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Math Help - Axiom of Induction Uh Oh

  1. #1
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    Axiom of Induction Uh Oh

    Hello MHF.

    Let's look at this sequence...
    1/[1*2] + 1/[2*3] + 1/[3*4] + ....

    Get the pattern?

    Yah its...
    1/[n(n+1)] = n/[n+1]

    So we have...
    1/[1*2] + 1/[2*3] + 1/[3*4] + .... + 1/[n(n+1)] = n/[n+1]

    I'm supposed to show that the statement holds for all positive integers n.

    This will require the use of mathematical induction.

    I haven't really learned this yet so be patient. I know I have to show that when n=1 the statement is true.

    So we have...

    n=1 means that we just do the first term (left side), which is 1/(1*2) = 1/2 or 0.5. The right hand side is 1/(1+1) = 1/2 = 0.5 SAME YAY so far so good!

    But then it wants me to show that they n=k and then it wants me to show that n=k+1.

    Can someone walk me through these steps?


    Thanks
    Last edited by UC151CPR; October 15th 2009 at 07:28 PM.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    South Coast of England
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    Hello UC151CPR
    Quote Originally Posted by UC151CPR View Post
    Hello MHF.

    Let's look at this sequence...
    1/[1*2] + 1/[2*3] + 1/[3*4] + ....

    Get the pattern?

    Yah its...
    1/[n(n+1)] = n/[n+1]

    So we have...
    1/[1*2] + 1/[2*3] + 1/[3*4] + .... + 1/[n(n+1)] = n/[n+1]

    I'm supposed to show that the statement holds for all positive integers n.

    This will require the use of mathematical induction.

    I haven't really learned this yet so be patient. I know I have to show that when n=1 the statement is true.

    So we have...

    n=1 means that we just do the first term (left side), which is 1/(1*2) = 1/2 or 0.5. The right hand side is 1/(1+1) = 1/2 = 0.5 SAME YAY so far so good!

    But then it wants me to show that they n=k and then it wants me to show that n=k+1.

    Can someone walk me through these steps?


    Thanks
    We set up the statement we want to prove as a propositional function P(n) (a statement about n that may or may not be true) as follows:

    P(n) is \sum_{i=1}^n\frac{1}{i(i+1)}=\frac{n}{n+1}

    We then show that if P(n) is true when n = k (for some integer value of k), then it's also true when n = k+1 (the next integer). In other words, we show that

    P(k) \Rightarrow P(k+1)

    OK. So let's look at what happens if P(k) is true.

    P(k) \Rightarrow \sum_{i=1}^k\frac{1}{i(i+1)}=\frac{k}{k+1}

    \Rightarrow \left(\sum_{i=1}^k\frac{1}{i(i+1)}\right)+\frac{1}  {(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}

    \Rightarrow \sum_{i=1}^{k+1}\frac{1}{i(i+1)}=\frac{k(k+2)+1}{(  k+1)(k+2)}

    =\frac{(k+1)^2}{(k+1)(k+2)}

    =\frac{k+1}{k+2}

    \Rightarrow\sum_{i=1}^{k+1}\frac{1}{i(i+1)}=\frac{  k+1}{(k+1)+1}

    \Rightarrow P(k+1), since we now have the proposition P(n) where n has the value (k+1)

    This, together with the fact that you have already proved that P(1) is true, completes the proof.

    Grandad
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  3. #3
    Member Jones's Avatar
    Joined
    Apr 2006
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    Norway
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    Hi Granddad,

    Yes my algebra is beyond bad, but could you explain what you did here ?:

    This one is fine
    <br />
\Rightarrow \sum_{i=1}^{k+1}\frac{1}{i(i+1)}=\frac{k(k+2)+1}{(  k+1)(k+2)}<br />

    How did you get this?
    <br />
=\color{red}\frac{(k+1)^2}{(k+1)(k+2)}<br />
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  4. #4
    Super Member
    Joined
    Aug 2009
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    Israel
    Posts
    976
    Quote Originally Posted by Jones View Post
    Hi Granddad,

    Yes my algebra is beyond bad, but could you explain what you did here ?:

    This one is fine
    <br />
\Rightarrow \sum_{i=1}^{k+1}\frac{1}{i(i+1)}=\frac{k(k+2)+1}{(  k+1)(k+2)}<br />

    How did you get this?
    <br />
=\color{red}\frac{(k+1)^2}{(k+1)(k+2)}<br />
    \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}
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  5. #5
    Super Member
    Joined
    Aug 2009
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    Israel
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    First, we must assume that the statement is correct for some positive integer n. That is, \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} + ... + \frac{1}{n(n+1)} = \frac{n}{n+1} is assumed to be correct.

    According to the principle of induction, it will be correct for all positive integers if we prove that, given the assumption that it is correct for some positive integer n, it is also correct for n+1. That is to say, we now need to prove that the following equation:

    \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+1+1)} = \frac{n+1}{n+1+1} (*)

    Holds.

    However, from the assumption we have, we know that

    \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} + ... + \frac{1}{n(n+1)} = \frac{n}{n+1}

    So we can substitute this expression from the LHS of (*), and then we get:

    \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} as the LHS.
    The RHS of (*) stays \frac{n+1}{n+2}. Now we need to prove both sides are equal. Let's play with the left hand side:

    \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n}{n+1}\cdot \frac{n+2}{n+2} + \frac{1}{(n+1)(n+2)} = \frac{n(n+2)+1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)} = \frac{n+1}{n+2}

    So the LHS of (*) is equal to \frac{n+1}{n+2}. But so is the RHS, so we conclude that (*) holds and therefore the equation is correct.
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