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Mathematical Induction

  1. Oct 19, 2005 #1
    Show taht for every natural numbers n>=2 [tex]n \geq 2 [/tex] the number 2^2^n - 6 [tex]2^{2^n} -6[/tex] is a multiple of 10. Using mathematical induction.
    Okay i got no clue how to start this question. Ahhh Is there a series where the x^n is a series? Well this stuff really sucks.
    Last edited: Oct 19, 2005
  2. jcsd
  3. Oct 19, 2005 #2


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    [tex]2^{2^{n+1}} = \left( 2^{2^n}\right)^2[/tex]
  4. Oct 19, 2005 #3
    Oh WOw... that helped a lot.. now i can show that its inductive hmm but how would i show its a multiple of 10?
    hmm maybe if i said 10^2 is 100 so thats a multiple of 10. [tex]10*n\leq2^{2^n}-6 [/tex] if i proved that.. hmm i wonder if that would be right... [tex]10*(n+1)\leq2^{2^(n+1)}[/tex] hmm if i proved that would i have solved the question?
    wow if hmm well ahaha thanks a LOT!!!!!!!! YOUR AWSOME. that little hint helped a lot but i got no clue if im actually doing it right.
    Last edited: Oct 19, 2005
  5. Oct 19, 2005 #4
    For Mathematical Induction, you assume that P_{k} is true, in this case for n greater or equal to 2. With this, you can immediately say that 2^(2^k) - 6 is equal to 10a, for some a, where a is a positive integer.
    Can you then use this fact, and the hint provided, to prove the desired result for 2^(2^(k+1)) - 6?
  6. Oct 19, 2005 #5
    hmm should could i say that

    [tex] 10(a+x)\leq (\left2^{2^n}\right )^2 - 6[/tex] where x is some positive number and should that it is inductive to prove that 10 is a multiple?
    [tex] 10(a)\leq2^{2^n} - 6 [/tex] then
    [tex] 10(a+x)\leq(\left 2^{2^n}\right)^2 - 6[/tex]
    then show that
    [tex] 10(a+x)\leq2^{2^{n+1}} -6 [/tex] ahhh im confusd now.. ahh
    Last edited: Oct 19, 2005
  7. Oct 19, 2005 #6


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    HINT: If M - 6 is a multiple of 10 then M ends in the digit 6! :)
  8. Oct 19, 2005 #7
    Hmmm... Why are there so many inequalitites in your working? From what I know, the only inequality to appear in your solution should be the fact that n is greater than or equal to 2, but this is just a specification, and should not appear in your proof.
  9. Oct 19, 2005 #8
    Steps in Mathematical Induction
    1) Let [tex]P_{n}[/tex] be the statement [tex]2^{2^n}-6[/tex] is divisible by 10, for n[tex]\geq[/tex] 2.
    2) Check that the result you want to prove is valid for n=2, so [tex]P_{2}[/tex] is true.
    3) Assume [tex]P_{k}[/tex] is true, for some n [tex]\geq[/tex] 2. So, [tex]2^{2^k}-6[/tex] = 10a, for some a, which is a positive integer.
    4) Using this result, you must somehow prove that [tex]2^{2^{k+1}}-6[/tex] is a multiple of 10. How would you go around doing it? Look at the first hint provided and observe... What has been done to the term [tex]2^{2^n}[/tex] ? USE BOTH THE RESULT FROM STEP 3 AND THE FIRST HINT

    5) Once you have proven step 4, give a conclusion. "Since [tex]P_{2}[/tex] is true, and for some n[tex]\geq[/tex]2, [tex]P_{k}[/tex] is true [tex]\Longrightarrow[/tex] [tex]P_{k+1}[/tex] is true. By Mathematical Induction, [tex]P_{n}[/tex] is true for all n[tex]\geq[/tex]2."
    Last edited: Oct 19, 2005
  10. Oct 19, 2005 #9
    Wow thanks a lot for your help. I will try to figure this out. You helped a lot pizzasky and Tide. :smile:
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