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 2008-12-15, 16:06 #1 flouran     Dec 2008 72·17 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!
2008-12-15, 18:37   #2
CRGreathouse

Aug 2006

597910 Posts

Quote:
 Originally Posted by flouran 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!
That's the definition of modular equivalence...

2008-12-15, 19:16   #3
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

164448 Posts

Quote:
 Originally Posted by flouran 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!

Huh? Awesome??? You have a strange notion of awesome.

It is the definition! A = B mod C if (A-B) is divisible by C. This is
basic.

Some advice: before tackling ANY mathematical problem, it is best
to know the underlying definitions.

2008-12-15, 20:02   #4

"Richard B. Woods"
Aug 2002
Wisconsin USA

170148 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 fully-knowledgeable, mature and armed like Athena?

Why be a wet blanket to a young mortal?

(BTW, are you sure you're up-to-date 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...in-plasticity/

Quote:
 Scientists have learned so much in the past ten years in the field of neurogenesis (how and why your brain creates new neuron networks). It is essential that we keep expanding these networks so that our brain’s life span equals our body’s life span. Guess what the worst thing is for expanding your brain’s neural networks? You guessed it! Staying in the same job too long! If you aren’t constantly challenging your brain to integrate new information and try new activities, your brain’s plasticity is suffering. Expanding your neural horizons is the key to successful brain health, especially with aging. We need to be constantly pushing ourselves to expand into new areas of interest, not just keep doing what we’re already good at. In other words, career change is good for your brain! Your brain is actually asking you to take up new challenges! Learning new things changes your brain chemistry, and all for the better.
Hint, hint ...

Last fiddled with by cheesehead on 2008-12-15 at 20:29 Reason: Hint, hint ...

 2008-12-15, 23:37 #5 flouran     Dec 2008 83310 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.
2008-12-16, 23:33   #6
CRGreathouse

Aug 2006

175B16 Posts

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by flouran 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!
That's the definition of modular equivalence...
Quote:
 Originally Posted by flouran 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.
I'm sorry my comment didn't help you. I thought it would be useful to point out that the property you mentioned is in fact the definition of modular equivalence. It seemed a useful point since if you had something contradictory in mind you could mention it and it could be discussed more fully.

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.

 2008-12-17, 00:03 #7 flouran     Dec 2008 83310 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.
2008-12-17, 01:49   #8
CRGreathouse

Aug 2006

597910 Posts

Quote:
 Originally Posted by flouran 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.
Your equation is just 2x mod x+1 = 9, right?

Assuming the modulus is positive (a usual assumption), you have
2x mod x+1 = x-1 mod x+1 = 9
so x-1 = 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.

2008-12-17, 01:52   #9
flouran

Dec 2008

72×17 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:
 Originally Posted by flouran Quick Question: Say I have (x!)mod(x+3) = 3, how would I use the Diophatine definition of the modular function to solve for x? My approach would be, 3 + k(x+3) = x!, and therefore, 3+kx+3k - x! = 0. Similarly, I could use the Gamma definition of the factorial function to say, $3+kx+3k - \Gamma(x+1) = 3+kx+3k - x \Gamma(x) = 0$ Then how would I solve for x from there? P.S. At least one of the answers for x should be 4.

2008-12-17, 03:01   #10
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×5×373 Posts

Quote:
 Originally Posted by flouran 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.

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?

2008-12-17, 03:04   #11
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22·5·373 Posts

Quote:
 Originally Posted by flouran 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.
And you still have not taken my advice and learned the DEFINITIONS.
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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