Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 months ago

Find in set builder notation the set of all positive integers a such that 137≡a (mod 5)?

I've tried looking in the text book, but i can't find the instructions for this question in my assignment. Thank you for the help in advance. :)

3 Answers

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  • Ian H
    Lv 7
    5 months ago

    The modular equation 137 ≡ a (mod 5) is equivalent to

    5n + a = 137

    The least value of a which solves it is a = 2, with n = 35

    (because 5*35 + 2 = 137)

    But adding or subtracting any multiples of 5 to a also produces valid options.

    For example a = 22, because 5*23 + 22 = 137

    In everyday terms a can take any value given by 137 – 5n, where n

    can be any positive or negative integer.

    Now you need to convert plain English to set builder notation.

    Something like “a is the set of real integers such that a = 137 – 5n, integer, n”

    This is a gentle introduction to set notation

    https://www.purplemath.com/modules/setnotn.htm

    Scroll down to “Give a solution using a rule”

    That example is sufficiently similar to guide your own effort at set building

  • 5 months ago

    { a : (a > 0) & (5|(a - 2)) }

    where the vertical bar means "is an exact divisor of"

  • ted s
    Lv 7
    5 months ago

    137 / 5 = 27 R2====> 137 ≡ a (mod 5) = 2 + 5 n...thus a = 2 or 7 or 12 or 17 ....{ n = 0 , 1,2 , ..., 27 }

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