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    Hi

    here is my solution and need a little help with part a a) Prove by induction that 5^n + 2.3^(n+1) + 1 is divisible by 4 for all n ∈ N.
    solution:

    Let p(n) be the statement
    then we have to prove p(n) is true for every natural number

    p(0) = 5^0 + 2 . 3^1 + = 8 which is devisible by 4

    p (1) = 5^1 + 2.3^2 + 1 = 24 which is again devisible by 4

    so p(n) is true for every natural number


    now I have to do the inductive hypothesis step which the most difficult bit in this question

    what I have done is :

    let K= N

    p(K) = p (N) AND that implies p ( k+1) is true


    5 ^ (k+1) + 2 . 3^(k+2) +1

    I don't know what to do after this step

    ---------------------------

    b) Prove that if 2^x = 5 then x is irrational.

    i used two methods for this one

    the first one is by calculating

    2^x = 5
    ln 2^x = ln 5
    x ln 2 = ln 5
    x = ln 5 / ln 2
    x= 2.3219

    so x is irrational


    OR
    by contradiction

    suppose that x is rational
    a,b are positive intergers

    2^x = 5
    2 ^ a/b = 5 raise both sides by the power (b)

    2^ (a/b) ^b = 5 ^b

    2 ^a = 5^b

    and from that we can say x is irrational





    cheers,
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    (Original post by EffieFlowers)
    :sexface:
    what :p:
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    (Original post by A.braham)
    Hi


    b) Prove that if 2^x = 5 then x is irrational.

    i used two methods for this one

    the first one is by calculating

    2^x = 5
    ln 2^x = ln 5
    x ln 2 = ln 5
    x = ln 5 / ln 2
    x= 2.3219

    so x is irrational
    Woah! How does this prove that x is irrational? I don't see it; you've just truncated a decimal to 4 dp and claimed the whole thing is irrational!

    OR
    by contradiction

    suppose that x is rational
    a,b are positive intergers

    2^x = 5
    2 ^ a/b = 5 raise both sides by the power (b)

    2^ (a/b) ^b = 5 ^b

    2 ^a = 5^b

    and from that we can say x is irrational
    You're pretty much there, but some explanation is needed to say that there are no integers solutions to 2^a = 5^b. (Why isn't there? Note that there would be if the equation was 2^a = 4^b.)
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    (Original post by A.braham)
    Hi

    here is my solution and need a little help with part a a) Prove by induction that 5^n + 2.3^(n+1) + 1 is divisible by 4 for all n ∈ N.
    solution:

    Let p(n) be the statement
    then we have to prove p(n) is true for every natural number

    p(0) = 5^0 + 2 . 3^1 + = 8 which is devisible by 4

    p (1) = 5^1 + 2.3^2 + 1 = 24 which is again devisible by 4

    so p(n) is true for every natural number

    ,
    All you have proved is that p(n) is divisible four in the limited cases that n=0 and n=1 How does this prove the general case?
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    lool you start confsuing me now guys


    ok for part a

    we have to do p(0) or p (1) and after that we have to do n+1

    so what is wrong about that ? :confused:
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    Prove by induction that 5^n + 2.3^{n+1} + 1 is divisible by 4 for all n \in \mathbb{N}.

    You rightly showed that:
    p(1) = 5^1 + 2.3^2 + 1 = 24 which is again divisible by 4

    And now, assume that p(k)=5^k+2 \times 3^{k+1} + 1 is divisible by 4.

    Then investigate k+1, and show that if p(k) is divisible by 4, then so p(k+1) must be.
    The first step you might like to consider:
    p(k+1)=5^{k+1} + 2 \times 3^{k+2} +1
    p(k+1)= 5^k \times 5 + 2 \times 3^{k+1} \times 3 +1

    After messing around with this, I think that taking out a factor of p(k) - which you know is a multiple of 4 - might help you.

    Once you've shown that it's true for the number 1 and, {if it's true for k, then it's true for k+1}, you can then say it's true for all of the natural numbers!

    Hope this helps
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    (Original post by A.braham)
    lool you start confsuing me now guys


    ok for part a

    we have to do p(0) or p (1) and after that we have to do n+1

    so what is wrong about that ?
    That sounds about right

    Just make sure that you're showing that n+1 works when any n works.
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    (Original post by placenta medicae talpae)
    Prove by induction that 5^n + 2.3^{n+1} + 1 is divisible by 4 for all n \in \mathbb{N}.

    You rightly showed that:
    p(1) = 5^1 + 2.3^2 + 1 = 24 which is again divisible by 4

    And now, assume that p(k)=5^k+2 \times 3^{k+1} + 1 is divisible by 4.

    Then investigate k+1, and show that if p(k) is divisible by 4, then so p(k+1) must be.
    The first step you might like to consider:
    p(k+1)=5^{k+1} + 2 \times 3^{k+2} +1
    p(k+1)= 5^k \times 5 + 2 \times 3^{k+1} \times 3 +1

    After messing around with this, I think that taking out a factor of p(k) - which you know is a multiple of 4 - might help you.

    Once you've shown that it's true for the number 1 and, {if it's true for k, then it's true for k+1}, you can then say it's true for all of the natural numbers!

    Hope this helps

    you see this
    p(k+1)=5^{k+1} + 2 \times 3^{k+2} +1
    p(k+1)= 5^k \times 5 + 2 \times 3^{k+1} \times 3 +1

    that is where I'm stuck at

    I couldn't do any factorisation

    I ended up with
    5^k . 5 + 18.3^k + 1 = 4a

    which is incorrect :mad:
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    (Original post by A.braham)
    you see this



    that is where I'm stuck at

    I couldn't do any factorisation

    I ended up with
    5^k . 5 + 18.3^k + 1 = 4a

    which is incorrect
    Oh right, okay, no worries!

    Does this help you:
    5 \times 5^k + 6 \times 3^{k+1} +1
    Is the same as:
    5 \times 5^k+5 \times 2 \times 3^{k+1} + 5 \times 1 \hspace{2mm} - \hspace{2mm} 4 \times 3^{k+1} - 4 \times 1

    Maybe some factorisation might work now?
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    (Original post by placenta medicae talpae)
    Oh right, okay, no worries!

    Does this help you:
    5 \times 5^k + 6 \times 3^{k+1} +1
    Is the same as:
    5 \times 5^k+5 \times 2 \times 3^{k+1} + 5 \times 1 \hspace{2mm} - \hspace{2mm} 4 \times 3^{k+1} - 4 \times 1

    Maybe some factorisation might work now?

    i got it now cheeers
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    (Original post by A.braham)
    i got it now cheeers
    No probs!
 
 
 
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