Question

Use mathematical induction to prove that the statement is true for every positive integer n.

2 is a factor of n2 - n + 2

I really need help on this one

Answer 1

@Loser66

Answer 2

Have you done mathematical induction before, like
-Base case
-hypothesis for case n
-proof for case n+1

Answer 3

i has not mate

Answer 4

so hows do i do that thingy majig your talking about

Answer 5

are you there my fellow good sir

Answer 6

Yes, I was looking for a good reference for you.
The following link shows us how to do mathematical inductions, with examples. Example 4 is particularly similar to your problem.
http://www.analyzemath.com/math_induction/mathematical_induction.html

Answer 7

ok thnx mate i will chat you up if i have any problems

Answer 8

Good, way to go! :)

Answer 9

ok cheerio

Answer 10

Statement P (n) is defined by

n 3 + 2 n is divisible by 3

STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

1 3 + 2(1) = 3

3 is divisible by 3

hence p (1) is true.

STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M , where M is a positive integer.

We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.

Answer 11

ok so mate i need more explanation on this it so short worded and down righ confusing

Answer 12

@mathmate could you help me for a sec mate

Answer 13

The first step is to establish that the statement you want to prove is true for some n.
So P(1)=n^3+2n=1^3+2(1)=3 is divisible by three.

In principle, if we have time to show this for all integers, we're done, but we cannot, because n goes to infinity

So the next best thing is to show that, if 3|P(n), then 3|P(n+1).
3|x reads 3 divides x.

Why ? Because if the hypothesis 3|P(n) is true, then 3|P(n+1) is true, then 3|P(n+2) is true, and so on.

Once we have achieved that, we go back to say that 3|P(1), therefore 3|P(2), 3|P(3)... without having to do ALL of the integers.

Answer 14

ok mate so i get that so how di i do part two of my fiding my answer mate

Answer 15

you there mate

Answer 16

@Ashleyisakitty

Answer 17

@sky425

Answer 18

@leon549

Answer 19

@MrNood

Answer 20

anybody of service here mate

Answer 21

@helpppppppppp

Answer 22

f n is even, then n = 2m for some integer m.
then substitute
n^2 - n + 2
= (2m)^2 - (2m) + 2
=4m^2 - 2m + 2
= 2 ( 2m^2 - m + 1) . and this is a factor of 2 by definition, since it is 2 times some integer ,

if n is odd , then n = 2r+1 for some integer r
n^2 - n + 2
= (2r + 1) ^2 - (2r + 1) + 2
= 4r^2 + 4r + 1 - 2r -1 + 2 by foiling and distributing
= 4r^2 + 2r + 2
= 2 ( 2r^2 + r + 1) . and this is a factor of 2 again. we are done :)
ok i tried my best and this is what i got is it true

Answer 23

This is a correct direct proof using "proof by cases".
The question asks for proof by mathematical induction, which involves the three steps:
1. Proof a base case (say, n=1).
2. Inductive hypothesis: assume statement is true for case n=k.
3. Inductive proof and conclusion: prove that statement is true for case n=k+1.
Done.

The steps are extremely similar to example 4 that you have studied. Just have to replace the statement with yours.

1. base case : 2 is a factor of f(n)=n2 - n + 2 when n=1.
Well, n^2-n+2 = f(1)= 1^2-1+2 = 2, so 2 is a factor of 2, hence statement is true for n=1 (base case).
2. Inductive hypothesis:
Assume that (or IF) statement is true for case n=k:
2 is a factor of n2 - n + 2 for n=k, therefore
2 is a factor of f(k)= k^2-k+2, or f(k)=2m, where m\(\in\)Z
3. Work on case n=k+1
f(k+1)=(k+1)^2-(k+1)+2
= k^2+2k+1-k-1+2
=(k^2-k+2) +2k
=(2m)+2k
=2(m+1), hence 2 is a factor of f(k+1), given that 2 is a factor of f(k).

we have just shown that 2 is a factor of f(k+1), which completes the proof by mathematical induction that 2 is a factor of f(n) for all n\(\in\)Z+.