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for two sets to be equal

if we have two sets, A and B,
If A={q is an integer : p=q+2}
and B={d is an integer : h=d+2}
does that mean A=B?
Original post by cooldudeman
if we have two sets, A and B,
If A={q is an integer : p=q+2}
and B={d is an integer : h=d+2}
does that mean A=B?


Is that the whole question? What are p,h?
Original post by ghostwalker
Is that the whole question? What are p,h?


Its not the whole question, I.was just asking in general. Because this is the question. 10a.
I find this well defined stufff so hard to understand.

I know u said that you can't remember rings too much but if u can, can u check this please.
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1414610613780.jpg92.1KB
1414610649413.jpg46.2KB
Original post by cooldudeman
Its not the whole question, I.was just asking in general.


Since p,h are not defined, and I can't see anything in your extra working that does so, then it's not even meaningful to talk about those sets, let alone compare them.

If you had "there exists p in Z such that p=q+2" in the predicate, and similarly to the other one, then yes, the two sets are equal, though it's rather a torturous way to define Z.


Because this is the question. 10a.
I find this well defined stufff so hard to understand.

I know u said that you can't remember rings too much but if u can, can u check this please.
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Again, rather torturous, but I think it's OK.

I would have gone:

Let [a]n=[b]n\text{Let } [a]_n=[b]_n

Then n|(a-b)

Since m|n it follows that m|(a-b) and so

[a]m=[b]m[a]_m=[b]_m

So, θ([a]n)=[a]m=[b]m=θ([b]n)\theta ([a]_n)=[a]_m=[b]_m=\theta ([b]_n)

and theta is well defined.
(edited 9 years ago)
Original post by ghostwalker
Since p,h are not defined, and I can't see anything in your extra working that does so, then it's not even meaningful to talk about those sets, let alone compare them.

If you had "there exists p in Z such that p=q+2" in the predicate, and similarly to the other one, then yes, the two sets are equal, though it's rather a torturous way to define Z.



Again, rather torturous, but I think it's OK.

I would have gone:

Let [a]n=[b]n\text{Let } [a]_n=[b]_n

Then n|(a-b)

Since m|n it follows that m|(a-b) and so

[a]m=[b]m[a]_m=[b]_m

So, θ([a]n)=[a]m=[b]m=θ([b]n)\theta ([a]_n)=[a]_m=[b]_m=\theta ([b]_n)

and theta is well defined.


OK no problem.

For part c, finding the kernel. Would it be ker(theta)={[r]€Z_n|r=sm for some s€Z}?
I have no idea what to do.for the image..
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(edited 9 years ago)
Original post by cooldudeman
OK no problem.

For part c, finding the kernel. Would it be ker(theta)={[r]€Z_n|r=sm for some s€Z}?
I have no idea what to do.for the image..
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Yep, looks good.

Re image. A clue is that it's onto.

Further very big hint in spoiler:

Spoiler

(edited 9 years ago)
Original post by ghostwalker
Yep, looks good.

Re image. A clue is that it's onto.

Further very big hint in spoiler:

Spoiler



Im sorry I still dont know. I know in general terms for this question im(theta)={b€Z_m|b=theta(p) for some p€Z_n}

But im not understanding what this is. The image in functions is the outcome when you input a value in the function. Is it the same for this case? So wouldnt the image just be [a]_m where a goes from 0 to n-1?. I really dont get it

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(edited 9 years ago)
Original post by cooldudeman
Im sorry I still dont know. I know in general terms for this question im(theta)={b€Z_m|b=theta(p) for some p€Z_n}

But im not understanding what this is. The image in functions is the outcome when you input a value in the function. Is it the same for this case?

Yes.


So wouldnt the image just be [a]_m where a goes from 0 to n-1?.


Yes.

I.e. the image is Zm\mathbb{Z}_m
(edited 9 years ago)
Original post by ghostwalker
Yes.



Yes.

I.e. the image is Zn\mathbb{Z}_n


Wait [a]_m where a is between 0 and n-1 is Z_n? Cuz I would have written the answer like {[h]€Z_m : 0<=h<=n-1}. This is the same as Z_n?

Can I also just ask what does [f] alone mean if f is an integer. Where it doesn't have the for eg _n next to it.
Original post by cooldudeman
Wait [a]_m where a is between 0 and n-1 is Z_n? Cuz I would have written the answer like {[h]€Z_m : 0<=h<=n-1}. This is the same as Z_n?


Yes.
Should be Z_m



Can I also just ask what does [f] alone mean if f is an integer. Where it doesn't have the for eg _n next to it.


If you were working solely in Z_n, then you might omit the n, and just use [f].

[ ] brackets tend to be used for equivalence classes, in general.
(edited 9 years ago)
Original post by ghostwalker
Yes.



Yes.

I.e. the image is Zn\mathbb{Z}_n

hi, isn't the image in fact Z_m? Because my lecturer was saying that.
Can you explain why its that because im assuming you just made a mistake.
Original post by cooldudeman
hi, isn't the image in fact Z_m? Because my lecturer was saying that.
Can you explain why its that because im assuming you just made a mistake.


Yep, my mistake, it's Z_m, the codomain - got the two mixed up.

I was thrown of by:

Original post by cooldudeman
Wait [a]_m where a is between 0 and n-1 is Z_n? Cuz I would have written the answer like {[h]€Z_m : 0<=h<=n-1}. This is the same as Z_n?



In bold, should be Z_m

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