Use mathematical induction to prove that the statement is true for every positive integer n.
2 is a factor of n2 - n + 2
basis case n=1 2 is a factor of 1^2 - 1 + 2 ? yes because 1^2 -1 + 2 = 2 if 2 is a factor of k^2 -k + 1 is it true that 2 is a factor of (k+1)^2 -(k+1) + 2 ? thats what we have to prove
ok?
okay sorry my computer shut down due to updates! @perl
hi
okay so how do we finish this?
so we need to prove the following statement: "if 2 is a factor of k^2 -k + 1 for some integer k *then* is it true that 2 is a factor of (k+1)^2 -(k+1) + 2 "
lets look at (k+1)^2 - (k+1) + 2 , expand that
proofs are hard :(
now let me ask quick why did you expand it?
@perl ?
I think he's away...
we want to use the fact that k^2 - k + 2 is divisible by 2
oh nevermind...
at some point
OK
okay than what
(k+1)^2 - (k+1) + 2 = k^2 + 2k + 1 - (k+1) + 2 = k^2 + 2k + 1 - k - 1 + 2 = k^2 +2k -k + 1 -1 + 2 =k^2 + 2k - k + 2 = k^2 -k + 2 + 2k = (k^2 - k + 2) + 2k, but we know that (k^2 -k + 2) is divisible by 2 since we assumed that
so we have two numbers both of which are divisible by 2. therefore the sum will be divisible by 2. and we are done
you kinda lost me now.
we assumed that k^2 + -k + 2 is divisible by 2. using this assumption we can prove that (k+1)^2 - (k+1) + 2 is divisible by 2
okay so basically (k+1)^2 -(k+1) +2 would be my answer i dont have to put a # in for k
you know the statement is true for n=1. and we prove that if the statement is true for n=k, then the statement is true for n=k+1. so the statment is true for n=1, then since it is true for n=1 it is true for n= 1+1 = 2. now since it is true for n=2 it is true for n= 2 + 1 = 3, etc
well am i suppose to just write that one equation though
Im confused on what to put/
????
he's working on more than one problem on os apparently
@UsukiDoll do you know exactally what i have to put
>_________<!
I haven't taken a proof writing class, so no sorry...but I would put what perl mentioned
i know but would it be the equation i wrote..
perhaps... do you know spherical coordinates?
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