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Math Help - prove by induction help

  1. #1
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    prove by induction help

    I'm doing homework and getting stuck at this problem

    I'm confused with the concept of induction and I don't know how to do this


    a) Prove by induction that
    5^n + 2.3^(n+1) + 1 is divisible by 4 for all n N.

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


    any idea guys ?

    Thanks,
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zisi View Post
    I'm doing homework and getting stuck at this problem

    I'm confused with the concept of induction and I don't know how to do this


    a) Prove by induction that
    5^n + 2.3^(n+1) + 1 is divisible by 4 for all n N.

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


    any idea guys ?

    Thanks,
    The first one is just plain old induction. What have you tried? Can you use non-induction methods? Namely, 5^n+2\cdot3^{n+1}+1\equiv 1+2\cdot\left(-1\right)^{n+1}+1=\begin{cases}0 & \mbox{if}\quad n\text{ is even}\\ 4 & \mbox{if} \quad n\text{ is odd }\end{cases}\equiv 0\text{ mod }4

    For the second one, what is the context of the problem? What are you doing in class right now?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    The first one is just plain old induction. What have you tried? Can you use non-induction methods? Namely, 5^n+2\cdot3^{n+1}+1\equiv 1+2\cdot\left(-1\right)^{n+1}+1=\begin{cases}0 & \mbox{if}\quad n\text{ is even}\\ 4 & \mbox{if} \quad n\text{ is odd }\end{cases}\equiv 0\text{ mod }4

    For the second one, what is the context of the problem? What are you doing in class right now?
    for the second one , I guess we have to prove it by contradiction
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  4. #4
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    Here is how I attempted the question

    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
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  5. #5
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    Prove by induction that  <br />
5^n+2\cdot3^{n+1}+1<br />
is divisble by 4 for all integer n\in \mathbb{N}.

    Proof:

    Base case:

    For n=1,  <br />
5^n+2\cdot3^{n+1}+1= 5^1+2\cdot3^2+1=5+18+1= 24= 6 \times 4.

    For induction hypothesis:

    Suppose for ingeter k that  5^k+2\cdot3^{k+1}+1 is divisible by 4. Then

    5^{k+1}+2\cdot3^{(k+1)+1}+1= 5\cdot5^{k}+2\cdot3\cdot3^{k+1}+1
    =(4+1)5^k+2(2+1)3^{k+1}+1
     <br />
= 4\cdot5^k+5^k+ 4\cdot3^{k+1} +2\cdot3^{k+1}+1<br />
    = 4(5^k+ 3^{k+1}) +(5^k+2\cdot3^{k+1}+1)<br />

    By induction hypothesis  <br />
(5^n+2\cdot3^{n+1}+1) is divisble by 4 for all integer n\in \mathbb{N}
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