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• Today, 01:54
Yes, I don't see how the answers given by your book could possibly be correct.
3 replies | 15 view(s)
• Today, 01:52
Because we both made the same mistake reading the problem. We are told: Ahmad has x marbles. He has 40 more marbles than Weiming This means...
4 replies | 16 view(s)
• Today, 01:45
Looks good! (Yes)
1 replies | 13 view(s)
• Today, 01:38
Yes, that's good. Here, they are telling you: m+4=40 So, you need to solve for $$m$$ to determine how many pupils were there in the...
3 replies | 15 view(s)
• Today, 01:35
That looks good! (Yes)
4 replies | 16 view(s)
• Yesterday, 21:51
Thanks Steenis ... I appreciate your help ... Peter
2 replies | 47 view(s)
• Yesterday, 02:54
I am reading Paul E. Bland's book, "Rings and Their Modules". I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need...
2 replies | 47 view(s)
• Yesterday, 01:53
Thanks for the help GJA ... But ... just a clarification ... I can verify that d \mid 1 and that d|1\Longrightarrow c|1 ... but I cannot follow...
3 replies | 108 view(s)
• August 14th, 2018, 22:26
Thanks GJA ... Appreciate your help ... Peter
4 replies | 96 view(s)
• August 14th, 2018, 12:36
Hello all, A friend of mine on another forum, knowing I am involved in the math help community, approached me regarding a question in statistics....
9 replies | 178 view(s)
• August 14th, 2018, 01:07
Thanks GJA ... OK ... then consider the ring R = \mathbb{Z}_{6} \equiv \mathbb{Z} / 6 \mathbb{Z} = \{ \overline{0}, \overline{1}, \overline{2},...
4 replies | 96 view(s)
• August 14th, 2018, 00:11
Thanks to Steenis and Opalg for clarifying Bland Proposition 4.3.5 ... Hmm... ... seems that Bland made a bit of a mess of that Proposition ... ...
16 replies | 272 view(s)
• August 14th, 2018, 00:06
I am reading Paul E. Bland's book, "Rings and Their Modules". I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need...
3 replies | 108 view(s)
• August 13th, 2018, 19:20
I would look at all factors present, and take the smaller power present in each: 2\cdot3^2\cdot7=126
3 replies | 86 view(s)
• August 13th, 2018, 11:34
The problem states:
5 replies | 119 view(s)
• August 13th, 2018, 00:43
Looks good! (Yes)
5 replies | 119 view(s)
• August 13th, 2018, 00:35
I just wanted to say, I'm really liking the way you title your threads usefully and show your work. (Yes)
3 replies | 61 view(s)
• August 13th, 2018, 00:14
You've got Kat and Nora right, but Devi would receive 24 + 2x (that's 2x more than Kate). And so the sum $$S$$ would be: ...
3 replies | 61 view(s)
• August 12th, 2018, 23:19
I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ......
1 replies | 65 view(s)
• August 12th, 2018, 23:13
I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ......
4 replies | 96 view(s)
• August 12th, 2018, 21:20
Thanks for the help Olinguito ... Your assistance is very much appreciated ... Peter
16 replies | 272 view(s)
• August 12th, 2018, 06:57
Hi Olinguito ... thanks again for your posts and help ... Now ... just a clarification ... In the post above, you write the following ... ...
16 replies | 272 view(s)
• August 11th, 2018, 09:02
In his very interesting post above, Olinguito indirectly implies (I think) that (up to an isomorphism) there is only one ring of two elements and one...
16 replies | 272 view(s)
• August 11th, 2018, 02:25
Thanks for a most interesting and VERY helpful post Olinguito ... I very much appreciate your help ... Thanks again ... Peter
16 replies | 272 view(s)
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1. Happy birthday, Bacterius!
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