Recent content by j0k3R_

hi again " Take an integer n and the numbers 0*x, 1*x, ..., n*x of which there are n+1. Now partition [0,1] into [0, 1/n), [1/n,2/n), ... ,[n-1/n,1] intervals of which there are n. By the pigeonhole principle, two of the fractional parts of the n+1 numbers must fall into on" thank you --...

"Can you see where to go from there? (Hint: think pigeonhole principle and try to show the numbers are dense on [0,1])." can you illustrate for me how ot use pigeonhole principle here

yes also one more thing, iam feeling little bit uneasy when you give "subtract then divide everything in interval"; it is valid to say we have x in (changed interval), then x(operation) in (interval * operation)? because maybe i am thinking "there exists some pair (m, n)"

okay i am understanding your argument. in previous problem i proved closure R - Q is R. so, we can choose 0 < delta < e', and if y' + delta is rational then we obtain simple contradiction. is it okay?

cellotim i think i am coming close to the solution basically we need to show there exists (y - (m + nx), y + (m + nx)) C (y - e/2, y + e/2) for any e > 0. So take y - e/2 which is in R. since x is in R as well, we can apply the standard result from R is an archimedean ordered field to see...

Homework Statement Let A = {m + nx | m, n in Z} and x irrational. Show cl(A) = R.

no, i think for going to mathematics grad school such a course like your "abstract mathematical" LA using maybe axler, halmos, or roman, is much much better than engineering one if math grad school, then read axler, halmos, roman, mathwonk notes, etc., for the pure abut yes he can also learn...

i got mthis solution in few second but am spending several hours thinking about it but i do not have confidence i am thinking "this is silly something i did wrong, this norm can be complex structure etc" thinking about something extra etc.etc okay thanks for this very much :) oof

suppose we have v satisfying the conditions. then |L(v) + M(v)| <= |L(v)| + |M(v)| <= ||L|| + ||M|| so sup(|L(v) + M(v)|) = ||L + M|| <= ||L|| + ||M||.

Homework Statement for L, M: V -> W, L, M, linear let||L|| = sup{|L(v)|: v in V, |v| <= 1} show ||L + M|| < ||L|| + ||M|| Homework Equations The Attempt at a Solution so is it true that if |L(x) + M(x)| defines a sup for L + M (x for which |L(x) + M(x)| is the sup), then it also defines a...

mathwonk, can you recommend to me a book for studying advanced group theory, after studying hungerford ch. 1, 2, etc.