Prove by induction n^2 > 5n+10 for n > 6. (Hint #1, use the weak principle of ma
ID: 3657310 • Letter: P
Question
Prove by induction n^2 > 5n+10 for n > 6. (Hint #1, use the weak principle of mathematical induction; Hint #2, see the Math Review on Inequalities) 1. Base case: 2. State the inductive hypothesis: 3. State what we have to show: 4. Proof proper: Please do step by step...Explanation / Answer
Base: 72 > 5(7) + 10 49 > 35 + 10 49 > 45 Thus: $n ( n > 6 Ù n2 > 5n + 10 ) Hypothesis: "n [ ( n > 6 ) ® ( n2 > 5n + 10 ) ] Inductive hypo: "k [ ( k > 6 ) ® ( k2 > 5k + 10 ) ] ® "k [ ( k > 6 ) ® ( (k + 1)2 > 5(k + 1) + 10 ) ] So from 1 below we must derive : "k [ ( k > 6 ) ® ( (k + 1)2 > 5(k + 1) + 10 ) ] 1. "k [ ( k > 6 ) ® ( k2 > 5k + 10 ) ] assume 2. k > 6 assume 3. ( k > 6 ) ® ( k2 > 5k + 10 ) 1, universal instantiation 4. k2 > 5k + 10 2, 3 modus ponens 5. (k + 1)2 = k2 + 2k + 1 fact 6. (k + 1)2 > (5k + 10) + 2k + 1 5 made unequal by substituting k2 as per 4 7. (k + 1)2 > 5k + 10 + 2(2) + 1 2 and since 6 > 2, a fortiori k > 2 [*] 8. (k + 1)2 > 5k + 10 + 4 + 1 7, multiplication 9. (k + 1)2 > 5k + 10 + 5 8, addition 10. (k + 1)2 > 5k + 5 + 10 9, commutativity 11. (k + 1)2 > 5(k + 1) + 10 10, factoring 12. ( k > 6 ) ® ( (k + 1)2 > 5(k + 1) + 10 ) 2 - 11 deduction theorem 13. "k [ ( k > 6 ) ® ( (k + 1)2 > 5(k + 1) + 10 ) ] 12, universal generalization [*] Note that the change of a k in step 6 to '2' in step 7 expresses the transitivity of the greater than relation (ie, '>' ): 1. "x"y"z [ (( x > y ) Ù ( y > z )) ® ( x > z ) ] by definition of transitive 2. "y"z [ (( k > y ) Ù ( y > z )) ® ( k > z ) ] 1, universal instantiation (ui) 3. "z [ (( k > 6 ) Ù ( 6 > z )) ® ( k > z ) ] 2, ui 4. (( k > 6 ) Ù ( 6 > 2 )) ® ( k > 2 ) 3, ui In words: if we know that a > b (and by step 2 we know that k > 6) and we also know b > c (as we know 6 > 2), then we know that a > c (and so we know that k > 2, and thus expressing a greater than relation, step 7 must be true).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.