Uni Stuff > Mathematics
Group Theory and Linear Algebra Thread
/0:
Well if you write and as a products of factors, since and share no common factors, you can multiply them together and they must still divide .
TrueTears:
--- Quote from: QuantumJG on July 29, 2010, 06:25:35 pm ---I saw my lecturer about this question today, but I forgot how he proved it.
Let with . Prove that if a|c and b|c then ab|c.
--- End quote ---
We have ax + by = 1, c = ae and c = bf for some integers x, y, e and f. Then c = cax + cby = (bf)ax + (ae)by = ab(fx + ey), so ab|c.
QuantumJG:
--- Quote from: TrueTears on July 29, 2010, 10:04:01 pm ---
--- Quote from: QuantumJG on July 29, 2010, 06:25:35 pm ---I saw my lecturer about this question today, but I forgot how he proved it.
Let with . Prove that if a|c and b|c then ab|c.
--- End quote ---
We have ax + by = 1, c = ae and c = bf for some integers x, y, e and f. Then c = cax + cby = (bf)ax + (ae)by = ab(fx + ey), so ab|c.
--- End quote ---
Yeah that's how he proved it. Thanks!
QuantumJG:
I'm having trouble with this:
Show that if with (mod 8), then the equation
has no solutions if
WTF!
brightsky:
Actually forget mod 4, do it in mod 8.
We know that given a whole number is odd, then we have .
Likewise (tweaked from mod 4 case), if is even, then we have two cases:
1. or
2.
From here, by brute force, we can work out that the possibilities for can be any number other than 7.
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