I am reading the book What is mathematics and I am stuck on this question. If anybody could help me understand better, they would be very kind. Thankyou.
#3
LukeD
354
3
One thing that might help is this:
2^0 \equiv 1 \pmod 5
2^1 \equiv 2 \pmod 5
2^2 \equiv -1 \pmod 5
2^3 \equiv -2 \pmod 5
2^4 \equiv 1 \pmod 5
2^5 \equiv 2 \pmod 5
2^6 \equiv -1 \pmod 5
2^7 \equiv -2 \pmod 5
see a pattern?
You should also know something about a nice way to simplify 2^0 + 2^1 + \cdots + 2^n
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#4
zazisr
3
0
oh ok . i see now. then the answer is 0 to that.
Thank You.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?