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Extra info for Algebra: Monomials and Polynomials

Sample text

B) ( g m )n = g mn . (C) g m g n = g m +n . The proof will justify this argument by applying the notation described at the beginning of this chapter. We have to be careful with this approach, because in the lemma we have m, n ∈ Z, but the notation was given under the assumption that n ∈ N+ . To make this work, we’ll have to consider the cases where m and n are positive or negative separately. We call this a case analysis. Proof. (A) Each claim follows by case analysis. −1 If m = 0, then g −m = g 0 = e = e −1 = g 0 = ( g m )−1 .

79. Show that E is an additive group under the addition defined above, with • P∞ as the zero element; and • for any P = ( p1 , p2 ) ∈ E, then −P = ( p1 , − p2 ) ∈ E. 80. Choose different values for a and b to generate another elliptic curve. Graph it, and illustrate each kind of addition. 81. 5 the elliptic curve E determined by the equation y 2 = x 3 − x. (a) Compute the cyclic group generated by (0, 0) in E. 2 + 1, 2 + 2 is a point on E. (b) Verify that (c) Compute the cyclic group generated by 2 + 1, 2 + 2 in E.

The proof will justify this argument by applying the notation described at the beginning of this chapter. We have to be careful with this approach, because in the lemma we have m, n ∈ Z, but the notation was given under the assumption that n ∈ N+ . To make this work, we’ll have to consider the cases where m and n are positive or negative separately. We call this a case analysis. Proof. (A) Each claim follows by case analysis. −1 If m = 0, then g −m = g 0 = e = e −1 = g 0 = ( g m )−1 . Otherwise, m = 0.