Discussion:
Primitive 10th-root
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unknown
2003-09-04 08:51:17 UTC
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Hi, everyone.
I have a question about method for finding primitive 10th-root.
since 2^i=/=1 for 1<=i<=9, and 2^10=1 in Z_11
I see 2 is a primitive 10th-root of unity.

but, in case more large number 6,7,8,
Except direct calculation,(<-this is too big number calculation to me)
How can I see 6,7,8 are also primitive 10th-root of unity?
Surely,
without direct calculation, by using Fermat's little theorem,
i see 6^10=1(mod 11), 7^10=1(mod 11), 8^10=1(mod 11).
but, without complex calculation,
how can I convinced that there is no integer i,j,k (from 1 to 9) s.t. 6^i=1,
7^j=1, 8^k=10 (mod 11)?

Thanks for any help.
William Elliot
2003-09-04 10:51:07 UTC
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Post by unknown
I have a question about method for finding primitive 10th-root.
since 2^i=/=1 for 1<=i<=9, and 2^10=1 in Z_11
I see 2 is a primitive 10th-root of unity.
but, in case more large number 6,7,8,
Except direct calculation,(<-this is too big number calculation to me)
How can I see 6,7,8 are also primitive 10th-root of unity?
Surely,
without direct calculation, by using Fermat's little theorem,
i see 6^10=1(mod 11), 7^10=1(mod 11), 8^10=1(mod 11).
but, without complex calculation,
how can I convinced that there is no integer i,j,k (from 1 to 9) s.t. 6^i=1,
7^j=1, 8^k=10 (mod 11)?
Let p be a prime. Then Z_p is a finite field with p elements.
Z_p^*, the multiplicative group of Z_p, has p-1 elements.

Now use theorem, if G is a finite group of size n,
then for all x in G, x^n = identity.

Thus Fermat's little theorem
x /= 0 ==> x^(p-1) = 1 (mod p)
Jon and Mary Miller
2003-09-04 18:36:02 UTC
Permalink
Post by William Elliot
Post by unknown
I have a question about method for finding primitive 10th-root.
since 2^i=/=1 for 1<=i<=9, and 2^10=1 in Z_11
I see 2 is a primitive 10th-root of unity.
but, in case more large number 6,7,8,
Except direct calculation,(<-this is too big number calculation to me)
How can I see 6,7,8 are also primitive 10th-root of unity?
Surely,
without direct calculation, by using Fermat's little theorem,
i see 6^10=1(mod 11), 7^10=1(mod 11), 8^10=1(mod 11).
but, without complex calculation,
how can I convinced that there is no integer i,j,k (from 1 to 9) s.t. 6^i=1,
7^j=1, 8^k=10 (mod 11)?
Let p be a prime. Then Z_p is a finite field with p elements.
Z_p^*, the multiplicative group of Z_p, has p-1 elements.
Which, in the case of p=11, means that the order of any element is 1, 2,
5, or 10. (Lagrange's Theorem) That certainly limits calculations.

For example, 6^2 = 36 = 3 (mod 11) . Now, 6^5 = (6^2)^2*6 (creative use
of parentheses) = 3^2*6 = 9*6 = 54 = 10 (mod 10). Not calculation-free,
but something you can easily do with pencil and paper. Or even in an
extremely limited (ASCII only) text processor.

Jon Miller
unknown
2003-09-05 07:59:28 UTC
Permalink
Post by Jon and Mary Miller
Post by William Elliot
Post by unknown
I have a question about method for finding primitive 10th-root.
since 2^i=/=1 for 1<=i<=9, and 2^10=1 in Z_11
I see 2 is a primitive 10th-root of unity.
but, in case more large number 6,7,8,
Except direct calculation,(<-this is too big number calculation to me)
How can I see 6,7,8 are also primitive 10th-root of unity?
Surely,
without direct calculation, by using Fermat's little theorem,
i see 6^10=1(mod 11), 7^10=1(mod 11), 8^10=1(mod 11).
but, without complex calculation,
how can I convinced that there is no integer i,j,k (from 1 to 9) s.t. 6^i=1,
7^j=1, 8^k=10 (mod 11)?
Let p be a prime. Then Z_p is a finite field with p elements.
Z_p^*, the multiplicative group of Z_p, has p-1 elements.
Which, in the case of p=11, means that the order of any element is 1, 2,
5, or 10. (Lagrange's Theorem) That certainly limits calculations.
For example, 6^2 = 36 = 3 (mod 11) . Now, 6^5 = (6^2)^2*6 (creative use
of parentheses) = 3^2*6 = 9*6 = 54 = 10 (mod 10). Not calculation-free,
but something you can easily do with pencil and paper. Or even in an
extremely limited (ASCII only) text processor.
Jon Miller
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