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

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- schrodinger

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- anonymous

\[*_k\] is the kth new operation in this progression

- anonymous

you know what, just use casio calculator lol

- anonymous

we can try to see if there is pattern with the lower operations like

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## More answers

- 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

wow.....

- nincompoop

dan, you're not telling us anything new

- anonymous

|dw:1440050985534:dw|

- 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

itd be an interesting question to find out if this does infact happen, if so at which operator?

- 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

I have to try again tomorrow
I'm sleepy

- anonymous

noo dont quit i need u -.-

- anonymous

oh okay :)

- freckles

Did you try his first problem though?

- anonymous

yeah

- 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

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

|dw:1440051564779:dw|

- anonymous

yep thats right

- freckles

I will continue the video tomorrow and look some more

- freckles

but it seems interesting I think

- anonymous

i dont like the way he uses s(k) for k+1 but other than that its pretty good!

- anonymous

|dw:1440052034629:dw|

- anonymous

|dw:1440052925677:dw|

- anonymous

that right there i am thinking \[2*_{98}2=4\]
do u want to confirm that 2 operator 2 is always 4

- anonymous

|dw:1440053172695:dw|

- anonymous

by induction

- anonymous

|dw:1440053515919:dw|

- 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

difference*

- 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

yes i quit too just after that
@freckles

- anonymous

|dw:1440053776717:dw|

- anonymous

|dw:1440054119253:dw|

- anonymous

kinda crazy to see it tip over that much

- anonymous

and yep u did it right ganeshie

- 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

but then again

- anonymous

i wonder if there is way to write sum kind of expression with summations and comparing those

- 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

|dw:1440054511816:dw|

- anonymous

|dw:1440060525252:dw|

- anonymous

this property looks interesting

- anonymous

i think its true

- anonymous

|dw:1440060687164:dw|

- anonymous

whenever u are distributing any operation over addition, this is happening the middle operation is shifted down by 1

- anonymous

|dw:1440061272416:dw|

- 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

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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