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 Post subject: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 16:22:37 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
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Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 16:34:09 UTC
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DgrayMan wrote:
Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

What is a'? What is "integers with x and a"?

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 17:01:35 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
a' means a inverse. and x and a are in G, which is a set of integers

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:07:32 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
DgrayMan wrote:
Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

I think you'd best get your notation fixed, as it stands you're asking to show that an integer, has a cube root if there is another integer such that , which I guess is true, but only because these conditions imply that and .

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:12:39 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
I wrote the problem wrong sorry its (x^2)ac=a' . its hard to write it here on the forum so that's the best i can do. the set is {G,*} the problem is x squared ac = a'

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:14:40 UTC
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DgrayMan wrote:
I wrote the problem wrong sorry its (x^2)ac=a' . its hard to write it here on the forum so that's the best i can do. the set is {G,*} the problem is x squared ac = a'

If you're having problems typesetting equations, check out our help board at this link.

There's still the problem of what is c in this expression now? Can you pick it? Is it fixed? Is it supposed to be an integer?

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:26:19 UTC
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Wait, is this supposed to be a group theory question? If so you need to specify the group operation, since you appear to be writing it multiplicatively. Since your last post was similar and on group theory I'm assuming this is the case and moving this to Algebra and Number Theory, if this really is just a basic algebra question, you should tell me to move it back.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:34:03 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
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ugh. i mean (x^2)ax=a'. no c in there.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:36:02 UTC
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DgrayMan wrote:
ugh. i mean (x^2)ax=a'. no c in there.

Again, my question is as to whether or not this is a group theory question, and what the group operation is. Integers under addition doesn't make sense, because then it's false just by other facts about integers.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:46:02 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
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yes its group theory, the operation is just any operation denoted by *.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:51:13 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
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im gonna scan the page this problem is on and post it so this confusion will end.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 18:54:14 UTC
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DgrayMan wrote:
im gonna scan the page this problem is on and post it so this confusion will end.

That's probably best. It's common to write for a general group operation, but you specified that the set was the integers, so generally this means you also want addition to be your group operation.

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 21:59:52 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
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Question #7 is the one I'm working on

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Mon, 30 Jan 2012 22:13:54 UTC
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DgrayMan wrote:
Question #7 is the one I'm working on

Oh, then this is easy.

, now using the fact that the group law is associative, and that we get:

On the other hand, we know:

So we get:

, but then so , hence is a cube root for .

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 Post subject: Re: if (x^3)ax=a' then a has a cube root.Posted: Tue, 31 Jan 2012 17:39:31 UTC
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Joined: Sat, 21 Jan 2012 03:59:22 UTC
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you gotta be kidding me?!?

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