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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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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\[*_k\] is the kth new operation in this progression
you know what, just use casio calculator lol
we can try to see if there is pattern with the lower operations like

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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
wow.....
dan, you're not telling us anything new
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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
itd be an interesting question to find out if this does infact happen, if so at which operator?
I got stuck on that one problem... I got as far as: \[4 *_3(4*_34^{4^{4^{4}}})\] then I quit :p
I have to try again tomorrow I'm sleepy
noo dont quit i need u -.-
oh okay :)
Did you try his first problem though?
yeah
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
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)))
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yep thats right
I will continue the video tomorrow and look some more
but it seems interesting I think
i dont like the way he uses s(k) for k+1 but other than that its pretty good!
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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
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by induction
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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
difference*
\[\begin{align} 4 *_4 4 &=4*_3(4*_43)\\~\\ &=4*_3(4*_3(4*_42))\\~\\ &=4*_3(4*_3(4*_3(4*_41)))\\~\\ &=4*_3(4*_3(4*_34))\\~\\ &=4*_3(4*_3(4*_2(4*_33)))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_32))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_2(4*_31)))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_24))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4^4))))\\~\\ &=4*_3(4*_3(4*_2(4^{(4^4)})))\\~\\ &=4*_3(4*_3(4^{(4^{(4^4)})}))\\~\\ \end{align}\] did i get this right ?
yes i quit too just after that @freckles
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kinda crazy to see it tip over that much
and yep u did it right ganeshie
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
but then again
i wonder if there is way to write sum kind of expression with summations and comparing those
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
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this property looks interesting
i think its true
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whenever u are distributing any operation over addition, this is happening the middle operation is shifted down by 1
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\[2 *_{100}3=2 *_{99}(2 *_{100} 2)=2 *_{99}(4)=2 *_{99}4 \\ \text{ so we are really comparing } 2 *_{99}4 \text{ and } 27*_{99}15 \\ \text{ isn't it enough just to say you know since we have the same operation } \\ \text{ that since } 27>2 \text{ and } 15>4 \\ \text{ then } \\ 2*_{100}3=2*_{99}4<27*_{99}15\]
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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