induction question (1 Viewer)

[5647382910]

prove, by induction, that n^2 - 11n + 30 >=0 for n>= 1
tried everything, no hope.
thanks in advance

 

[Trebla]

prove, by induction, that n^2 - 11n + 30 >=0 for n>= 1
tried everything, no hope.
thanks in advance

This is all I could come up with atm, though it can be proved without induction...
n = 1
LHS = 1 - 11 +30
= 20
≥ 0
= RHS
Hence true for n = 1
Assume it is true for n = k
i.e. k² - 11k + 30 ≥ 0
Need to prove true for n = k + 1
i.e. (k + 1)² - 11(k + 1) + 30 ≥ 0
Case 1
LHS = (k + 1)² - 11(k + 1) + 30
= k² + 2k + 1 - 11k - 11 + 30
= k² - 11k + 30 + 2k - 10
≥ 2k - 10 (by assumption)
≥ 0 (if integer k ≥ 5)
Case 2 (if k is an integer such that 1 ≤ k ≤ 4)
LHS = (k + 1)² - 11(k + 1) + 30
= k² - 9k + 20
= (k - 4)(k - 5)
≥ 0 when 1 ≤ k ≤ 4 (not sure how to kick the assumption in :S)
= RHS
If statement is true for n=k it is also true for n=k+1
Since the statement is true for n=1, it is true for all positve integers n

Without induction, you can just graph the thing and say its non-negative for all positive integers of n. (NB: it dips below zero between n = 5 and n = 6, but they are for non-integer n)

 

[gurmies]

yeah, induction seems a little silly, plus by the looks of it, it proves to be a difficult one (as inequalities tend to be). I'd just say that the leading coefficient is > 0, and the discriminant < 0, thus there are no real roots, and thus it's a positive definite function...

 

[5647382910]

This is all I could come up with atm, though it can be proved without induction...
n = 1
LHS = 1 - 11 +30
= 20
≥ 0
= RHS
Hence true for n = 1
Assume it is true for n = k
i.e. k² - 11k + 30 ≥ 0
Need to prove true for n = k + 1
i.e. (k + 1)² - 11(k + 1) + 30 ≥ 0
Case 1
LHS = (k + 1)² - 11(k + 1) + 30
= k² + 2k + 1 - 11k - 11 + 30
= k² - 11k + 30 + 2k - 10
≥ 2k - 10 (by assumption)
≥ 0 (if integer k ≥ 5)
Case 2 (if k is an integer such that 1 ≤ k ≤ 4)
LHS = (k + 1)² - 11(k + 1) + 30
= k² - 9k + 20
= (k - 4)(k - 5)
≥ 0 when 1 ≤ k ≤ 4 (not sure how to kick the assumption in :S)
= RHS
If statement is true for n=k it is also true for n=k+1
Since the statement is true for n=1, it is true for all positve integers n

Without induction, you can just graph the thing and say its non-negative for all positive integers of n.

so are we allowed to make those assumptions in induction? are we allowed to divide it into two cases? thanks for the help btw

 

[youngminii]

This is all I could come up with atm, though it can be proved without induction...
n = 1
LHS = 1 - 11 +30
= 20
≥ 0
= RHS
Hence true for n = 1
Assume it is true for n = k
i.e. k² - 11k + 30 ≥ 0
Need to prove true for n = k + 1
i.e. (k + 1)² - 11(k + 1) + 30 ≥ 0
Case 1
LHS = (k + 1)² - 11(k + 1) + 30
= k² + 2k + 1 - 11k - 11 + 30
= k² - 11k + 30 + 2k - 10
≥ 2k - 10 (by assumption)
≥ 0 (if integer k ≥ 5)
Case 2 (if k is an integer such that 1 ≤ k ≤ 4)
LHS = (k + 1)² - 11(k + 1) + 30
= k² - 9k + 20
= (k - 4)(k - 5)
≥ 0 when 1 ≤ k ≤ 4 (not sure how to kick the assumption in :S)
= RHS
If statement is true for n=k it is also true for n=k+1
Since the statement is true for n=1, it is true for all positve integers n

Without induction, you can just graph the thing and say its non-negative for all positive integers of n. (NB: it dips below zero between n = 5 and n = 6, but they are for non-integer n)

Can you explain how you did case 1 and case 2? I don't understand =X
Also, it dips below 0 when 4 < n < 5

 

[madsam]

induction isnt that hard, first you solve for n = 1, then you assume when you sub in k its true, then you prove that k + 1 is true, and if k + 1 is true, that means the number 2 is true, which means 3 is true etc...

Assume it is true for n = k
i.e. k² - 11k + 30 ≥ 0
Need to prove true for n = k + 1
i.e. (k + 1)² - 11(k + 1) + 30 ≥ 0
Case 1
LHS = (k + 1)² - 11(k + 1) + 30
= k² + 2k + 1 - 11k - 11 + 30
= k² - 11k + 30 + 2k - 10
≥ 2k - 10 (by assumption)
≥ 0 (if integer k ≥ 5)

if you look carefully, he merely subed in k for n in the original equation, then (k + 1) for k

then he solved it and got k ≥ o (when k ≥ 5)

 

[Trebla]

The problem lies in the fact that the equation becomes negative when 5 < n < 6 of the original equation (and 4 < n < 5 for the (k + 1) expression).
It can be easily proved for integers k ≥ 5 (case 1), but the problem lies in using the assumption for integers 1 ≤ k ≤ 4 which I didn't do. The inequality is still true though, because it is non-negative for INTEGERS n (there are no integers n where the function dips below 0 at 5 < n < 6).