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Thread: 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 $\displaystyle n$. For each particular $\displaystyle n$ it is either true or false, but until $\displaystyle n$ is fixed, it does not have a truth value.

    Statements proved by induction usually have the form "For all $\displaystyle n$, $\displaystyle P(n)$" for some expression $\displaystyle P$. This $\displaystyle P$ that mentions $\displaystyle n$ is the inductive statement. Writing it explicitly: "$\displaystyle P(n)$ 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 $\displaystyle a_1 = 1$ and $\displaystyle a_{n+1}=a_n - \frac{1}{n(n+1)}$, then $\displaystyle a_n = \frac1n$.
    The thing you have to prove here is that $\displaystyle a_n = \frac1n$. This forms our 'inductive assertion' or 'induction hypothesis', that we can denote by $\displaystyle P(n)$.

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


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


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

    $\displaystyle =\frac{1}{n+1}$

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

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

    Grandad
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