Recent content by shamus390

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    Basic proof of units in a ring with identity.

    Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary... So if we let a, b and c be elements of R such that ab = ac, then: ab-ac=0R = a(b-c) = 0R, and a is not a zero divisor \Rightarrow (b - c) = 0R \Rightarrow (b-c) =...
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    Basic proof of units in a ring with identity.

    Homework Statement Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units. Homework Equations If ab is a unit then (ab)c=1=c(ab) for some c in R. The Attempt at a Solution Assume both a and b are...
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    Proof of zero divisor existence.

    [b]1. Let a != 0 and b be elements of the integers mod n. If the equation ax=b has no solution in Zn then a is a zero divisor in Zn The Attempt at a Solution Not sure where to start on this proof, I keep trying to find something using the properties of modular arithmetic but am coming up empty
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    Equation of a plane multipled by a constant.

    Ah, dropped the negative sign, is x-y=-1 correct?
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    Equation of a plane multipled by a constant.

    So essentially he is using ∏ to name the plane? Either I'm misunderstanding or this was a strange question (its from a review sheet for an exam Thursday).
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    Equation of a plane multipled by a constant.

    Homework Statement Given the points A(1,2,3) B(0,1,2) and C(2,3,-1) find: a.) a vector perpendicular to the plane pi(A,B,C) b.) the equation of the plane pi(A,B,C)The Attempt at a Solution a.) ∏<5,-5,0> b.)∏(x-y)=∏ Am I incorrect in assuming that I would find the normal vector and plane...
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    Unit Vector with Angle θ=-3∏/4 from Positive X-Axis

    Find the unit vector that makes an angle θ=-3∏/4 with the positive x-axis I know to find a unit vector you divide the given vector by its magnitude, so I guess my problem is finding any vector that makes that angle with the positive x axis. I figured if that angle was the slope of a line, then...
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