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Bee_see

  • one year ago

Prove by induction that 3n < n! when n is an integer and n >= 7. I have proven that when n=7, the proof is true... So I did n=k, 3^k < k! then 3^k+1 < (k+1)! I added the term k+1 on both sides...3^k+(k+1) < k! + (k+1). How do go to 3^k+1 < (k+1)! ?

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  1. Bee_see
    • one year ago
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    3^n < n!*

  2. zzr0ck3r
    • one year ago
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    \(k>6\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

  3. Bee_see
    • one year ago
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    where did you get k> 6 from?

  4. zzr0ck3r
    • one year ago
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    same thing as \(k\ge 7\)

  5. Bee_see
    • one year ago
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    how?

  6. zzr0ck3r
    • one year ago
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    k is an integer. \(k>6\implies 7,8,9,10,11,12...\) \(k\ge 7 \implies 7,8,9,10,11,12...\)

  7. zzr0ck3r
    • one year ago
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    \(k\ge7\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

  8. zzr0ck3r
    • one year ago
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    \((k+1)!=(k+1)k!\) always

  9. Bee_see
    • one year ago
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    I have 3^k + (k+1) < k! + (k+1) thiough

  10. zzr0ck3r
    • one year ago
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    You used addition to try and get there, I used multiplication. I am not saying you cant do it your way, but this is the only way I can do it.

  11. Bee_see
    • one year ago
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    multiplied what?

  12. Bee_see
    • one year ago
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    where does the 3 < k+1 and the rest come from?

  13. freckles
    • one year ago
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    3<k+1 for all integer k>=7

  14. Bee_see
    • one year ago
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    I don't have 3 < k+1 anywhere...

  15. freckles
    • one year ago
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    \[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\]

  16. freckles
    • one year ago
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    we do have k+1>3 though for k>=7 7+1>3 8+1>3 9+1>3 10+1>3 and so on...

  17. Bee_see
    • one year ago
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    it isn't 3^k(3) < (k+1)!?

  18. Bee_see
    • one year ago
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    I'm not sure from where you are starting and ending.

  19. freckles
    • one year ago
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    ? are you asking about my application of law of exponents ?

  20. freckles
    • one year ago
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    \[x^{a+b}=x^{a} \cdot x^{b} \] is the first thing I used

  21. freckles
    • one year ago
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    \[\text{ we wanted \to show } 3^{k+1} <(k+1)!\]

  22. freckles
    • one year ago
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    I started with the left hand side

  23. freckles
    • one year ago
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    \[3^{k+1}=3^k 3^{1}=3^k (3)\]

  24. freckles
    • one year ago
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    now using the inductive step we have: \[3^k(3)<k!(3)\] now we want to show this is <(k+1)!=k!*(k+1) and we know since k>=7 that 3 is going to be less than (k+1)

  25. Bee_see
    • one year ago
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    so you are going to make 3^k+1 < (k+1)! into 3^k (k+1) < k! +(k+1?

  26. freckles
    • one year ago
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    What are you doing ?

  27. freckles
    • one year ago
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    where does all of that come from?

  28. freckles
    • one year ago
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    we just wanted to show 3^(k+1) < (k+1)!

  29. Bee_see
    • one year ago
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    ok..how did you make (k+1)! into k!(3)?

  30. freckles
    • one year ago
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    do you understand that we have 3<k+1?

  31. freckles
    • one year ago
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    3<k+1 holds for any integer k>2 and we have k>=7 so this inequality 3<k+1 is true for our k

  32. freckles
    • one year ago
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    k!*(k+1) is equal to (k+1)!

  33. freckles
    • one year ago
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    just like 6!*7 is equal to 7!

  34. Bee_see
    • one year ago
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    ok

  35. freckles
    • one year ago
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    \[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\] I wrote this like this to show my reasons for each steps you can write this without all the reasons...you know just in one like: \[3^{k+1}=3^{k} \cdot 3^1 =3^{k} 3<k! (3)<k! \cdot (k+1)=(k+1)!\]

  36. freckles
    • one year ago
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    just in one line like*

  37. Bee_see
    • one year ago
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    ok

  38. freckles
    • one year ago
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    I think I see what you were trying to do

  39. freckles
    • one year ago
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    instead of adding (k+1) on both sides it probably might be better if you multiply both sides by (k+1)

  40. freckles
    • one year ago
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    \[3^{k}<k! \\ \text{ multiply both sides by } (k+1) \\ 3^k(k+1)<k! \cdot (k+1) \\ \text{ recall } k!(k+1)=(k+1)! \\ \text{ so we have } 3^k(k+1) <(k+1)! \\ \text{ but we still kind of need } 3<k+1 \text{ which is true since } k \ge 7 \\ 3^{k}(3)<3^k(k+1)<k!(k+1)=(k+1)! \\ \text{ and we know by law of exponents } 3^k(3)=3^{k+1} \\ \text{ so add this \to our little line } \\ 3^{k+1}=3^k(3)<3^k(k+1)<k!(k+1)=(k+1)!\] still the same line we had above

  41. freckles
    • one year ago
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    I just decided to work from the side that had 3^(k+1) to show this was less than (k+1)!

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