math brain teaser.... are you brain enough?

[UA_Iron]

those are just the equalled amounts. they are not computated in the problem.

the biproducts of the operation, they are not limited to the numbers listed, only the operatations on them are.

 

[UA_Iron]

did you want to see who could do it in the least amount of operations?

 

[jackangel]

i think he wants people to keep churning away like goodmonkeys, results irrelevant.

 

[hanselthecaretaker]

the biproducts of the operation, they are not limited to the numbers listed, only the operatations on them are.

Yes, thanks for clarifying.

 

[Beachbum1546]

I tried doing it with only using the number once, unlike what UA_Iron did. I could go on for an eternity if I used the numbers over and over again.

 

[samoth]

add them, subtract them, square them, square root, cube, cube root, mulitply, divide.... you name it

Well, that took all the fun out of it...

I define my own operator and product space such that any action on those numbers is confinded to the set of R which I define here to be a subset of the integer 9.

QED

 

[redguru]

Well, that took all the fun out of it...

I define my own operator and product space such that any action on those numbers is confinded to the set of R which I define here to be a subset of the integer 9.

QED

How many different operators? I declare an infinite number of operators that when used with the numbers given by Chef is also confined to the set noted by R

R being a subset of the integer 9. Therefore the number of operations using those numbers is without boundary.

 

[hamstershaver]

alert!!!!!!!!!!!!

 

[samoth]

How many different operators? I declare an infinite number of operators that when used with the numbers given by Chef is also confined to the set noted by R

R being a subset of the integer 9. Therefore the number of operations using those numbers is without boundary.

Damn you!

 

[redguru]

Damn you!

I was too damn lazy to set up a Euclidian proof table. Almost wanted to make it an elaborate non-answer.