1. A

    Modulus proof involving least positive residues

    I'm a bit stuck on this question: Let p be a prime number. Prove that (2p−1)(2p−2)···(p + 1) ≡−1(mod p). Hint: find the least positive residues of 2p−1,2p−2,...,p + 1 (modulo p). I think the least positive residues are {0,1,2,...,p} but not entirely sure, and I don't see how it's going to help...
  2. V

    Modulus Function Help

    Sketch, on seperate diagrams, the graphs of y=|x|, y=|x-3|, y=|x-3| + |x-3|. Find the solution set of the equation |x-3| + |x-3| = 6
  3. E

    modulus (2)

    it is a if and only if question 1<(4x-4)/(2x+3)<3 if 1<(4x-4)/(2x+3)<3 then |x+3/2|>5 2x+3<4x-4<6x+9 7<2x 4x-4<6x+9 7/2<x -13<2x 5<x+3/2 -13/2<x -5<x+3/2 then does it consider as |x+3/2|>5?
  4. E


    it is a if and only if type of question if |x-3|<|12x+5| then |x+13/3|>11/3 how do u guys solve this? my steps is as below (x-3)^2<(12x+5)^2 x^2-6x+9<144x^2+120x+25 0<143x^2+126x+25 and..... i get the wrong ans because it doesn't get |x+13/3|>11/3
  5. E

    modulus proving

    prove that |x-y|=|y-x| for all x and y can we square both side for them ? then we get 2x=2y, x=y? B)|a+b|<_|a|+|b| to prove |x-y|>_ |x|-|y| how about this one ? my teacher gave me a hint that consider |x-y|+|y| how do we actually add two modulus together ? do we need to open up the modulus...
  6. S

    modulus of hermite polynomial

    How to prove that |H_n(x)| \leq |H_n(ix)| where H_n(x) is the Hermite polynomial.
  7. J

    Modulus and inequalities question

    Given |x-2| + |x-1|>1, how do I find the values of x that satisfy this equation? I know the normal way of solving this is to square both sides, but that doesn't seem to work in this case. Thank you!
  8. Z

    cryptography maths

    Dear all, been stuck at this problem for awhile. I am doing a cryptography implementation with the following equation: m + xr = m1 + xr1 mod q The objective is to solve r1 from this equation such that q is a prime number and m, r and x is an element of q with some help, i derive at: r1 = (m...
  9. A

    integral of modulus function

    how to integrate |x*cos(Pi*x)| within the limits -1 to 1? I do not know how to proceed. kindly help me. Thanks with warm regards, Aranga
  10. T

    Solving a pattern regarding modulus

    Hi everybody, i'm studying for my discrete math final next week and i'm having an issue with a certain problem. I have to solve this pattern regarding modulus: 5 8 56 45 23 34 67 2 3 3 3 5 4 23 4 5 5 ? ? ? ? (filling in the ?s) N = 3 I'm pretty sure I need to form a certain...
  11. Z

    Probability question

    Can someone give me a hint on how to calculate P(|Z|∈[3/2 ; 5/2])? Y(t)=tZ^2; t=4 Z~U(-2;2) U - continuous uniform distribution. Sorry for the images but I am new here and I thought it would be faster. Do you have any tutorials how to post maths symbols?
  12. B

    solve for x and y :modulus

  13. B

    solve for x; modulus inequalities

    (a). -2|x-2|<-7 (b). |2x+1|>|-x+2|
  14. B

    modulus inequalitys

    (a). -|-x - 2|>1 (b). 1<|x - 2|<3 (c). 6/|x+1|>/=3
  15. A


    -abs(x+1)+abs(1-x)=2 can anyone tell me how to solve this with a sound approach i dont have any clue what to start with abs is mod(modulus) answer is x is less than or equal to -1
  16. A

    modulus aka absolute value inequality

    my attempt i squared both sides led me to 4x^2+20x+25 is greater than x^2+4x+4 3x^2+16x+21 is greater than 0 this lead to x is greater than -3 and less than -7/3 answer is x less than -3 and greater than -7/3
  17. A

    absolute Value aka modulus

    mod(a+b) is greater than or equal to mod(mod(a)+mod(b)) i dont get it
  18. leibnitz


  19. A

    Modulus graph transformations

    How would you draw the following graphs; Given f(x) = |x| a) y = 2 - f(x-1) Would it be the original modulus graph turned upside down (like a sharp n) and the top is point (1,2)? b) y = f(1-x) -1 Not too sure on this one :/ Cheers
  20. S

    complex number in polar form argument and modulus doubt

    Pls share ur ideas i wish to knw were i have gone wrong or is the text answr is wrong pls help(Headbang)(Headbang)(Headbang)