Dear Numbers&Polys Enthusiast-Scholars Problem A1: Sinamon-David Lemma. If 1: x = x+x then x = 0. Proof: By adding the additive inverse of x to each side of (1), x - x = [x + x] - x = x + [x - x] , by associativity of addition, = x + 0 = x. Hence 0 = x, as desired. QED Proof of A1: The idea is to use distributivity, as follows: c·0 = c·[0+0] = [c·0] + [c·0] , by DMA. Hence by the SimDav-Lemma, c·0 equals 0. QED ================================================================ Proof of A2: By uniqueness of additive inverse, ISTShow (It Suffices To Show) that 2: b + [[-1]·b] = 0 But note that 0 = 0·b , by (A1) and commutativity of multiplication, = [1 + -1]·b , by AID = [1·b] + [[-1]·b] , by DMA. Courtesy of our multiplicative-identity axiom, this yields 0 = LHS(2), as desired. QED ================================================================ What follows is a rewording of Claire's proof. Proof of A3: Consider this inequality: 3: |x+y|² =< [|x| + |y|]² . Note that each of |x+y| and [|x| + |y|] is non-negative, courtesy (1.20a) and (1.15a). Consequently we can apply (1.16k) and (1.17a), by having let a := |x+y| and b := |x| + |y| , to conclude from (3) that 4: |x+y| =< [|x| + |y|] holds, which is the triangle inequality. SECOND STEP: Courtesy of (1.20c), we can restate our goal, (3), thusly: 3': [x+y]² =< [|x| + |y|]² . Now LHS(3') = x² + y² + 2·x·y . Observe that RHS(3') = |x|² + |y|² + 2·|x|·|y| = x² + y² + 2·|xy| , by (1.20c) and (1.20f). Courtesy of OA, then, our goal (3') reduces to 2·x·y =< 2·|xy| . and hence, thanks to OM, to 3'': x·y =< |xy| . But this follows from (1.20d) applied to a := xy . QED ================ Here is Claire's proof, lightly edited: ==== /------------------------------------------------------------\ / \ A3: PROOF: Firstly, |ab| >= ab by Thm 1.20d. So, 2|ab| >= 2ab by Thm OM and using that 2>0 by Thm 1.5f. Thus a^2+[[2|ab|+b^2] >= a^2+[[2[ab]]+b^2] by OA twice. Consequently |a|^2+[[2|ab|]+|b|^2 >= a^2+[[2[ab]]+b^2] by Thm 1.20c. Note that |a|^2+[[2|a||b|]+|b|^2 >= a^2+[[2[ab]]+b^2] by Thm 1.20f This yields [|a|+|b|]^2 >= [a+b]^2 by Thm 1.10b used twice. Consequently Thm 1.20c yields that [|a|+|b|]^2 >= |a+b|^2 . Therefore, |a|+|b| >= |a+b| by Thm 1.15k, since |a|+|b|>= 0 by thm 1.15a and |a+b| >= 0 by Thm 1.20a. \ / \____________________________________________________________/ ================================================================ Problem A4: Use the QF to find the zeros of the polynomial x²+8x+15. Graph the poly and list the intervals where it is positive. ================================================================ Problem A5: Addition does NOT distribute over multiplication. The other three statements are true. ================================================================ ;;;; End of file "a.soln.txt" ;;;;