subrings of rational numbers

[logistic_guy]

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.
D. The set of squares of rational numbers.
E. The set of all rational numbers with odd numerators.
F. The set of all rational numbers with even numerators.

 

[pka]

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.
D. The set of squares of rational numbers.
E. The set of all rational numbers with odd numerators.
F. The set of all rational numbers with even numerators.

Once again you have shown none of your work.
The book The Theory of RINGS by Neal H. McCoy can help you,
On page 2 of that book is stated the following:
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 \(\displaystyle a-b \in S\).
Use that and try to answer the question. Post your efforts.

 

[logistic_guy]

thank pka

what is \(\displaystyle a\) and \(\displaystyle b\)?

how is \(\displaystyle a\) and \(\displaystyle b\) related to the rational number?

i'll take a random rational number with odd denomator \(\displaystyle \frac{1}{3}\). it don't have \(\displaystyle a\) and \(\displaystyle b\)

i don't understand the relation between \(\displaystyle \frac{1}{3}\) and \(\displaystyle a\) and \(\displaystyle b\)

 

[fresh_42]

thank pka

what is \(\displaystyle a\) and \(\displaystyle b\)?

how is \(\displaystyle a\) and \(\displaystyle b\) related to the rational number?

i'll take a random rational number with odd denomator \(\displaystyle \frac{1}{3}\). it don't have \(\displaystyle a\) and \(\displaystyle b\)

i don't understand the relation between \(\displaystyle \frac{1}{3}\) and \(\displaystyle a\) and \(\displaystyle b\)

A subring [imath] \emptyset \neq S\subseteq \mathbb{Q} [/imath] has to be a ring in the first place. This means that it has to contain [imath] a-b [/imath] which automatically proves that [imath] 0\in S [/imath] and [imath] -a\in S [/imath] for any elements [imath] a,b\in S. [/imath] Furthermore, it has to be closed under multiplication, i.e. [imath] a\cdot b\in S [/imath] for any elements [imath] a,b\in S. [/imath]

You need placeholders [imath] a,b\in S [/imath] since you cannot check these properties for all elements of [imath] S. [/imath]

This means for your first example where [imath] S=\left\{\dfrac{r}{s}\in \mathbb{Q}\,|\,s \text{ is odd }\right\} [/imath] are the quotients with odd denominators, that you have to check whether [imath] \dfrac{r}{s}-\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'}, [/imath] and whether [imath] \dfrac{r}{s}\cdot\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where [imath] s [/imath] and [imath] s' [/imath] or odd in both cases.

E.g. [imath] \dfrac{4}{3}-\dfrac{6}{7}=\dfrac{10}{21} [/imath] and [imath] \dfrac{4}{3}\cdot\dfrac{6}{7}=\dfrac{8}{7} [/imath] have odd denominators then we still won't know if that will be also the case for, let's say [imath] \frac{22254896}{1681613241} [/imath] and [imath] \frac{79812630503}{9473127} .[/imath] That's why you have to show it for arbitrary elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where we only know that [imath] s,s' [/imath] are odd.

The logical reasoning is called deduction: if it is true for [imath] a,b\in S [/imath] and [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} \in S[/imath] then it is true for [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} .[/imath] It is true here means if [imath] a-b\in S [/imath] AND [imath] a\cdot b\in S [/imath] then [imath] \frac{22254896}{1681613241} - \frac{79812630503}{9473127} \in S[/imath] and [imath] \frac{22254896}{1681613241} \cdot \frac{79812630503}{9473127} \in S.[/imath] By proving it for arbitrary [imath] a,b [/imath] of the given shape, we have proven it for all elements in [imath] S. [/imath]

 

[fresh_42]

A famous example of why examples don't prove general statements is the polynomial [imath]p(n)= n^2-n+41. [/imath] The claim is that all values of [imath] p(n) [/imath] with [imath] n\in \mathbb{N} [/imath] are prime.

 

[khansaheb]

p(n)=n2−n+41.p(n)= n^2-n+41. p(n)=n2−n+41. The claim is that all values of p(n) p(n) p(n) with n∈N n\in \mathbb{N} n∈N are prime.

@logistic_guy - do you see where does the above claim fall apart?

 

[logistic_guy]

A subring [imath] \emptyset \neq S\subseteq \mathbb{Q} [/imath] has to be a ring in the first place. This means that it has to contain [imath] a-b [/imath] which automatically proves that [imath] 0\in S [/imath] and [imath] -a\in S [/imath] for any elements [imath] a,b\in S. [/imath] Furthermore, it has to be closed under multiplication, i.e. [imath] a\cdot b\in S [/imath] for any elements [imath] a,b\in S. [/imath]

You need placeholders [imath] a,b\in S [/imath] since you cannot check these properties for all elements of [imath] S. [/imath]

This means for your first example where [imath] S=\left\{\dfrac{r}{s}\in \mathbb{Q}\,|\,s \text{ is odd }\right\} [/imath] are the quotients with odd denominators, that you have to check whether [imath] \dfrac{r}{s}-\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'}, [/imath] and whether [imath] \dfrac{r}{s}\cdot\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where [imath] s [/imath] and [imath] s' [/imath] or odd in both cases.

E.g. [imath] \dfrac{4}{3}-\dfrac{6}{7}=\dfrac{10}{21} [/imath] and [imath] \dfrac{4}{3}\cdot\dfrac{6}{7}=\dfrac{8}{7} [/imath] have odd denominators then we still won't know if that will be also the case for, let's say [imath] \frac{22254896}{1681613241} [/imath] and [imath] \frac{79812630503}{9473127} .[/imath] That's why you have to show it for arbitrary elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where we only know that [imath] s,s' [/imath] are odd.

The logical reasoning is called deduction: if it is true for [imath] a,b\in S [/imath] and [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} \in S[/imath] then it is true for [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} .[/imath] It is true here means if [imath] a-b\in S [/imath] AND [imath] a\cdot b\in S [/imath] then [imath] \frac{22254896}{1681613241} - \frac{79812630503}{9473127} \in S[/imath] and [imath] \frac{22254896}{1681613241} \cdot \frac{79812630503}{9473127} \in S.[/imath] By proving it for arbitrary [imath] a,b [/imath] of the given shape, we have proven it for all elements in [imath] S. [/imath]

thank fresh_42

this is too much information

give me some time to read it

@logistic_guy - do you see where does the above claim fall apart?

yes

 

[khansaheb]

yes

You are a person of few words - please share your work!

What is the "breaking point" of the proposition?

 

[logistic_guy]

A subring [imath] \emptyset \neq S\subseteq \mathbb{Q} [/imath] has to be a ring in the first place. This means that it has to contain [imath] a-b [/imath] which automatically proves that [imath] 0\in S [/imath] and [imath] -a\in S [/imath] for any elements [imath] a,b\in S. [/imath] Furthermore, it has to be closed under multiplication, i.e. [imath] a\cdot b\in S [/imath] for any elements [imath] a,b\in S. [/imath]

You need placeholders [imath] a,b\in S [/imath] since you cannot check these properties for all elements of [imath] S. [/imath]

This means for your first example where [imath] S=\left\{\dfrac{r}{s}\in \mathbb{Q}\,|\,s \text{ is odd }\right\} [/imath] are the quotients with odd denominators, that you have to check whether [imath] \dfrac{r}{s}-\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'}, [/imath] and whether [imath] \dfrac{r}{s}\cdot\dfrac{r'}{s'} [/imath] is again a quotient with an odd denominator for any two elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where [imath] s [/imath] and [imath] s' [/imath] or odd in both cases.

E.g. [imath] \dfrac{4}{3}-\dfrac{6}{7}=\dfrac{10}{21} [/imath] and [imath] \dfrac{4}{3}\cdot\dfrac{6}{7}=\dfrac{8}{7} [/imath] have odd denominators then we still won't know if that will be also the case for, let's say [imath] \frac{22254896}{1681613241} [/imath] and [imath] \frac{79812630503}{9473127} .[/imath] That's why you have to show it for arbitrary elements [imath] a=\dfrac{r}{s} [/imath] and [imath] b=\dfrac{r'}{s'} [/imath] where we only know that [imath] s,s' [/imath] are odd.

The logical reasoning is called deduction: if it is true for [imath] a,b\in S [/imath] and [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} \in S[/imath] then it is true for [imath] \frac{22254896}{1681613241} , \frac{79812630503}{9473127} .[/imath] It is true here means if [imath] a-b\in S [/imath] AND [imath] a\cdot b\in S [/imath] then [imath] \frac{22254896}{1681613241} - \frac{79812630503}{9473127} \in S[/imath] and [imath] \frac{22254896}{1681613241} \cdot \frac{79812630503}{9473127} \in S.[/imath] By proving it for arbitrary [imath] a,b [/imath] of the given shape, we have proven it for all elements in [imath] S. [/imath]

i think i'm understand

i'll start with multiplication

\(\displaystyle \frac{r}{s}\frac{r'}{s'} = \frac{rr'}{ss'}\)

is multiply odd x odd = odd?

\(\displaystyle \frac{r}{s} - \frac{r'}{s'} = \frac{rs'}{ss'} - \frac{rs}{ss'} = \frac{rs' - rs}{ss'}\)

same denomator i get here

i've a question fresh_42 you tell me \(\displaystyle s\) is odd and i think \(\displaystyle s'\) a different odd but you don't tell me about \(\displaystyle r\) and \(\displaystyle r'\). what are they?

You are a person of few words - please share your work!

What is the "breaking point" of the proposition?

the closes number to 2024 is 2021 not prime

 

[fresh_42]

i think i'm understand

i'll start with multiplication

\(\displaystyle \frac{r}{s}\frac{r'}{s'} = \frac{rr'}{ss'}\)

is multiply odd x odd = odd?

\(\displaystyle \frac{r}{s} - \frac{r'}{s'} = \frac{rs'}{ss'} - \frac{r's}{ss'} = \frac{rs' - r's}{ss'}\)

same denomator i get here

Yes, that's how it works. [I corrected your typo in the quote.]

It might be worth a thought why the resulting quotients cannot cancel in a way such that an even denominator remains left, especially with the other example of even denominators in mind.

i've a question fresh_42 you tell me \(\displaystyle s\) is odd and i think \(\displaystyle s'\) a different odd but you don't tell me about \(\displaystyle r\) and \(\displaystyle r'\). what are they?

We do not have any information about [imath] r,r' [/imath]. They can be any numbers since this example of a supposed subring was defined only by denominators. E.g. a subring [imath] S [/imath] requires that [imath] 0\in S. [/imath] If we write [imath] 0=\dfrac{0}{1} [/imath] then we get [imath] 0\in S. [/imath] That also [imath] 0=\dfrac{0}{2} [/imath] doesn't matter. By the same argument, we get [imath] \mathbb{Z}\subseteq S. [/imath]

the closes number to 2024 is 2021 not prime

How in the world did you find these numbers? And are they examples for [imath] n [/imath] or for [imath] p(n) [/imath]?

The goal to learn from the polynomial [imath] p(n)=n^2-n+41 [/imath] was to check low numbers:
[math]\begin{array}{lll} n=1\, : \,p(n)=41 \longrightarrow \text{ prime }\\ n=2\, : \,p(n)=43 \longrightarrow \text{ prime }\\ n=3\, : \,p(n)=47 \longrightarrow \text{ prime }\\ n=4\, : \,p(n)=53 \longrightarrow \text{ prime }\\ n=5\, : \,p(n)=61 \longrightarrow \text{ prime }\\ n=7\, : \,p(n)=83 \longrightarrow \text{ prime }\\ \text{etc.} \end{array}[/math]You will have to go a long way if you proceed like this in order to find a non-prime [imath] p(n). [/imath] On the other side, you could immediately see for which number [imath] p(n) [/imath] isn't prime without any calculation. Can you name this number [imath] n [/imath]?

 

[logistic_guy]

Yes, that's how it works. [I corrected your typo in the quote.]

thank

It might be worth a thought why the resulting quotients cannot cancel in a way such that an even denominator remains left, especially with the other example of even denominators in mind.

this i don't understand. do you mean \(\displaystyle \frac{r}{s}\) and \(\displaystyle \frac{r'}{s'}\) won't cancel because they're different quotients? we won't have an even domator because you say odd x odd = odd so \(\displaystyle ss'=s''\)

We do not have any information about [imath] r,r' [/imath]. They can be any numbers since this example of a supposed subring was defined only by denominators.

if they can be anything then you've information about them. they can be anything is enough information

E.g. a subring [imath] S [/imath] requires that [imath] 0\in S. [/imath] If we write [imath] 0=\dfrac{0}{1} [/imath] then we get [imath] 0\in S. [/imath] That also [imath] 0=\dfrac{0}{2} [/imath] doesn't matter. By the same argument, we get [imath] \mathbb{Z}\subseteq S. [/imath]

\(\displaystyle S\) is supposed to be not empty. if \(\displaystyle 0 \in S\) how to gaurantee \(\displaystyle S\) is not empty? maybe all quotients with odd denomator are \(\displaystyle 0 = \frac{0}{s}\).

How in the world did you find these numbers? And are they examples for [imath] n [/imath] or for [imath] p(n) [/imath]?

2024 is the current years and \(\displaystyle p(45) = 2021\) is the closest non prime number to the current year

On the other side, you could immediately see for which number [imath] p(n) [/imath] isn't prime without any calculation. Can you name this number [imath] n [/imath]?

i see what you're aiming at. you want \(\displaystyle n = 41\)

 

[fresh_42]

this i don't understand. do you mean \(\displaystyle \frac{r}{s}\) and \(\displaystyle \frac{r'}{s'}\) won't cancel because they're different quotients? we won't have an even domator because you say odd x odd = odd so \(\displaystyle ss'=s''\)

I meant that [imath] \dfrac{r}{s}-\dfrac{r'}{s'}=\dfrac{rs'-r's}{ss'} [/imath] and we don't have any information about the numerator because [imath] r,r' [/imath] are arbitrary. So if the numerator can be any number, how can we be certain that [imath] rs'-r's [/imath] hasn't all odd primes in it that [imath] s\cdot s' [/imath] has, too, so that they cancel and only [imath] 2 [/imath] will be left in the denominator? Why can't this happen, or if it happens, why don't we have to care?

if they can be anything then you've information about them. they can be anything is enough information

In a way, yes.

\(\displaystyle S\) is supposed to be not empty. if \(\displaystyle 0 \in S\) how to gaurantee \(\displaystyle S\) is not empty?

What does it mean to be empty?

If [imath] S=\emptyset [/imath] then the statement: "[imath] q\not\in S [/imath] for all [imath] q\in \mathbb{Q} [/imath]" is true. If on the other hand, [imath] 0\in \mathbb{Q} [/imath] and [imath] 0\in S [/imath] then [imath] q=0 [/imath] would be a counterexample. This means that our assumptions had to be wrong. So either [imath] \mathbb{Q}=\emptyset [/imath] which is obviously false, or [imath] S\neq \emptyset .[/imath]

i see what you're aiming at. you want \(\displaystyle n=41\)

The fascinating fact with that example is, that every single number [imath] p(n) [/imath] is prime for all [imath] n=1,2,3,\ldots,40. [/imath]

 

[logistic_guy]

I meant that [imath] \dfrac{r}{s}-\dfrac{r'}{s'}=\dfrac{rs'-r's}{ss'} [/imath] and we don't have any information about the numerator because [imath] r,r' [/imath] are arbitrary. So if the numerator can be any number, how can we be certain that [imath] rs'-r's [/imath] hasn't all odd primes in it that [imath] s\cdot s' [/imath] has, too, so that they cancel and only [imath] 2 [/imath] will be left in the denominator? Why can't this happen, or if it happens, why don't we have to care?

why can't this happen, i don't know. but i think we don't care because we got what we want an odd denmator

What does it mean to be empty?

i'm not sure but i guess it mean it is \(\displaystyle S = \{ \ \}\)

If [imath] S=\emptyset [/imath] then the statement: "[imath] q\not\in S [/imath] for all [imath] q\in \mathbb{Q} [/imath]" is true. If on the other hand, [imath] 0\in \mathbb{Q} [/imath] and [imath] 0\in S [/imath] then [imath] q=0 [/imath] would be a counterexample. This means that our assumptions had to be wrong. So either [imath] \mathbb{Q}=\emptyset [/imath] which is obviously false, or [imath] S\neq \emptyset .[/imath]

i don't understand this well but i think you're trying to say it's absolutely \(\displaystyle S\) isn't empty

The fascinating fact with that example is, that every single number [imath] p(n) [/imath] is prime for all [imath] n=1,2,3,\ldots,40. [/imath]

i think that scientist who discover this formula was disappointed when he find out when \(\displaystyle n > 40\) some numbers won't be prime

i'll assume we're agree i proof (A) is a subring. for (B) do i make \(\displaystyle a = \frac{r}{s}\) and i say \(\displaystyle s\) is even

\(\displaystyle \frac{r}{s}\frac{r'}{s'} = \frac{rr'}{ss'}\) i'm sure even x even = even

\(\displaystyle \frac{r}{s} - \frac{r'}{s'} = \frac{rs' - r's}{ss'}\) i get also even domator

is this mean (B) is also a subring?

 

[fresh_42]

why can't this happen, i don't know. but i think we don't care because we got what we want an odd denmator

Both.

If [imath] 2\,|\,s\cdot s' [/imath] then [imath] 2\,|\,s [/imath] or [imath] 2\,|\,s' [/imath] (by definition of a prime number) and since neither [imath] s [/imath] nor [imath] s' [/imath] are even, this is impossible.

And, yes, we do not care. We look at all possible representations of rational numbers. [imath] \dfrac{2}{6}=\dfrac{1}{3}, [/imath] and one is in [imath] S [/imath] and the other one is not. We do not consider "numbers", we consider their representations as quotients. If we would consider quotients as equivalence classes, i.e. [imath] \dfrac{2}{6}=\dfrac{1}{3}, [/imath] then we would run into problems, since the definition of [imath] S [/imath] is related to a particular representation. Otherwise, we would have to define [imath] S [/imath] more carefully, e.g.
[math] S=\left\{q\in \mathbb{Q}\,|\,\exists \,r,s\in \mathbb{Z} \, : \,q=\dfrac{r}{s}\text{ and }s\text{ is odd }\right\} \text{ or }S=\left\{\dfrac{r}{s}\,|\,r/s\text{ is already canceled and }s\text{ is odd }\right\}[/math].

i'm not sure but i guess it mean it is \(\displaystyle S = \{ \ \}\)

This is only a notation. I have written a logical statement about what this notation means: All rational numbers are not in [imath] S. [/imath]

i don't understand this well but i think you're trying to say it's absolutely \(\displaystyle S\) isn't empty

As soon as we find an element in [imath] S [/imath] it cannot be empty.

i think that scientist who discover this formula was disappointed when he find out when \(\displaystyle n > 40\) some numbers won't be prime

I am pretty sure that Euler knew that right from the beginning. However, there are a lot of prime numbers even beyond [imath] 40, [/imath] e.g. [imath] 43^2-43+41 [/imath] and [imath] 44^2-44+41 [/imath] are both prime!

i'll assume we're agree i proof (A) is a subring. for (B) do i make \(\displaystyle a = \frac{r}{s}\) and i say \(\displaystyle s\) is even

\(\displaystyle \frac{r}{s}\frac{r'}{s'} = \frac{rr'}{ss'}\) i'm sure even x even = even

\(\displaystyle \frac{r}{s} - \frac{r'}{s'} = \frac{rs' - r's}{ss'}\) i get also even domator

is this mean (B) is also a subring?

Yes, with the logical implications I mentioned above. Your definition is a bit too sloppy since it distinguishes between [imath] \dfrac{1}{3} [/imath] and [imath] \dfrac{2}{6}. [/imath] If you use canceled quotients instead, then you'll have an additional property that must be checked!

 

[logistic_guy]

Yes, with the logical implications I mentioned above. Your definition is a bit too sloppy since it distinguishes between [imath] \dfrac{1}{3} [/imath] and [imath] \dfrac{2}{6}. [/imath] If you use canceled quotients instead, then you'll have an additional property that must be checked!

what's the additional property?

why i need an additional property when the denomator is even but i don't need an additional property when the denomator is odd?

 

[fresh_42]

what's the additional property?

why i need an additional property when the denomator is even but i don't need an additional property when the denomator is odd?

The problem is: will we distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] or not? If yes, then we have no problems, except that [imath] 1/3 \in S [/imath] but [imath] 2/6 \not\in S.[/imath] This is a matter of how we look at rational numbers: by their numerical value or by single representations as quotients.

If we don't want to distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] then we must find a way to decide whether [imath] 1/3 \in S[/imath] or whether [imath] 2/6 \not\in S[/imath] is true. It cannot be both if we consider them as the same number. The usual requirement would be to demand that any quotient [imath] \dfrac{r}{s} [/imath] is fully canceled. This eliminates the possibility [imath] 2/6 \not\in S.[/imath] On the other hand, fully canceled isn't automatically true anymore for [imath] \dfrac{r}{s}-\dfrac{r'}{s'}=\dfrac{rs'-r's}{ss'}.[/imath] So we must make sure that [imath] \dfrac{rs'-r's}{ss'}\in S [/imath] even after all possible cancellations. This is an additional property.

The fact, that we have to make this choice of how we interpret [imath] \dfrac{r}{s}\in \mathbb{Q} [/imath] is due to the sloppy question. It depends on a given fraction. Numerical values of rational numbers do not depend on whether the quotient is canceled or not.

And, we get the next problem in the case of even denominators. A fully canceled [imath] 0 [/imath] would be [imath] \dfrac{0}{1} [/imath] which does not have an even denominator.

 

[logistic_guy]

The problem is: will we distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] or not? If yes, then we have no problems, except that [imath] 1/3 \in S [/imath] but [imath] 2/6 \not\in S.[/imath] This is a matter of how we look at rational numbers: by their numerical value or by single representations as quotients.

If we don't want to distinguish between [imath] 1/3 [/imath] and [imath] 2/6 [/imath] then we must find a way to decide whether [imath] 1/3 \in S[/imath] or whether [imath] 2/6 \not\in S[/imath] is true. It cannot be both if we consider them as the same number. The usual requirement would be to demand that any quotient [imath] \dfrac{r}{s} [/imath] is fully canceled. This eliminates the possibility [imath] 2/6 \not\in S.[/imath] On the other hand, fully canceled isn't automatically true anymore for [imath] \dfrac{r}{s}-\dfrac{r'}{s'}=\dfrac{rs'-r's}{ss'}.[/imath] So we must make sure that [imath] \dfrac{rs'-r's}{ss'}\in S [/imath] even after all possible cancellations. This is an additional property.

The fact, that we have to make this choice of how we interpret [imath] \dfrac{r}{s}\in \mathbb{Q} [/imath] is due to the sloppy question. It depends on a given fraction. Numerical values of rational numbers do not depend on whether the quotient is canceled or not.

And, we get the next problem in the case of even denominators. A fully canceled [imath] 0 [/imath] would be [imath] \dfrac{0}{1} [/imath] which does not have an even denominator.

i don't think the question is sloppy. it give a hint for (A) (B) (E) (F): when written in lowest terms

i don't mention this hint because i thought it's not important.

can this hint help you?

what is the meaning of this hint? do it distinguish between \(\displaystyle \frac{1}{3}\) and \(\displaystyle \frac{2}{6}\) or not?

 

[fresh_42]

i don't think the question is sloppy. it give a hint for (A) (B) (E) (F): when written in lowest terms

i don't mention this hint because i thought it's not important.

can this hint help you?

what is the meaning of this hint? do it distinguish between \(\displaystyle \frac{1}{3}\) and \(\displaystyle \frac{2}{6}\) or not?

I think when written in lowest terms means the same as fully canceled.

In this case, we have this additional property of elements of [imath] S, [/imath] namely when written in lowest terms. [imath] \dfrac{rs'-r's}{ss'} [/imath] isn't written in lowest terms. You have to make sure that [imath] ss' [/imath] is odd even after any possible cancelation.

And if the denominators are even, then [imath] S [/imath] isn't a subring because [imath] 0=\dfrac{0}{2}=\dfrac{0}{1} [/imath] when written in lowest terms, and [imath] 1 [/imath] is not even.

 

[Dr.Peterson]

i don't think the question is sloppy. it give a hint for (A) (B) (E) (F): when written in lowest terms

i don't mention this hint because i thought it's not important.

can this hint help you?

what is the meaning of this hint? do it distinguish between \(\displaystyle \frac{1}{3}\) and \(\displaystyle \frac{2}{6}\) or not?

That's not a hint! It's an essential part of the problem. That's an example of why it's important to copy a problem completely, rather than including only what you think is important.

It says that the denominators referred to are not just any denominator with which a fraction happens to have been written, but the smallest possible denominator that can be used. Any rational number can be written with an even denominator; but only some have an even denominator when written in lowest terms.

A hint would be a suggestion for how to solve a problem, not a part of the problem itself without which the problem is incomplete. A hint can help; this makes the problem valid.

 

[logistic_guy]

I think when written in lowest terms means the same as fully canceled.

In this case, we have this additional property of elements of [imath] S, [/imath] namely when written in lowest terms. [imath] \dfrac{rs'-r's}{ss'} [/imath] isn't written in lowest terms. You have to make sure that [imath] ss' [/imath] is odd even after any possible cancelation.

And if the denominators are even, then [imath] S [/imath] isn't a subring because [imath] 0=\dfrac{0}{2}=\dfrac{0}{1} [/imath] when written in lowest terms, and [imath] 1 [/imath] is not even.

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

That's not a hint! It's an essential part of the problem. That's an example of why it's important to copy a problem completely, rather than including only what you think is important.

It says that the denominators referred to are not just any denominator with which a fraction happens to have been written, but the smallest possible denominator that can be used. Any rational number can be written with an even denominator; but only some have an even denominator when written in lowest terms.

A hint would be a suggestion for how to solve a problem, not a part of the problem itself without which the problem is incomplete. A hint can help; this makes the problem valid.

thank

yes it's very essential and i fail to include it