subrings of rational numbers

[fresh_42]

so we've an additional property and we can't say it's a subring or not a subring when the denomator is even because the fraction isn't written in lowest term

It means we have to check it. E.g. if we consider the odd denominators, then we have to look at [imath] s\cdot s' [/imath] that occurs in [imath] \dfrac{r\cdot r'}{s\cdot s'} [/imath] as well as in [imath] \dfrac{r\cdot s' -r'\cdot s}{s\cdot s'} .[/imath] The typical reasoning in abstract algebra goes along the lines: What if not?

Assume [imath] s\cdot s' [/imath] is even after cancelation, i.e. when written in lowest terms. Written in the lowest terms means, that [imath] r\cdot r' =q\cdot t[/imath] and [imath] s\cdot s'=q\cdot t' [/imath] for some coprime numbers [imath] t,t'[/imath] such that
[math] \dfrac{r\cdot r'}{s\cdot s'}=\dfrac{q\cdot t}{q\cdot t'}=\dfrac{t}{t'}\quad\text{or}\quad\dfrac{r\cdot s'- r'\cdot s}{s\cdot s'}=\dfrac{q\cdot t}{q\cdot t'}=\dfrac{t}{t'}. [/math]Assume that [imath] t' [/imath] is even. Then [imath] 2\,|\,t'\,|\,q\cdot t' = s\cdot s'. [/imath] Therefore, by the definition of prime numbers, such as [imath] 2, [/imath] we get that [imath] 2\,|\,s [/imath] or [imath] 2\,|\,s' [/imath] which is impossible since they are both odd. This contradiction means that [imath] t' [/imath] is odd and not even, and [imath] S [/imath] with odd denominators is a subring.


The same argument fails if we consider even denominators. Then assuming [imath] t' [/imath] would be odd doesn't tell us anything about [imath] q\cdot t'=s\cdot s' .[/imath] E.g.
[math] \dfrac{5}{6}-\dfrac{1}{6}=\dfrac{5-1}{6}=\dfrac{2\cdot 2}{2\cdot 3}=\dfrac{2}{3}. [/math]

This shows how you should proceed with the other examples: prove it or find a counterexample. In order to prove it, you have to list all properties that define [imath] S [/imath] and show that [imath] a-b \in S [/imath] and [imath] a\cdot b \in S[/imath] whenever [imath] a,b\in S. [/imath]


The reason why @Dr.Peterson said that written in lower terms was an essential property and not just a hint is the following: If we distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] then we look at the ring of pairs [imath]\left[ \mathbb{Z}\times (\mathbb{Z}\setminus \{0\})\right] [/imath] with many ambiguities such as [imath] 1/3\in S [/imath] and [imath] 2/6\not \in S. [/imath] In order to avoid them, we consider the ring [imath] \left[\mathbb{Z}\times (\mathbb{Z}\setminus \{0\}) / \sim \right][/imath] where [imath] \sim [/imath] means that we consider all quotients with the same numerical value as equivalent and do not distinguish among them. In this ring we have [imath] 1/3=2/6 .[/imath] However, it means that we have to make a decision of which element we want to work with when defining [imath] S. [/imath] Will it be [imath] 1/3 [/imath] or will it be [imath] 2/6[/imath]? Written in lower terms or fully canceled is a rule that determines one unique element in every equivalence class of quotients with the same numerical value (up to [imath] \pm 1 [/imath] to be correct).

You said rational numbers, which is
[math] \mathbb{Q}= \mathbb{Z}\times (\mathbb{Z}\setminus \{0\}) / \sim [/math]and unique but defined [imath] S [/imath] by its representations which is ambiguous.

 

[logistic_guy]

\(\displaystyle 2|t'|q.t' = s.s'\)

what this mean. say it in words. i guess it mean \(\displaystyle 2\) is divided by coprime \(\displaystyle t'\) and coprime \(\displaystyle t'\) is divided by \(\displaystyle q.t'\) equal \(\displaystyle s.s'\)

i assume \(\displaystyle t'\) is even. why \(\displaystyle 2\) is divided by \(\displaystyle s\) and \(\displaystyle 2\) is divided by \(\displaystyle s'\) is impossible? what is \(\displaystyle s\) and \(\displaystyle s'\)?

your example here \(\displaystyle \frac{5}{6} - \frac{1}{6}\) has two same denomators. isn't the idea i start with two different numbers in the denomators and see if their lowest term is odd? like \(\displaystyle \frac{5}{4} - \frac{1}{6} = \frac{30}{24} - \frac{4}{24} = \frac{29}{24}\) lowest is even not odd

This shows how you should proceed with the other examples: prove it or find a counterexample. In order to prove it, you have to list all properties that define [imath] S [/imath] and show that [imath] a-b \in S [/imath] and [imath] a\cdot b \in S[/imath] whenever [imath] a,b\in S. [/imath]

still it's not clear

The reason why @Dr.Peterson said that written in lower terms was an essential property and not just a hint is the following: If we distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] then we look at the ring of pairs [imath]\left[ \mathbb{Z}\times (\mathbb{Z}\setminus \{0\})\right] [/imath] with many ambiguities such as [imath] 1/3\in S [/imath] and [imath] 2/6\not \in S. [/imath] In order to avoid them, we consider the ring [imath] \left[\mathbb{Z}\times (\mathbb{Z}\setminus \{0\}) / \sim \right][/imath] where [imath] \sim [/imath] means that we consider all quotients with the same numerical value as equivalent and do not distinguish among them. In this ring we have [imath] 1/3=2/6 .[/imath] However, it means that we have to make a decision of which element we want to work with when defining [imath] S. [/imath] Will it be [imath] 1/3 [/imath] or will it be [imath] 2/6[/imath]? Written in lower terms or fully canceled is a rule that determines one unique element in every equivalence class of quotients with the same numerical value (up to [imath] \pm 1 [/imath] to be correct).

You said rational numbers, which is
[math] \mathbb{Q}= \mathbb{Z}\times (\mathbb{Z}\setminus \{0\}) / \sim [/math]and unique but defined [imath] S [/imath] by its representations which is ambiguous.

this i don't understand

 

[fresh_42]

\(\displaystyle 2|t'|q.t' = s.s'\)

what this mean. say it in words. i guess it mean \(\displaystyle 2\) is divided by coprime \(\displaystyle t'\) and coprime \(\displaystyle t'\) is divided by \(\displaystyle q.t'\) equal \(\displaystyle s.s'\)

It only means: two divides [imath] t' [/imath] and [imath] t' [/imath] and [imath] t' [/imath] divides [imath] q'\cdot t'. [/imath] Coprime doesn't matter here. It is like [imath] 4\,|\,28\,|\,84. [/imath] Four divides twenty-eight divides eighty-four, so four divides eighty-four, too.

still it's not clear


this i don't understand

I'm sorry for that. It's a bit like I was asking you to explain the word sophisticated without using English words. You asked whether certain subsets of the rational numbers are a subring of the rational numbers or not. To answer it, we must speak about rational numbers, arbitrary numbers, addition and multiplication in [imath] \mathbb{Q}. [/imath]

1. What is a rational number? Firstly, they contain the integers. However, integers are not quotients. So how will we write them? Will it be [imath] 2=\dfrac{2}{1} [/imath] or [imath] 2=\dfrac{4}{2} [/imath]? One has an even, the other one an odd denominator. This leads us immediately to the second point: When do two rational numbers be equal? This property is defined by
   [math] \dfrac{a}{b}=\dfrac{c}{d} \Longleftrightarrow a\cdot d = b\cdot c\quad({}^*).[/math]
   Your question didn't say whether we distinguish between [imath] \dfrac{2}{1} [/imath] and [imath] \dfrac{4}{2} [/imath] or not. This only changed when you added "in its lowest representation". This is more than a hint because it invoked the definition [imath] ({}^*) [/imath] and ruled out the case [imath] 2=\dfrac{4}{2} [/imath] and left us with the case [imath] 2=\dfrac{2}{1}. [/imath] It says, that we do not speak about different quotients being the same number, only that we speak about fully canceled numbers as [imath] 2=\dfrac{2}{1} [/imath] is. Only this "hint" makes the question unambiguous and which is why @Dr.Peterson said that it is more than a hint but rather a crucial requirement.
   
   
2. We need arbitrary numbers since we cannot check the entire subset. E.g., the subset A contains all numbers with odd denominators, i.e. all numbers [imath] \left\{\left.q=\dfrac{a}{b}\right| \;b\text{ is odd }\right\}. [/imath] We cannot check all of them of whether e.g. addition in A stays in A. We need to prove it for any, i.e. arbitrary two quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] with odd [imath] b,d. [/imath]
   
   
3. Addition is generally defined as
   [imath] \dfrac{a}{b} +\dfrac{c}{d}=\dfrac{ad+bc}{bd}.[/imath]
   However, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even again. We need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
4. Multiplication is generally defined as
   [math] \dfrac{a}{b}\cdot \dfrac{c}{d}=\dfrac{ac}{bd}. [/math]Again, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even and need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
5. When is a number odd or even? The distinction is found by a division by two: odd if two is not a factor of the number, even if it is a factor. But how do we know whether two is a factor of [imath] b\cdot d [/imath]? The best way to write this with rigour is to use the fact that two is a prime number. A prime number is defined to be a number that is not one or minus one, and that if it divides a product of two numbers, then it has to divide one of its factors. Examples: Seven is prime and if it divides one-hundred and eighty-two , which is the product of fourteen and thirteen then seven has to divide thirteen or fourteen (or both). Six is not a prime and it divides thirty-six which is a product of four and nine. However, six neither divides four nor nine.
   
   In our situation, we need to figure out whether the product [imath] b\cdot d [/imath] is odd or even, i.e. whether two divides a product. Since two is prime, we know: If two divides [imath] b\cdot d [/imath] then it has to divide either [imath] b [/imath] or [imath] d [/imath] (or both). Hene, if both [imath] b [/imath] and [imath] d [/imath] are odd, then two cannot divide [imath] b\cdot d [/imath] and [imath] b\cdot d [/imath] is odd, too. If both are even then [imath] b=2\cdot b' [/imath] and [imath] d=2\cdot d'[/imath] so
   [math] 2\,|\,2^2\cdot b' \cdot d'=b\cdot d [/math]and [imath] b\cdot d [/imath] is even, too. However, we need to prove this for canceled quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] and neither [imath] \dfrac{ad+bc}{bd} [/imath] nor [imath] \dfrac{ac}{bd} [/imath] are automatically canceled, i.e. written in lowest terms. IN the case of odd denominators, everything is clear: If it is odd, then the canceled version must also be odd because an additional factor two cannot come out of nowhere by the process of cancellation. However, if both are even, i.e. have a factor two, then cancelation of [imath] \dfrac{ad+bc}{bd} [/imath] could eliminate the factors two:
   [math] \dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=\dfrac{4}{1}. [/math]
   In this case, an example is sufficient, because one counterexample is enough to disprove a general statement. But this is only valid for counterexamples, not to prove a general statement by examples.

 

[logistic_guy]

It only means: two divides [imath] t' [/imath] and [imath] t' [/imath] and [imath] t' [/imath] divides [imath] q'\cdot t'. [/imath] Coprime doesn't matter here. It is like [imath] 4\,|\,28\,|\,84. [/imath] Four divides twenty-eight divides eighty-four, so four divides eighty-four, too.


I'm sorry for that. It's a bit like I was asking you to explain the word sophisticated without using English words. You asked whether certain subsets of the rational numbers are a subring of the rational numbers or not. To answer it, we must speak about rational numbers, arbitrary numbers, addition and multiplication in [imath] \mathbb{Q}. [/imath]

1. What is a rational number? Firstly, they contain the integers. However, integers are not quotients. So how will we write them? Will it be [imath] 2=\dfrac{2}{1} [/imath] or [imath] 2=\dfrac{4}{2} [/imath]? One has an even, the other one an odd denominator. This leads us immediately to the second point: When do two rational numbers be equal? This property is defined by
   [math] \dfrac{a}{b}=\dfrac{c}{d} \Longleftrightarrow a\cdot d = b\cdot c\quad({}^*).[/math]
   Your question didn't say whether we distinguish between [imath] \dfrac{2}{1} [/imath] and [imath] \dfrac{4}{2} [/imath] or not. This only changed when you added "in its lowest representation". This is more than a hint because it invoked the definition [imath] ({}^*) [/imath] and ruled out the case [imath] 2=\dfrac{4}{2} [/imath] and left us with the case [imath] 2=\dfrac{2}{1}. [/imath] It says, that we do not speak about different quotients being the same number, only that we speak about fully canceled numbers as [imath] 2=\dfrac{2}{1} [/imath] is. Only this "hint" makes the question unambiguous and which is why @Dr.Peterson said that it is more than a hint but rather a crucial requirement.
   
   
2. We need arbitrary numbers since we cannot check the entire subset. E.g., the subset A contains all numbers with odd denominators, i.e. all numbers [imath] \left\{\left.q=\dfrac{a}{b}\right| \;b\text{ is odd }\right\}. [/imath] We cannot check all of them of whether e.g. addition in A stays in A. We need to prove it for any, i.e. arbitrary two quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] with odd [imath] b,d. [/imath]
   
   
3. Addition is generally defined as
   [imath] \dfrac{a}{b} +\dfrac{c}{d}=\dfrac{ad+bc}{bd}.[/imath]
   However, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even again. We need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
4. Multiplication is generally defined as
   [math] \dfrac{a}{b}\cdot \dfrac{c}{d}=\dfrac{ac}{bd}. [/math]Again, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even and need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
5. When is a number odd or even? The distinction is found by a division by two: odd if two is not a factor of the number, even if it is a factor. But how do we know whether two is a factor of [imath] b\cdot d [/imath]? The best way to write this with rigour is to use the fact that two is a prime number. A prime number is defined to be a number that is not one or minus one, and that if it divides a product of two numbers, then it has to divide one of its factors. Examples: Seven is prime and if it divides one-hundred and eighty-two , which is the product of fourteen and thirteen then seven has to divide thirteen or fourteen (or both). Six is not a prime and it divides thirty-six which is a product of four and nine. However, six neither divides four nor nine.
   
   In our situation, we need to figure out whether the product [imath] b\cdot d [/imath] is odd or even, i.e. whether two divides a product. Since two is prime, we know: If two divides [imath] b\cdot d [/imath] then it has to divide either [imath] b [/imath] or [imath] d [/imath] (or both). Hene, if both [imath] b [/imath] and [imath] d [/imath] are odd, then two cannot divide [imath] b\cdot d [/imath] and [imath] b\cdot d [/imath] is odd, too. If both are even then [imath] b=2\cdot b' [/imath] and [imath] d=2\cdot d'[/imath] so
   [math] 2\,|\,2^2\cdot b' \cdot d'=b\cdot d [/math]and [imath] b\cdot d [/imath] is even, too. However, we need to prove this for canceled quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] and neither [imath] \dfrac{ad+bc}{bd} [/imath] nor [imath] \dfrac{ac}{bd} [/imath] are automatically canceled, i.e. written in lowest terms. IN the case of odd denominators, everything is clear: If it is odd, then the canceled version must also be odd because an additional factor two cannot come out of nowhere by the process of cancellation. However, if both are even, i.e. have a factor two, then cancelation of [imath] \dfrac{ad+bc}{bd} [/imath] could eliminate the factors two:
   [math] \dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=\dfrac{4}{1}. [/math]
   In this case, an example is sufficient, because one counterexample is enough to disprove a general statement. But this is only valid for counterexamples, not to prove a general statement by examples.

thank

this is too much information

give me some time to understand it

 

[logistic_guy]

It only means: two divides [imath] t' [/imath] and [imath] t' [/imath] and [imath] t' [/imath] divides [imath] q'\cdot t'. [/imath] Coprime doesn't matter here. It is like [imath] 4\,|\,28\,|\,84. [/imath] Four divides twenty-eight divides eighty-four, so four divides eighty-four, too.


I'm sorry for that. It's a bit like I was asking you to explain the word sophisticated without using English words. You asked whether certain subsets of the rational numbers are a subring of the rational numbers or not. To answer it, we must speak about rational numbers, arbitrary numbers, addition and multiplication in [imath] \mathbb{Q}. [/imath]

1. What is a rational number? Firstly, they contain the integers. However, integers are not quotients. So how will we write them? Will it be [imath] 2=\dfrac{2}{1} [/imath] or [imath] 2=\dfrac{4}{2} [/imath]? One has an even, the other one an odd denominator. This leads us immediately to the second point: When do two rational numbers be equal? This property is defined by
   [math] \dfrac{a}{b}=\dfrac{c}{d} \Longleftrightarrow a\cdot d = b\cdot c\quad({}^*).[/math]
   Your question didn't say whether we distinguish between [imath] \dfrac{2}{1} [/imath] and [imath] \dfrac{4}{2} [/imath] or not. This only changed when you added "in its lowest representation". This is more than a hint because it invoked the definition [imath] ({}^*) [/imath] and ruled out the case [imath] 2=\dfrac{4}{2} [/imath] and left us with the case [imath] 2=\dfrac{2}{1}. [/imath] It says, that we do not speak about different quotients being the same number, only that we speak about fully canceled numbers as [imath] 2=\dfrac{2}{1} [/imath] is. Only this "hint" makes the question unambiguous and which is why @Dr.Peterson said that it is more than a hint but rather a crucial requirement.
   
   
2. We need arbitrary numbers since we cannot check the entire subset. E.g., the subset A contains all numbers with odd denominators, i.e. all numbers [imath] \left\{\left.q=\dfrac{a}{b}\right| \;b\text{ is odd }\right\}. [/imath] We cannot check all of them of whether e.g. addition in A stays in A. We need to prove it for any, i.e. arbitrary two quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] with odd [imath] b,d. [/imath]
   
   
3. Addition is generally defined as
   [imath] \dfrac{a}{b} +\dfrac{c}{d}=\dfrac{ad+bc}{bd}.[/imath]
   However, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even again. We need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
4. Multiplication is generally defined as
   [math] \dfrac{a}{b}\cdot \dfrac{c}{d}=\dfrac{ac}{bd}. [/math]Again, we no longer have any control over whether the new denominator [imath] b\cdot d [/imath] is odd or even and need an argument for that. Examples do not count since we require arbitrary elements of A.
   
   
5. When is a number odd or even? The distinction is found by a division by two: odd if two is not a factor of the number, even if it is a factor. But how do we know whether two is a factor of [imath] b\cdot d [/imath]? The best way to write this with rigour is to use the fact that two is a prime number. A prime number is defined to be a number that is not one or minus one, and that if it divides a product of two numbers, then it has to divide one of its factors. Examples: Seven is prime and if it divides one-hundred and eighty-two , which is the product of fourteen and thirteen then seven has to divide thirteen or fourteen (or both). Six is not a prime and it divides thirty-six which is a product of four and nine. However, six neither divides four nor nine.
   
   In our situation, we need to figure out whether the product [imath] b\cdot d [/imath] is odd or even, i.e. whether two divides a product. Since two is prime, we know: If two divides [imath] b\cdot d [/imath] then it has to divide either [imath] b [/imath] or [imath] d [/imath] (or both). Hene, if both [imath] b [/imath] and [imath] d [/imath] are odd, then two cannot divide [imath] b\cdot d [/imath] and [imath] b\cdot d [/imath] is odd, too. If both are even then [imath] b=2\cdot b' [/imath] and [imath] d=2\cdot d'[/imath] so
   [math] 2\,|\,2^2\cdot b' \cdot d'=b\cdot d [/math]and [imath] b\cdot d [/imath] is even, too. However, we need to prove this for canceled quotients [imath] \dfrac{a}{b},\dfrac{c}{d} [/imath] and neither [imath] \dfrac{ad+bc}{bd} [/imath] nor [imath] \dfrac{ac}{bd} [/imath] are automatically canceled, i.e. written in lowest terms. IN the case of odd denominators, everything is clear: If it is odd, then the canceled version must also be odd because an additional factor two cannot come out of nowhere by the process of cancellation. However, if both are even, i.e. have a factor two, then cancelation of [imath] \dfrac{ad+bc}{bd} [/imath] could eliminate the factors two:
   [math] \dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=\dfrac{4}{1}. [/math]
   In this case, an example is sufficient, because one counterexample is enough to disprove a general statement. But this is only valid for counterexamples, not to prove a general statement by examples.

your explanation is very rich. i think i'm understand some of it

A is subring as odd denomators are friendly

B isn't a subring as the counterexample

C i think isn't a subring as it've mixed denomators. it sometimes have odd and sometimes even denomators, so we've A and B in set C

is my thinking correct?

 

[fresh_42]

your explanation is very rich. i think i'm understand some of it

A is subring as odd denomators are friendly

B isn't a subring as the counterexample

C i think isn't a subring as it've mixed denomators. it sometimes have odd and sometimes even denomators, so we've A and B in set C

is my thinking correct?

Let me quote the question as a reminder of how A, B, and C are defined.

here is the question

Which of the following are subrings of \(\displaystyle \mathbb{Q}\)?

A. The set of all rational numbers with odd denominators.
B. The set of all rational numbers with even denominators.
C. The set of nonnegative rational numbers.

Now to answer your question: Partly, and I wouldn't use the word friendly.

A is a subring because addition and multiplication of quotients with odd denominators have again odd denominators, even after a possible cancelation.

B isn't a subring because the argument we used for A isn't true any longer. Cancellation after addition of quotients with even denominators can create denominators with odd denominators: [imath] 1/2+1/2=1/1. [/imath] So addition isn't possible within the set of numbers with even denominators. Addition can lead outside of B. That is the reason.

C has a different problem. Whether denominators are odd or even isn't relevant here any longer. This property cannot be used in any way, so your argument fails. It's simply not important for C. To answer the question for C we have to go back to the very beginning.

A nonempty subset \(\displaystyle S\) of a ring \(\displaystyle R\) is a subring if for all \(\displaystyle a,b \in S\) it is true that \(\displaystyle ab \in S\) and \(\displaystyle a-b \in S\).

We need to consider the definition of a ring in order to answer the question for C. This is because there is a qualitative distinction between addition and multiplication in a ring. Addition requires that we can always solve equations [imath] a+x=b. [/imath] This implies that we have a [imath] 0 [/imath] (as a solution to [imath] a+x=a [/imath]) and additively inverse elements (as a solution to [imath] a+x=0 [/imath]). It automatically creates the requirement of being non-empty as we need the existence of the [imath] 0. [/imath] This property does not necessarily hold for multiplication. E.g. the integers are a ring since we can always solve [imath] a+x=b [/imath] although we can not solve [imath] a\cdot x=b [/imath] within the integers as the example [imath] 2\cdot x= 1 [/imath] shows: [imath] x=1/2 [/imath] isn't an integer.

This leads to the answer of C. [imath] 0 [/imath] and [imath] 1 [/imath] are non-negative rational numbers, so [imath] 0,1\in C [/imath] and C is non-empty. But how should we solve [imath] 1+x=0 [/imath] within the set of non-negative rational numbers? We can't, and that's why C isn't a ring and thus no subring either.

In the wording of @pka's answer: [imath] 0,1\in C [/imath] but [imath] 0-1\not\in C. [/imath] We say that C isn't closed under addition.

 

[logistic_guy]

Let me quote the question as a reminder of how A, B, and C are defined.


Now to answer your question: Partly, and I wouldn't use the word friendly.

A is a subring because addition and multiplication of quotients with odd denominators have again odd denominators, even after a possible cancelation.

B isn't a subring because the argument we used for A isn't true any longer. Cancellation after addition of quotients with even denominators can create denominators with odd denominators: [imath] 1/2+1/2=1/1. [/imath] So addition isn't possible within the set of numbers with even denominators. Addition can lead outside of B. That is the reason.

C has a different problem. Whether denominators are odd or even isn't relevant here any longer. This property cannot be used in any way, so your argument fails. It's simply not important for C. To answer the question for C we have to go back to the very beginning.


We need to consider the definition of a ring in order to answer the question for C. This is because there is a qualitative distinction between addition and multiplication in a ring. Addition requires that we can always solve equations [imath] a+x=b. [/imath] This implies that we have a [imath] 0 [/imath] (as a solution to [imath] a+x=a [/imath]) and additively inverse elements (as a solution to [imath] a+x=0 [/imath]). It automatically creates the requirement of being non-empty as we need the existence of the [imath] 0. [/imath] This property does not necessarily hold for multiplication. E.g. the integers are a ring since we can always solve [imath] a+x=b [/imath] although we can not solve [imath] a\cdot x=b [/imath] within the integers as the example [imath] 2\cdot x= 1 [/imath] shows: [imath] x=1/2 [/imath] isn't an integer.

This leads to the answer of C. [imath] 0 [/imath] and [imath] 1 [/imath] are non-negative rational numbers, so [imath] 0,1\in C [/imath] and C is non-empty. But how should we solve [imath] 1+x=0 [/imath] within the set of non-negative rational numbers? We can't, and that's why C isn't a ring and thus no subring either.

In the wording of @pka's answer: [imath] 0,1\in C [/imath] but [imath] 0-1\not\in C. [/imath] We say that C isn't closed under addition.

i don't understand C well, but let us go to D

set of squares of rational number. do it mean something like \(\displaystyle \frac{5^2}{7^2}\)

do this belong to the set? should we care if the squares are prime or not?

 

[fresh_42]

i don't understand C well, but let us go to D

set of squares of rational number. do it mean something like \(\displaystyle \frac{5^2}{7^2}\)

That's how I would read it.

do this belong to the set? should we care if the squares are prime or not?

No, being prime is irrelevant. In this case, closeness under addition is the crucial property. Can you find a square that if added to itself is no longer a square? You don't even need quotients.

 

[logistic_guy]

Can you find a square that if added to itself is no longer a square? You don't even need quotients.

yes with fraction but remove itself

\(\displaystyle \frac{3^2}{2^2} + \frac{4^2}{2^2} = \frac{9+16}{2^2} = \frac{5^2}{2^2}\)

 

[fresh_42]

yes with fraction but remove itself

\(\displaystyle \frac{3^2}{2^2} + \frac{4^2}{2^2} = \frac{9+16}{2^2} = \frac{5^2}{2^2}\)

That's why we need variables to represent elements and not examples. You have found a Pythagorean triple, i.e. an example where [imath] a^2+b^2=c^2. [/imath] However, we need to have this property for all elements of D if D was a subring. So given any two arbitrary squared rational numbers, we must show that the sum (or even better the difference) will be a squared rational number again. But this cannot be shown since it is wrong. A single counterexample is sufficient to demonstrate the impossibility, e.g. [imath] \dfrac{4}{1}-\dfrac{1}{1}=\dfrac{3}{1} [/imath] isn't a square although [imath] 4/1 [/imath]and [imath] 1/1 [/imath] are elements of D, squared rational numbers.

A subring [imath] S [/imath] of a ring [imath] R [/imath] is a set such that

1. [imath] S\neq \emptyset[/imath]
2. [imath] S\subseteq R [/imath]
3. [imath] \forall\;s,t\in S\, : \,s-t\in S [/imath]
4. [imath] \forall\;s,t\in S\, : \,s\cdot t\in S [/imath]

We have [imath] R=\mathbb{Q} [/imath] and the first two properties are more or less automatically given by the phrase

The set of all rational numbers with ...

and the fact that neither of these sets is obviously empty. That's why we are discussing only the last two properties in this thread. Property three is a difference because we want to make sure that [imath] 0\in S [/imath] and [imath] -s\in S [/imath] whenever [imath] s\in S. [/imath] This additional (hidden) property is not necessary for multiplication since rings are not required to provide multiplicative inverse elements. So we need only property four without any division.

If we want to prove that a set [imath] S [/imath] is a subring, we must ensure that all properties are fulfilled, especially the third and fourth. Some of the sets you have listed fail to fulfill property three. Now, what does this mean? We have to show that all elements [imath] s-t [/imath] are in [imath] S [/imath] whenever [imath] s [/imath] and [imath] t [/imath] are in [imath] S. [/imath] This makes the considerations a bit more detailed in case we defined [imath] S [/imath] by the kind of its denominators. If we want to demonstrate that not all elements [imath] s-t [/imath] are in [imath] S [/imath] whenever [imath] s [/imath] and [imath] t [/imath] are in [imath] S[/imath] it is sufficient to find a single counterexample. This is because
[math] \lnot \;(\;\forall\;x\in S\, : \,P(x)\;)\Longleftrightarrow \exists\, x_0\in S\, : \,\lnot\;P(x_0) [/math]This means in words: The statement that ( all elements [imath] x [/imath] of [imath] S [/imath] fulfill property [imath] P(x) [/imath] ) is wrong, if and only if the statement ( there is (at least) one element [imath] x_0 [/imath] of [imath] S [/imath] such that the property [imath] P(x_0) [/imath] is wrong ) is true. That's why we need a counterexample [imath] x_0\in S [/imath] if we want to disprove a statement, and we need variable element names [imath] x\in S [/imath] if we want to prove a statement.

Your example of a Pythagorean triple was one example [imath] x [/imath] when we needed the statement for all [imath] x [/imath] for a proof.
My example of [imath] x_0=4-1 [/imath] was one example when I needed only the existence of a single counterexample [imath] x_0 [/imath] for a disproof.

 

[logistic_guy]

That's why we need variables to represent elements and not examples. You have found a Pythagorean triple, i.e. an example where [imath] a^2+b^2=c^2. [/imath] However, we need to have this property for all elements of D if D was a subring. So given any two arbitrary squared rational numbers, we must show that the sum (or even better the difference) will be a squared rational number again. But this cannot be shown since it is wrong. A single counterexample is sufficient to demonstrate the impossibility, e.g. [imath] \dfrac{4}{1}-\dfrac{1}{1}=\dfrac{3}{1} [/imath] isn't a square although [imath] 4/1 [/imath]and [imath] 1/1 [/imath] are elements of D, squared rational numbers.

A subring [imath] S [/imath] of a ring [imath] R [/imath] is a set such that

1. [imath] S\neq \emptyset[/imath]
2. [imath] S\subseteq R [/imath]
3. [imath] \forall\;s,t\in S\, : \,s-t\in S [/imath]
4. [imath] \forall\;s,t\in S\, : \,s\cdot t\in S [/imath]

We have [imath] R=\mathbb{Q} [/imath] and the first two properties are more or less automatically given by the phrase

and the fact that neither of these sets is obviously empty. That's why we are discussing only the last two properties in this thread. Property three is a difference because we want to make sure that [imath] 0\in S [/imath] and [imath] -s\in S [/imath] whenever [imath] s\in S. [/imath] This additional (hidden) property is not necessary for multiplication since rings are not required to provide multiplicative inverse elements. So we need only property four without any division.

If we want to prove that a set [imath] S [/imath] is a subring, we must ensure that all properties are fulfilled, especially the third and fourth. Some of the sets you have listed fail to fulfill property three. Now, what does this mean? We have to show that all elements [imath] s-t [/imath] are in [imath] S [/imath] whenever [imath] s [/imath] and [imath] t [/imath] are in [imath] S. [/imath] This makes the considerations a bit more detailed in case we defined [imath] S [/imath] by the kind of its denominators. If we want to demonstrate that not all elements [imath] s-t [/imath] are in [imath] S [/imath] whenever [imath] s [/imath] and [imath] t [/imath] are in [imath] S[/imath] it is sufficient to find a single counterexample. This is because
[math] \lnot \;(\;\forall\;x\in S\, : \,P(x)\;)\Longleftrightarrow \exists\, x_0\in S\, : \,\lnot\;P(x_0) [/math]This means in words: The statement that ( all elements [imath] x [/imath] of [imath] S [/imath] fulfill property [imath] P(x) [/imath] ) is wrong, if and only if the statement ( there is (at least) one element [imath] x_0 [/imath] of [imath] S [/imath] such that the property [imath] P(x_0) [/imath] is wrong ) is true. That's why we need a counterexample [imath] x_0\in S [/imath] if we want to disprove a statement, and we need variable element names [imath] x\in S [/imath] if we want to prove a statement.

Your example of a Pythagorean triple was one example [imath] x [/imath] when we needed the statement for all [imath] x [/imath] for a proof.
My example of [imath] x_0=4-1 [/imath] was one example when I needed only the existence of a single counterexample [imath] x_0 [/imath] for a disproof.

if a counter example like \(\displaystyle \frac{2^2}{1^2} - \frac{1^2}{1^2} = \frac{3}{1^2}\) tell me squares of rational numbers isn't subring, then i'll always search for counter example instead of using other meothds because it's easier

(e)
is it correct to say this is counter example

\(\displaystyle \frac{1}{7} + \frac{5}{7} = \frac{6}{7}\) so (e) isn't subring

 

[fresh_42]

if a counter example like \(\displaystyle \frac{2^2}{1^2} - \frac{1^2}{1^2} = \frac{3}{1^2}\) tell me squares of rational numbers isn't subring, then i'll always search for counter example instead of using other meothds because it's easier

(e)
is it correct to say this is counter example

\(\displaystyle \frac{1}{7} + \frac{5}{7} = \frac{6}{7}\) so (e) isn't subring

That's correct.