Question

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

2 is a factor of n2 - n + 2

Answer 1

we want to prove that the statement is true for all positive integer n.
Let me rewrite the statement as a type of propositional function.
P(n): 2 is a factor of n^2 - n + 2
so
P(1): 2 is a factor of 1^2 - 1 + 2
P(2) : 2 is a factor of 2^2 - 2 + 2
etc...
and P(1) is true or false, P(2) is true or false, etc. we would never finish being able to check P(n) for positive integers n since there are more numbers to check, an infinite number . so to get around this we use mathematical induction. it says roughly that if P(1) is true and if P(k) implies P(k+1) , then P(n) is true for any n positive integer.

basis case : is P(1) true?
P(1) : 2 is a factor of 1^2 -1 + 2 = 2 ,
why yes because 2 is a factor of 2 , 2 = 2*1

now prove the statement 'P(k) implies P(k+1)'.

Answer 2

thats the part im confused on most is proving that statement i have no idea how

Answer 3

ok to prove a conditional statement "if , then " , we assume the 'if' part. and then proceed to prove the 'then' part using what we assumed in the 'if' part.

Answer 4

we want to prove that P(k) -> P(k+1) , where -> means 'implies'.
so assume P(k) is true, where k is some positive integer.
now we want to prove that P(k+1) is true

Answer 5

so lets write out these statements
P(k) : 2 is a factor of k^2 - k +2
we assume this is true for some integer k. on the basis of this does it follow that P(k+1) is true?
lets write out P(k+1)
P(k+1) : 2 is a factor of (k+1)^2 - (k+1) + 2

Answer 6

ok so we are going to show that P(k+1) is true, given P(k) is true. that is what we need to demonstrate

Answer 7

ok i understand so far

Answer 8

lets expand P(k+1)

Answer 9

P(k+1) : 2 is a factor of (k+1)^2 - (k+1) + 2
rewriting
P(k+1): 2 is a factor of (k+1)(k+1) - (k+1) + 2

Answer 10

P(k+1) : 2 is a factor of (k+1)^2 - (k+1) + 2
P(k+1): 2 is a factor of (k+1)(k+1) - (k+1) + 2
P(k+1): 2 is a factor of k^2 + 2k + 1 - (k+1) + 2 (i just foiled)

Answer 11

any questions so far

Answer 12

so far im just rewriting or simplifying P(k+1)

Answer 13

so since you wrote all that out how do you see if its true

Answer 14

P(k+1) : 2 is a factor of (k+1)^2 - (k+1) + 2
P(k+1): 2 is a factor of (k+1)(k+1) - (k+1) + 2
P(k+1): 2 is a factor of k^2 + 2k + 1 - (k+1) + 2
P(k+1): 2 is a factor of k^2 + 2k + 1 - k-1 + 2
P(k+1): 2 is a factor of k^2 + k + 2
, ok so far we have rewritten P(k+1) , this equivalent form of P(k+1) is probably easier to show it is a factor of 2 .
now we are given P(k) , 2 is a factor of k^2 - k + 2 .
so can we show that
2 is a factor of k^2 - k + 2 IMPLIES 2 is a factor of k^2 + k + 2
or to say another way
can we show that
'2 is a factor of k^2 + k + 2' follows from '2 is a factor of k^2 - k + 2'

Answer 15

oh ok that makes sense