By Joseph V. Collins

Excerpt from An effortless Exposition of Grassmann's Ausdehnungslehre, or conception of Extension

The sum qf any variety of vectors is located by means of becoming a member of the start element of the second one vector to the top element of the 1st, the start aspect of the 3rd to the tip aspect of the second one. etc; the vector from the start aspect of the 1st vector to the tip aspect of the final is the sum required.

The sum and distinction of 2 vectors are the diagonals of the parallelogram whose adjoining aspects are the given vectors.

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**Extra info for An elementary exposition of Grassmann's Ausdehnungslehre, or Theory of extension**

**Example text**

23] D. H. Fremlin. Consequences Of Martin’s Axiom. Cambridge University Press, 1984. [24] G. Gruenhage. Perfectly normal compacta, cosmic spaces, and some partition problems. In Open Problems in Topology, pages 85–95. North-Holland, Amsterdam, 1990. [25] G. Gruenhage, J. Tatch Moore. Perfect compacta and basis problems in topology. In Elliott Pearl, editor, Open Problems in Topology II. Elsevier, 2007. [26] K. P. Hart. Efimov’s problem. In Elliott Pearl, editor, Open Problems in Topology II. Elsevier, 2007.

His tests formalize computable betting strategies. A test in Schnorr’s sense is a computable function M from {0, 1}∗ to the non-negative rationals. When the player has seen z = Z n , she can make a bet q where 0 ≤ q ≤ M (z) on the next bit Z(n). If she is right she wins q, otherwise she loses q. Thus M must satisfy the fairness condition M (z0) + M (z1) = 2M (z) for each string z. She wins on Z if M (Z n ) is unbounded. We call a set Z computably random if no computable betting strategy wins. Choose c ∈ N such that the start capital M (∅) is at most 2c .

3. For sets A and B, let A ⊕ B denote the set 2A ∪ 2B + 1, namely the set which is A on the even bit positions and B on the odd positions. The K-trivial sets are closed under ⊕ by the following result of Downey, Hirschfeldt, Nies and Stephan. 4 ([9]). If A and B are K-trivial via b, then A ⊕ B is K-trivial via 3b + O(1). Proof. It is sufficient to describe each string A⊕B 2n with K(n)+3b+O(1) bits. To do this, we need to describe n only once; if we have a shortest description of n we also know its length r = K(n).