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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[*_k\] is the kth new operation in this progression

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know what, just use casio calculator lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can try to see if there is pattern with the lower operations like

anonymous
 one year ago
Best ResponseYou've already chosen the best response.027 + 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

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dan, you're not telling us anything new

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440050985534:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0even 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0itd be an interesting question to find out if this does infact happen, if so at which operator?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I got stuck on that one problem... I got as far as: \[4 *_3(4*_34^{4^{4^{4}}})\] then I quit :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I have to try again tomorrow I'm sleepy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0noo dont quit i need u .

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Did you try his first problem though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oh 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)))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440051564779:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I will continue the video tomorrow and look some more

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but it seems interesting I think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont like the way he uses s(k) for k+1 but other than that its pretty good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440052034629:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440052925677:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that right there i am thinking \[2*_{98}2=4\] do u want to confirm that 2 operator 2 is always 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440053172695:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440053515919:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align} 4 *_4 4 &=4*_3(4*_43)\\~\\ &=4*_3(4*_3(4*_42))\\~\\ &=4*_3(4*_3(4*_3(4*_41)))\\~\\ &=4*_3(4*_3(4*_34))\\~\\ &=4*_3(4*_3(4*_2(4*_33)))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_32))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_2(4*_31)))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_24))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4^4))))\\~\\ &=4*_3(4*_3(4*_2(4^{(4^4)})))\\~\\ &=4*_3(4*_3(4^{(4^{(4^4)})}))\\~\\ \end{align}\] did i get this right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0yes i quit too just after that @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440053776717:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440054119253:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0kinda crazy to see it tip over that much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and yep u did it right ganeshie

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0though 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if there is way to write sum kind of expression with summations and comparing those

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440054511816:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440060525252:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this property looks interesting

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440060687164:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whenever u are distributing any operation over addition, this is happening the middle operation is shifted down by 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440061272416:dw

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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and 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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