20081215, 16:06  #1 
Dec 2008
7^{2}×17 Posts 
So basically one can solve modular functions using a Diophantine equation? Wow, I did not know that. Thus, (a)mod(b) = c, then we can equate it to the equation, a = c+kb, where k is an integer? Correct me if I am wrong, but that is awesome. Thanks!

20081215, 18:37  #2 
Aug 2006
3^{2}·5·7·19 Posts 
That's the definition of modular equivalence...

20081215, 19:16  #3  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Huh? Awesome??? You have a strange notion of awesome. It is the definition! A = B mod C if (AB) is divisible by C. This is basic. Some advice: before tackling ANY mathematical problem, it is best to know the underlying definitions. 

20081215, 20:02  #4  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Oh, c'mon, Dr. Silverman. CRGreathouse simply informed flouran about the definition of modular equivalence without any putdown. You didn't need to add anything.
A math novice has just learned something for the first time and is excited about it! Can't you remember being there yourself sometime in the past? ... or were you born fullyknowledgeable, mature and armed like Athena? Why be a wet blanket to a young mortal? (BTW, are you sure you're uptodate on colloquial uses of "awesome"?)    FYI: A local PBS station is featuring "Brain Fitness Program and Neuroplasticity". Someone who discovered it a year ago wrote: http://midlifecrisisqueen.com/2007/1...inplasticity/ Quote:
Last fiddled with by cheesehead on 20081215 at 20:29 Reason: Hint, hint ... 

20081215, 23:37  #5 
Dec 2008
7^{2}×17 Posts 
For your information, R.D. Silverman and C.R. Greathouse, I never knew that the mod function could be represented by a Diophantine equation. BTW, I'm not a novice in general mathematics; I am a novice in number theory. If you really want to know, I am taking Linear Algebra and will take Differential Equations next year. Therefore, your comments certainly did not help either the forum or my investigation. Thus, they were both futile, immature, and not to mention stupid. But, I'll give you "masters" a piece of advice: when you want to post something, make sure it actually helps someone. Otherwise, you're not only wasting my time, but your time as well (and I am sure you guys do a lot in your free time). And, thanks, cheesehead for pointing out to someone the power of free speech.

20081216, 23:33  #6  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:
Quote:
And I don't want to speak for Dr. Silverman (a true expert), but I suspect he *would* call a student taking Linear Algebra and (soon) DiffEq a novice. 

20081217, 00:03  #7 
Dec 2008
7^{2}×17 Posts 
Yeah, CRGreathouse and Dr. Silverman, it's fine. Do either of you guys know how to solve the above equation I posted though? That would really be an excellent help.

20081217, 01:49  #8  
Aug 2006
3^{2}·5·7·19 Posts 
Quote:
Assuming the modulus is positive (a usual assumption), you have 2x mod x+1 = x1 mod x+1 = 9 so x1 = 9 and x = 10. There's not much to solve. If you let the modulus be negative you get another solution, but I'm not sure you'd count that. 

20081217, 01:52  #9  
Dec 2008
7^{2}·17 Posts 
No, that was not the equation I was asking for; that one is easy to solve. This is the equation I wanted to be solved:
Quote:


20081217, 03:01  #10  
Nov 2003
2^{2}×5×373 Posts 
Quote:
And you did not post an equation. You posted a congruence. If you don't even know what the mathematical objects are that you are manipulating, how will you solve anything? 

20081217, 03:04  #11  
Nov 2003
2^{2}×5×373 Posts 
Quote:
A modular relation is NOT A FUNCTION. It is an equivalence relation. You *are* a novice to the extent that you have not yet learned that problems begin with precise definitions. 

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