implies that x by Theorem 117 set y 0 in the theorem x and x means x 0 6 a x x

# Implies that x by theorem 117 set y 0 in the theorem

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implies that x 0 by Theorem 11.7 (set y = 0 in the theorem). x 0 and x 0 means x = 0. 6. (a) | x | = | x - y + y | ≤ | x - y | + | y | . Therefore | x | - | y | ≤ | x - y | | y | = | y - x + x | ≤ | y - x | + | x | = | x - y | + | x | . Therefore | y |-| x | ≤ | x - y | and | x |-| y | ≥ -| x - y | . That is -| x - y | ≤ | x | - | y | ≤ | x - y | which implies that vextendsingle vextendsingle | x | - | y | vextendsingle vextendsingle ≤ | x - y | . (b) By (a) | x | - | y | ≤ | x - y | . Therefore, if | x - y | < c , then | x | - | y | < c so | x | < c + | y | . (c) | x - y | ≥ 0. If | x - y | < epsilon1 for every epsilon1 > 0, then | x - y | = 0 by Pb. 4. | x - y | = 0 implies x = y . 3
7. Use induction. We know | x 1 | ≤ | x 1 | and | x 1 + x 2 | ≤ | x 1 | + | x 2 | so the inequality holds for n = 1 , n = 2. Assume that the inequality holds for any set of k real numbers. Then, for any set of k + 1 real numbers, | x 1 + x 2 + · · · + x k + x k +1 | = | ( x 1 + x 2 + · · · + x k )+ x k +1 | ≤ | x 1 + x 2 + · · · + x k | + | x k +1 | ≤ | x 1 | + x 2 | + · · · + | x k | + | x k +1 | R B. (a) Assume that the positive integer k S . That is, assume 1 + 2 + 3 + · · · + k = ( k + 2)( k - 1) 2 Prove that k + 1 S . 1 + 2 + 3 + · · · + k + ( k + 1) = ( k + 2)( k - 1) 2 + ( k + 1) = ( k 2 + k - 2 + 2( k + 1) 2 = k 2 + 3 k 2 = ( k + 3)( k ) 2 Therefore, k + 1 S . (b) Since 1 + 2 + 3 + · · · + n = n ( n + 1) 2 negationslash = ( n + 2)( n - 1) 2 for all n , there is n o positive integer for which this equation holds. C. (a) Show S is closed with respect to addition and multiplication. (b) 1 2 - 5 2 = 1 2 - 5 2 · 2 + 5 2 2 + 5 2 = 2 + 5 2 - 46 = - 1 23 - 5 46 2 4

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• Fall '08
• Staff
• Integers, Natural number, Prime number