R & R directsum Q are vector spaces over Q.Are they isomorphic?R=real,Q=rational?

Answer 1

If you use the Axiom of Choice to show there there is a Hamel basis for R as a vector space over Q, then R directsum Q has a Hamel basis with "one more element". Again using AC the two infinite sets have the same cardinality. A bijection between them induces a vector space isomorphism between the two vector spaces.

Answer 2

There are a gaggle of axioms defining a vector area, yet R+ (assuming you recommend genuine numbers > 0) isn't a vector area over R. For one difficulty, it does not incorporate 0. whether you improve R+ to be the genuine numbers >=0, it nonetheless isn't by way of fact if r is in R+, -r isn't. you may might desire to apply distinctive operations than prevalent addition and multiplication to make it artwork. Edit: David, sure, of course it is superb, however the way the question is reported, that's no longer. whether he had pronounced "prepare R is a vector area over R," this remains technically incorrect, when you consider that he quite might desire to declare the operations whilst defining the area, or a minimum of say "make R+ right into a vector area over R." wager I each so often get slightly snippy with questions that are actually not properly reported.