## 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

1. anonymous

$*_k$ is the kth new operation in this progression

2. anonymous

you know what, just use casio calculator lol

3. anonymous

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

4. 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

5. anonymous

wow.....

6. nincompoop

dan, you're not telling us anything new

7. anonymous

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8. 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

9. anonymous

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

10. freckles

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

I have to try again tomorrow I'm sleepy

12. anonymous

noo dont quit i need u -.-

13. anonymous

oh okay :)

14. freckles

Did you try his first problem though?

15. anonymous

yeah

16. 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

17. 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)))

18. anonymous

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19. anonymous

yep thats right

20. freckles

I will continue the video tomorrow and look some more

21. freckles

but it seems interesting I think

22. anonymous

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

23. anonymous

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24. anonymous

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25. anonymous

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

26. anonymous

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27. anonymous

by induction

28. anonymous

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29. 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

30. anonymous

difference*

31. 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 ?

32. ganeshie8

yes i quit too just after that @freckles

33. anonymous

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34. anonymous

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35. anonymous

kinda crazy to see it tip over that much

36. anonymous

and yep u did it right ganeshie

37. 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

38. anonymous

but then again

39. anonymous

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

40. 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

41. anonymous

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42. anonymous

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43. anonymous

this property looks interesting

44. anonymous

i think its true

45. anonymous

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46. anonymous

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

47. anonymous

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48. freckles

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

49. freckles

and yeah 2 any operation 2=4 ... $2 *_{S(k)}2 =2 *_k(2*_{S(k)}1) =2 *_{k}2 = \text{ following this ... until we get } = \\ 2 *_12=2 \times 2=4$