Ordinal arithmetic

Jan 2013
50
1
Croatia
Solve (w^k+n)^w where k € w\{0} and n € w. I managed to show that w^w <= (w^k+n)^w < w^(w^2).
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey kicma.

What constraints can you have on your ordinal? Are you just looking at w^(f(x)) where x is some whole positive number?
 
Jan 2013
50
1
Croatia
Sorry, I don't understand what you mean.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
What do you mean by "solve" here? To calculate the given value?
 
Jan 2013
50
1
Croatia
I'll post solution here if anyone needs in future.

w^w=w^(k*w)=(w^k)^w<=(w^k+n)^w<=(w^k+w^k)^w<=((w^k)*w)^w=(w^(k+1))^w=w^((k+1)*w)=w^w