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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    Please see attachment.
    I know that math induction is when you prove the first term and then you have to prove for all the other terms with some sort of k+1? It's in a little bit different format than I'm used to so maybe that's why I am getting confused.
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    The first step in an inductive proof is to form an inductive statement. It is an expression that depends on a variable n. For each particular n it is either true or false, but until n is fixed, it does not have a truth value.

    Statements proved by induction usually have the form "For all n, P(n)" for some expression P. This P that mentions n is the inductive statement. Writing it explicitly: " P(n) is ..." is an indispensable first step.
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    where do the k's and k+1's come into play?
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  4. #4
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    Induction

    Hello WartonMorton
    Quote Originally Posted by WartonMorton View Post
    Please see attachment.
    I know that math induction is when you prove the first term and then you have to prove for all the other terms with some sort of k+1? It's in a little bit different format than I'm used to so maybe that's why I am getting confused.
    The problem is this:
    Use mathematical induction to prove the following assertion:
    If a_1 = 1 and a_{n+1}=a_n - \frac{1}{n(n+1)}, then a_n = \frac1n.
    The thing you have to prove here is that a_n = \frac1n. This forms our 'inductive assertion' or 'induction hypothesis', that we can denote by P(n).

    So P(n) is the statement: a_n = \frac1n.


    We have no means of telling whether P(n) is true yet. It is just a hypothesis. What we can do, though, is to make use of the definition of a_{n+1} to deduce another statement that will follow if P(n) is true. From this we hope to prove that P(n) \Rightarrow P(n+1) (where \Rightarrow means 'implies'), which is the basis of a proof by induction.


    So we use the 'implies' sign, when we make this deduction. Like this:
    P(n) \Rightarrow a_n = \frac1n
    \Rightarrow a_{n+1} = \frac1n-\frac{1}{n(n+1)}, using the definition of a_n+1 that we've been given.
    =\frac{(n+1) - 1}{n(n+1)}

    =\frac{1}{n+1}

    \Rightarrow P(n+1)
    Do you understand that last step? We have shown that if P(n) is true, then a_{n+1} = \frac{1}{n+1}, which is simply P(n+1).

    Finally we check that P(1) is true. Which it is:
    a_1 = 1 = \frac{1}{1}
    This completes the proof.

    Grandad
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