anonymous
  • anonymous
What exactly is a mutually incongruent solution of a congruence? Can you give an example?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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ganeshie8
  • ganeshie8
\(1\) and \(6\) are not mutually incongruent in mod \(5\) because \(1\equiv 6\pmod{5}\)
ganeshie8
  • ganeshie8
\(1\) and \(3\) are mutually incongruent in mod \(5\) because \(1\not\equiv 3\pmod{5}\)
ganeshie8
  • ganeshie8
incongruent is just an opposite of congruent

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anonymous
  • anonymous
ok, so two numbers m and n are mutually incongruent mod c means m not congruent n (mod c)
ganeshie8
  • ganeshie8
Yep!
anonymous
  • anonymous
For this example: Find a complete set of mutually incongruent solutions of 7x≡5 (mod 11)
anonymous
  • anonymous
x = 7, so all solutions are x = 7 + 11k, k an integer
ganeshie8
  • ganeshie8
the answer should be just \(\{7\}\)
anonymous
  • anonymous
why {7} ? and not, say {18} ??
ganeshie8
  • ganeshie8
sure, you can put any single integer that is congruent to 7
ganeshie8
  • ganeshie8
but the numbers 0 to n-1 are more convenient and look better
anonymous
  • anonymous
But how does the number 7 or any number congruent to 7 mod 11 capture the idea of "mutually incongruent" solution? if 7 not congruent m (mod 11), what numbers play in the role of m according to the definition of mutually incongruent?
ganeshie8
  • ganeshie8
there are no other integer solutions, it is like asking for all the mutually perpendicular vectors that are solutions, if there is only one vector, you give only that.
ganeshie8
  • ganeshie8
every other solution is congruent to 7, so there are no other incongruent solutions
anonymous
  • anonymous
huhm... ok, let me give another example where it has more than 1 mutually incongruent solution. 9x ≡ 12 (mod 15). Since gcd(9,15) = 3, there will be 3 mutually incongruent solutions
ganeshie8
  • ganeshie8
looks good
ganeshie8
  • ganeshie8
no wait, how do you know there exists a solution ?
anonymous
  • anonymous
because gcd(9,15) = 3 | 12
anonymous
  • anonymous
I see x = 3 is one of the solution so all solution is x = 3 + k * 15/gcd(9,15) x = 3 + 5k
ganeshie8
  • ganeshie8
"If a solution exists", then there will be exactly 3 incongruent solutions
ganeshie8
  • ganeshie8
consider below congruence : \[9x\equiv 2\pmod{15}\] how many solutions ?
anonymous
  • anonymous
none, gcd(9,15) does not divide 2
anonymous
  • anonymous
right?
ganeshie8
  • ganeshie8
Yep!
anonymous
  • anonymous
so.....???
anonymous
  • anonymous
let's get back to 9x ≡ 12 (mod 15). There will be exactly 3 mutually incongruent solutions. And all solutions are x = 3 + 5k
anonymous
  • anonymous
what are the mutually incongruent solutions?
ganeshie8
  • ganeshie8
let k = 0,1,2
ganeshie8
  • ganeshie8
the incongruent solutions are \(\{3,8,13\}\) they are mutually incongruent in mod \(15\)
anonymous
  • anonymous
how do you know to let k = 0,1,2??
ganeshie8
  • ganeshie8
they are the most convenient ones
ganeshie8
  • ganeshie8
letting k = 222, 223,224 is technically fine but you're not keeping things simple here
anonymous
  • anonymous
ok, let 0 <= <= 10, then x = 3,8,13, 18, 23, 28, 33, 38, 43, 48, 53
ganeshie8
  • ganeshie8
3 and 18 are not incongruent
anonymous
  • anonymous
right. I was trying to understand how to find the incongruent solutions by listing a few solutions
anonymous
  • anonymous
what i notice is 3 + 15 = 18 8 + 15 = 23 13 + 15 = 28
anonymous
  • anonymous
it's somehow like {3,8,15}, {18,23,28}
ganeshie8
  • ganeshie8
notice that any integer greater than or equal to 15 is congruent to some integer in the set \(\{0,1,2,\ldots,14\}\)
ganeshie8
  • ganeshie8
so we don't need to bother about anything greater than 14
ganeshie8
  • ganeshie8
and we don't need to bother about anything less than 0
ganeshie8
  • ganeshie8
the set of residues \(\{0, 1, 2, \ldots, n-1\}\) is our most favorite one, when somebody asks you for incongruent solutions, they expect you to give them from this set
anonymous
  • anonymous
So {3,8,15} are the set of mutually incongruent because 3 + 5k not congruent 8 + 5s not congruent 15 + 5t, for all integer k,s,t Is this what the incongruent solutions really mean?
anonymous
  • anonymous
mod 15 btw
ganeshie8
  • ganeshie8
{3, 8, 13} right
anonymous
  • anonymous
opps yeah. Typo :D
anonymous
  • anonymous
3 + 5k not congruent 8 + 5s not congruent 13 + 5t (mod 15), for all integer k,s,t
ganeshie8
  • ganeshie8
3 + 15k not congruent 8 + 15s not congruent 13 + 15t, for all integer k,s,t in mod 15
ganeshie8
  • ganeshie8
notice i have changed 5k to 15k
ganeshie8
  • ganeshie8
3 + 5k refers to mod 5 3 + 15k refers to mod 15
anonymous
  • anonymous
I think I understand it better now although I still need to do some examples to clear up any confusion. @ganeshie8 thank you :)
ganeshie8
  • ganeshie8
basically we're "lifting" the solutions from mod5 to mod15, so \(x\equiv 3\pmod{5}\) gives 3 incongruent solutions in mod 15 : \(\{3,8,13\}\)
anonymous
  • anonymous
yeah, that's what I noticed earlier when I said 3 + 15 = 18 8 + 15 = 23 13 + 15 = 28
anonymous
  • anonymous
so instead of expressing all solutions in the form of 3 + 5k, we can express all solutions in terms of mutually incongruent soltuions. I.e {x | x = 3 + 15k or x = 8 + 15k or x = 13 + 15k }
ganeshie8
  • ganeshie8
you may simply say, the solutions are \(x\equiv 3\pmod{5}\)
ganeshie8
  • ganeshie8
you could also say \(x\equiv 3,8,13\pmod{15}\)
anonymous
  • anonymous
right right. That's exactly what I was trying to say :D
ganeshie8
  • ganeshie8
if you're not explicitly asked for the form of solutions, leaving it as \(x\equiv 3\pmod{5}\) looks neat
anonymous
  • anonymous
awesome. Thanks

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