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    ive been asked a question on proof by induction. ive been asked to show that when a^n is divided by a-1, its leaves a remainder of 1. where n ∈ ℕ and a ∈ ℕ\{1}.

    does ℕ\{1} mean, natural number - 1?

    if so does this mean that 'a' can be any whole number bigger than and equal to 0?
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    No, it means any natural number which isn't 1
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    (Original post by SimonM)
    No, it means any natural number which isn't 1
    oh right, so its any whole number bigger than 1. thanks so much.
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    i dont suppose you know how to solve this question?
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    Induction:

    Base case is trivial: a = 1(a-1) + 1
    Inductive step: a^(n+1) = a^(n)a = (k(a-1)+1)a = ...
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    I believe that if B is a subset of A, then {A}\{B} is the set of elements of A which are not elements of B.
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    (Original post by tazarooni89)
    I believe that if B is a subset of A, then {A}\{B} is the set of elements of A which are not elements of B.
    B doesn't need to be a subset of A.
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    (Original post by SimonM)
    Induction:

    Base case is trivial: a = 1(a-1) + 1
    Inductive step: a^(n+1) = a^(n)a = (k(a-1)+1)a = ...
    i dont understand how you got these equations and i am unsure where to go from here :confused:
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    (Original post by SimonM)
    B doesn't need to be a subset of A.
    Yeah, good point

    I actually meant to write "If B intersects A". But then strictly speaking it probably doesn't even need to do that either
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    (Original post by sarah_vickers)
    i dont understand how you got these equations and i am unsure where to go from here :confused:
    Expand it out, and try to write it in the form m(a-1) +1 for some integer m. (This is what it means to say that it has remainder 1 on division by a-1)
 
 
 
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