Recent content by raross

  1. R

    Solve 2^27841 mod 34 by Hand: Discrete Math Problem Solution

    hrm yeah that works. How would you solve it with modular exponentiation?
  2. R

    Solve 2^27841 mod 34 by Hand: Discrete Math Problem Solution

    could someone show me how u would solve 2^27841 mod 34 by hand? I know what theorm to use, I am just having trouble using it? Thanks
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    Solve Easy Precal Questions: Quadrant II, Sin 2(0\), Trig Expressions

    Can someone help me with a few simple problems: These 0's have lines through them, and are slanted. I used 0\ to represent that. 1.) If sin0\ = 3/5 and 0\ is in quadrant II, find sin 2(0\). (Hint: sin 2(0\) = 2 sin 0\ cos 0\)? 2.) The trigonometric expression, sin 0\(cot 0\ + tan 0\)...
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    Cantor Expansion Addition algorithm?

    How come no one can help me ? =/
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    Cantor Expansion Addition algorithm?

    Could someone help me and write an algorithm to add 2 Cantor expansions. The algorithm to get a decimal number to cantor expansion is: procedure decimal-to-cantor(x: positive integer) n := 1 y := x fy is a temporary variable used so that this procedure won't destroy the original value of...
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    Cantor Expressions: Solving A-F (2,7,19,87,1000,1M)

    2 = 1*1! + 1 7 = 2*2! + 1*1! + 2 19 = 3*3! + 1 87 = 4*4! - 9 Hrmm, how about these answers? Anyone have a clue?
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    Cantor Expressions: Solving A-F (2,7,19,87,1000,1M)

    Im not sure how many digits you can put at the end, but it only makes sense I guess. Anyone help?
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    Cantor Expressions: Solving A-F (2,7,19,87,1000,1M)

    So do these answers make sense or am I doing it wrong? 2 = 1*1! + 1 7 = 2*2! + 1*1! + 2 19 = 2*2! + 1*1! + 14 87 = 3*3! + 2*2! + 1*1! + 64 Thanks for your help!
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    Cantor Expressions: Solving A-F (2,7,19,87,1000,1M)

    What is the cantor expansion of: A. 2 B. 7 C. 19 D. 87 E. 1000 F. 1,000,000 The algorithm to solve these small problems is the most difficult for me. The algorithm that I came up with states: Asub(n) N! + Asub(n-1) (n-1)! +...+ Asub(2)2! + Asub(1)1!, where Asub1 is an integer with...
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    Is f(x) = 2x Onto, One-to-One, or a Bijection for Different Domains?

    Hey if anyone could help me with this I would be sooo grateful. I am trying to grasp the idea of onto, one-to-one and bijection(both) functions. A sample problem is: If f(x) = 2x . What is f(Z), all integers. What is f(N), all naturals. What is f(R), all real. These are 3 different problems...
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