anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[*_k\] is the kth new operation in this progression
anonymous
  • anonymous
you know what, just use casio calculator lol
anonymous
  • anonymous
we can try to see if there is pattern with the lower operations like

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anonymous
  • anonymous
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
anonymous
  • anonymous
wow.....
nincompoop
  • nincompoop
dan, you're not telling us anything new
anonymous
  • anonymous
|dw:1440050985534:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
itd be an interesting question to find out if this does infact happen, if so at which operator?
freckles
  • freckles
I got stuck on that one problem... I got as far as: \[4 *_3(4*_34^{4^{4^{4}}})\] then I quit :p
freckles
  • freckles
I have to try again tomorrow I'm sleepy
anonymous
  • anonymous
noo dont quit i need u -.-
anonymous
  • anonymous
oh okay :)
freckles
  • freckles
Did you try his first problem though?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
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
freckles
  • freckles
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)))
anonymous
  • anonymous
|dw:1440051564779:dw|
anonymous
  • anonymous
yep thats right
freckles
  • freckles
I will continue the video tomorrow and look some more
freckles
  • freckles
but it seems interesting I think
anonymous
  • anonymous
i dont like the way he uses s(k) for k+1 but other than that its pretty good!
anonymous
  • anonymous
|dw:1440052034629:dw|
anonymous
  • anonymous
|dw:1440052925677:dw|
anonymous
  • anonymous
that right there i am thinking \[2*_{98}2=4\] do u want to confirm that 2 operator 2 is always 4
anonymous
  • anonymous
|dw:1440053172695:dw|
anonymous
  • anonymous
by induction
anonymous
  • anonymous
|dw:1440053515919:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
difference*
ganeshie8
  • ganeshie8
\[\begin{align} 4 *_4 4 &=4*_3(4*_43)\\~\\ &=4*_3(4*_3(4*_42))\\~\\ &=4*_3(4*_3(4*_3(4*_41)))\\~\\ &=4*_3(4*_3(4*_34))\\~\\ &=4*_3(4*_3(4*_2(4*_33)))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_32))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_2(4*_31)))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4*_24))))\\~\\ &=4*_3(4*_3(4*_2(4*_2(4^4))))\\~\\ &=4*_3(4*_3(4*_2(4^{(4^4)})))\\~\\ &=4*_3(4*_3(4^{(4^{(4^4)})}))\\~\\ \end{align}\] did i get this right ?
ganeshie8
  • ganeshie8
yes i quit too just after that @freckles
anonymous
  • anonymous
|dw:1440053776717:dw|
anonymous
  • anonymous
|dw:1440054119253:dw|
anonymous
  • anonymous
kinda crazy to see it tip over that much
anonymous
  • anonymous
and yep u did it right ganeshie
anonymous
  • anonymous
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
anonymous
  • anonymous
but then again
anonymous
  • anonymous
i wonder if there is way to write sum kind of expression with summations and comparing those
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1440054511816:dw|
anonymous
  • anonymous
|dw:1440060525252:dw|
anonymous
  • anonymous
this property looks interesting
anonymous
  • anonymous
i think its true
anonymous
  • anonymous
|dw:1440060687164:dw|
anonymous
  • anonymous
whenever u are distributing any operation over addition, this is happening the middle operation is shifted down by 1
anonymous
  • anonymous
|dw:1440061272416:dw|
freckles
  • freckles
\[2 *_{100}3=2 *_{99}(2 *_{100} 2)=2 *_{99}(4)=2 *_{99}4 \\ \text{ so we are really comparing } 2 *_{99}4 \text{ and } 27*_{99}15 \\ \text{ isn't it enough just to say you know since we have the same operation } \\ \text{ that since } 27>2 \text{ and } 15>4 \\ \text{ then } \\ 2*_{100}3=2*_{99}4<27*_{99}15\]
freckles
  • freckles
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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