addition and multiplication that do not work like regular addition and multiplication. We turn addition and multiplication into abstractions called binary operators. Binary operators are functions on two variables that return one variable. The phrase a set is "closed" under a binary operator means that if we apply the binary operator to any two elements of a set, we get another element of that set. So, we define a binary operator that works like I showed in the post above. That is what $:=$ means. It means it is defined to be equal. So, rather than defining addition as the normal integer addition, we can define a "variant" addition. That turns the element $2(0)+1$ into the "zero" element and $2(1)+1$ into the "unit" of the ring. Notice that I used a different symbol for the addition and multiplication operators. I used $\bigoplus$ and $\bigotimes$ rather than $+$ and $\times$. That was to differentiate between "normal" addition and multiplication from the addition and multiplication that I defined for this problem.

To see how I format math symbols, you can Reply With Quote to one of my replies and it will show you how I formatted it.
That makes soooo much more sense. You a amazing. Thanks for taking the time to explain that. I would love it if you show me how to write an equation

9. ## Re: Set, Ring

LaTex Tutorial

That is a good tutorial. They changed the tags. Instead of $[\text{math}]$ and $[/\text{math}]$, you can use dollar signs. For example, $\$ 2 x \_ 1 + 1\ will display as $2x_1+1$. So will $[\text{tex}] 2 x \_ 1 + 1 [/\text{tex} ]$.