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CTK Exchange
Tibi
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Sep-02-05, 06:32 AM (EST) |
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"Gaussian factorisation"
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There is a unique factorisation theorem for Gaussian integers.
I factorised 20 in two different ways:(1) -1*(2+i)*(2-i)*(1+i)^4, and
(2) -1*(2+i)*(2-i)*(1-i)^4. If I understood the concept of Gaussian primes correctly, then all the factors in (1) and (2) are primes (and -1 is a unit). The two factorisations are different. Then how does the unique factorisation theorem hold? |
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alexb
Charter Member
1636 posts |
Sep-02-05, 07:43 AM (EST) |
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1. "RE: Gaussian factorisation"
In response to message #0
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>(1) -1*(2+i)*(2-i)*(1+i)^4, and
>(2) -1*(2+i)*(2-i)*(1-i)^4.
>
>Then how does the unique factorisation theorem hold? The factorization is unique up to the factors of unity. i is one such, and since i·(1 - i) = 1 + i, there is no contradiction.
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Tibi
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Sep-02-05, 10:39 AM (EST) |
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2. "RE: Gaussian factorisation"
In response to message #1
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Thank you.
Does it mean that 1+i is not a Gaussian prime?Something else that I would be grateful for : if somebody could please suggest a Gaussian factorisation algorithm. Thanks |
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