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anonymous

  • one year ago

understanding very big numbers Interesting question which number is bigger \[27 *_{99} 15 ~~or~~ 2*_{100} 3\] every operation can be defined recursively in terms of its previous operations (1)The successive operation S(n)=n+1 (2)Addition m+n can be defined by m+1=S(m) m+S(n)=S(n+m) (3) Multiplication m*n m*1=m m*S(n)=m+(m*n) (4) exponentiation m^n m^1=m m^S(n)=m*(m^n) (5)NewOperation lets say we define operations in this manner then which of these numbers is bigger \[27 *_{99} 15~~ or~~ 2*_{100} 3\] a link where this question and this process is explained https://www.youtube.com/watch?v=wPEYoW0Mj1U&list=PL5A714C94D40392AB&index=139

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  1. anonymous
    • one year ago
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    \[*_k\] is the kth new operation in this progression

  2. anonymous
    • one year ago
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    you know what, just use casio calculator lol

  3. anonymous
    • one year ago
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    we can try to see if there is pattern with the lower operations like

  4. anonymous
    • one year ago
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    27 + 15 and 2 * 3, 2+2+2 27*15 and 2^3 , 2*2*2 27^15 and 2^2^2 <--- this is the new operation following exponentiation

  5. anonymous
    • one year ago
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    wow.....

  6. nincompoop
    • one year ago
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    dan, you're not telling us anything new

  7. anonymous
    • one year ago
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    |dw:1440050985534:dw|

  8. anonymous
    • one year ago
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    even though it looks like on the lower levels 27 operator 15 is winning out , u dont know if at some point the higher operation will begin to tip the scales

  9. anonymous
    • one year ago
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    itd be an interesting question to find out if this does infact happen, if so at which operator?

  10. freckles
    • one year ago
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    I got stuck on that one problem... I got as far as: \[4 *_3(4*_34^{4^{4^{4}}})\] then I quit :p

  11. freckles
    • one year ago
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    I have to try again tomorrow I'm sleepy

  12. anonymous
    • one year ago
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    noo dont quit i need u -.-

  13. anonymous
    • one year ago
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    oh okay :)

  14. freckles
    • one year ago
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    Did you try his first problem though?

  15. anonymous
    • one year ago
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    yeah

  16. anonymous
    • one year ago
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    the rules he gives is so u can keep defining it until u get to the base case of recursion, but he shows a simpler way to do it if u keep going on

  17. freckles
    • one year ago
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    oh I was doing the base case that is why I stopped I didn't know how to get to the base case quick for the 4 star 3 thing 4^(4^4^(4)))

  18. anonymous
    • one year ago
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    |dw:1440051564779:dw|

  19. anonymous
    • one year ago
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    yep thats right

  20. freckles
    • one year ago
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    I will continue the video tomorrow and look some more

  21. freckles
    • one year ago
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    but it seems interesting I think

  22. anonymous
    • one year ago
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    i dont like the way he uses s(k) for k+1 but other than that its pretty good!

  23. anonymous
    • one year ago
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    |dw:1440052034629:dw|

  24. anonymous
    • one year ago
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    |dw:1440052925677:dw|

  25. anonymous
    • one year ago
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    that right there i am thinking \[2*_{98}2=4\] do u want to confirm that 2 operator 2 is always 4

  26. anonymous
    • one year ago
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    |dw:1440053172695:dw|

  27. anonymous
    • one year ago
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    by induction

  28. anonymous
    • one year ago
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    |dw:1440053515919:dw|

  29. anonymous
    • one year ago
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    from here its clear to see, but i wonder how big of a different it wud be have been if it was 2,4 instead of 2,3

  30. anonymous
    • one year ago
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    difference*

  31. ganeshie8
    • one year ago
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    \[\begin{align} 4 *_4 4 &=4*_3(4*_43)\\~\\ &=4*_3(4*_3(4*_42))\\~\\ &=4*_3(4*_3(4*_3(4*_41)))\\~\\ &=4*_3(4*_3(4*_34))\\~\\ &=4*_3(4*_3(4*_2(4*_33)))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_32))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_2(4*_31)))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_24))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4^4))))\\~\\ &=4*_3(4*_3(4*_2(4^{(4^4)})))\\~\\ &=4*_3(4*_3(4^{(4^{(4^4)})}))\\~\\ \end{align}\] did i get this right ?

  32. ganeshie8
    • one year ago
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    yes i quit too just after that @freckles

  33. anonymous
    • one year ago
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    |dw:1440053776717:dw|

  34. anonymous
    • one year ago
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    |dw:1440054119253:dw|

  35. anonymous
    • one year ago
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    kinda crazy to see it tip over that much

  36. anonymous
    • one year ago
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    and yep u did it right ganeshie

  37. anonymous
    • one year ago
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    though my argument for why the 2nd expression is so much more isnt really all that valid, like for example intuitively 2*2*2*2*2......*2 a billiontrillion times has to be much more than 27*27*27*27...*27 ---->15 timers no matter what the operation

  38. anonymous
    • one year ago
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    but then again

  39. anonymous
    • one year ago
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    i wonder if there is way to write sum kind of expression with summations and comparing those

  40. anonymous
    • one year ago
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    these numbers are just so big to deal with, it feels like u are just working with algebra, cant even begin to think about them as numbers its kind of cool

  41. anonymous
    • one year ago
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    |dw:1440054511816:dw|

  42. anonymous
    • one year ago
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    |dw:1440060525252:dw|

  43. anonymous
    • one year ago
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    this property looks interesting

  44. anonymous
    • one year ago
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    i think its true

  45. anonymous
    • one year ago
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    |dw:1440060687164:dw|

  46. anonymous
    • one year ago
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    whenever u are distributing any operation over addition, this is happening the middle operation is shifted down by 1

  47. anonymous
    • one year ago
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    |dw:1440061272416:dw|

  48. freckles
    • one year ago
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    \[2 *_{100}3=2 *_{99}(2 *_{100} 2)=2 *_{99}(4)=2 *_{99}4 \\ \text{ so we are really comparing } 2 *_{99}4 \text{ and } 27*_{99}15 \\ \text{ isn't it enough just to say you know since we have the same operation } \\ \text{ that since } 27>2 \text{ and } 15>4 \\ \text{ then } \\ 2*_{100}3=2*_{99}4<27*_{99}15\]

  49. freckles
    • one year ago
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    and yeah 2 any operation 2=4 ... \[2 *_{S(k)}2 =2 *_k(2*_{S(k)}1) =2 *_{k}2 = \text{ following this ... until we get } = \\ 2 *_12=2 \times 2=4\]

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