Recent content by General_Sax

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    (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)

    Thanks for the additional effort/attention Fredrik, but I've already used the geometric "series" -- perhaps it's more accurate to use the term "sum" -- in a proof. http://en.wikipedia.org/wiki/Geometric_series#Formula that's the one I used. Just split it up w/ some algebra. It's for a...
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    (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)

    Thanks for the help people. I think I've got it. Just used formula for geometric series and did some algebra -- hope it's good enough.
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    (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)

    So, there is no theorem to use? @Dickfore I'm confused as to the next step -- yes I've been trying to work it out. Should I try to factor (x-1) out of the expression?
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    (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)

    (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1) Where does this factorization come from? I need to know so I can use it in a proof. Thanks.
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    Prime Numbers: (2^n - 1) and (2^n + 1)

    Where does this factorization come from? I just need a link or something. Thanks.
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    Help me understand a homework solution - intro to ring theory - ideals

    Thanks. I appreciate it.
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    Help me understand a homework solution - intro to ring theory - ideals

    Help me understand a homework solution -- intro to ring theory -- ideals problem: solution: The first paragraph is just saying the ideals generated by the units in the ring is the whole ring correct? Also, the principal ideals generated by 2, 4 and 8 are all the same correct? So...
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    Show that a f: Z -> R , n -> n*1(subr) is a homomorphism of rings

    Homework Statement Show that a f: Z → R , n → n*1R is a homomorphism of rings Homework Equations The Attempt at a Solution I'm not sure how to exactly go about answering this question, but I'm going to try to start with the definition: f(a+b) = f(a) + f(b) f(a*b) = f(a) * f(b)...
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    Mod(prime) is a field mod(non-prime) is not

    Thanks
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    Mod(prime) is a field mod(non-prime) is not

    Not only that, but I don't truly understand why this is.
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    Mod(prime) is a field mod(non-prime) is not

    Now, I can show that if n is prime then Z/Zn is a field a = a b = an-2 a*b = an-1 = 1 (mod n) --> Fermat's little theorem However, I can't really seem to show that there is no multiplicative inverse for Z/Zn where n is not prime. First question: a =/=b correct? i know that there...
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    Calculus III or Thermodynamics: Which is the Best Fit for Next Semester?

    I would take thermo this term, calc3 next and PDE after that.
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    Question related to congruence class equations

    that supposed to read a!= 0 and b != 0 ... srry just copy and pasted.
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    Question related to congruence class equations

    Homework Statement 3) Let a be an integer = 0 and 6 n a natural number. Show that if gcd(a, n) = 1 then 6 there exists b ∈ Z, such that [a] · [b] = [0] and [b] = [0] in 6 Z/Zn Homework Equations The Attempt at a Solution Ok, so I'm still trying to digest the question and so...
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    Schools University of Toronto: Bad for undergrad?

    If the UofT is like the UofA, this doesn't matter much, because the class average is given along with your grade on your transcripts. So, if the class avg is a C+ and you get a B+, then grad admissions should see that you're pretty good student.
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