Loser66
  • Loser66
Find a reduced residue system modulo \(2^m\) where m is a positive integer Please, help.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
I know that 2^m is an even number and its reduce residue is all odd number in the set \(\{1,2, 3, ......,2^{m-1}\} \). That is \(\{2k -1| 1< k < 2^{m-1}\}\) All I want is finding it by \(\phi (2^m)\) but I don't know how.
ganeshie8
  • ganeshie8
I don't get the question... could you please share full details...
Loser66
  • Loser66
Is my answer correct? I meant reduce residue modulo \(2^m = \{2k-1 | 1

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More answers

imqwerty
  • imqwerty
yes
Loser66
  • Loser66
And we have some formula to find \(\phi (2^m)\), right?
imqwerty
  • imqwerty
but i don't know what is ϕ(2^m)
ganeshie8
  • ganeshie8
sorry i still don't understand the question...
Loser66
  • Loser66
\(\phi (2^m)= \) number of the terms in residue of 2^m, namely a, such that (a, 2^m) =1
ikram002p
  • ikram002p
ur answer is correct for the reduced residues
ganeshie8
  • ganeshie8
\(\phi(p^k) = p^k - p^{k-1}\)
Loser66
  • Loser66
Like this. if number is 5, then \(\phi (5) = 4\) because in {0,1,2,3,4}, (1,5)=1, (2,5) =1, (3,5) =1 and (4,5) =1 we have 4 number "a" satisfy (a,5) =1 @imqwerty
ikram002p
  • ikram002p
but in mean while phi(2^m) it counted terms in ur set =D
ganeshie8
  • ganeshie8
Notice that the integers that are "not" relatively prime to \(2^m\) are : \[\{1*2, 2*2, 3*2, \ldots, 2^{m-1}*2\}\] that means there are exactly \(2^{m-1}\) integers less than \(2^m\) that are "not" relatively prime to \(2^m\)
ikram002p
  • ikram002p
what is good about about 2^m is that the odd set less that 2^m satisfy the reduced residue which is what u did @Loser66 .
Loser66
  • Loser66
@ganeshie8 I got you
Loser66
  • Loser66
@ikram002p I want to solve it by arithmetic way,
ikram002p
  • ikram002p
=)
ganeshie8
  • ganeshie8
\(\phi\) is a multiplicative function, so we can evaluate it for any integer using : \[\phi(ab) = \phi(a)*\phi(b)\] where \((a,b)=1\)
ganeshie8
  • ganeshie8
for example : \(\phi(15) = \phi(3*5) = \phi(3)*\phi(5) = (3-1)(5-1)=8\) that means, there are exactly 8 positive integers less than 15 that are coprime with 15.
ganeshie8
  • ganeshie8
since every positive integer less than a prime is relatively prime, we also have : \(\phi(p)=p-1\) for a prime, \(p\)
ganeshie8
  • ganeshie8
those two properties allow you to compute \(\phi(n)\) for any integer \(n\)
Loser66
  • Loser66
how? if n is not prime, how can you compute phi (n) ?

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