mathematical induction?

Question

this is for my logic proof class... my professor asked us to explain why an even number of negative signs always makes that statement true. and we have to do this through mathematical induction.... and i have no clue what that is. can some one show me an example?

Answer 1

Mathematical induction is used to prove a statment P(n), where n is a natural number. For example, P(n) could be the statement that the sum of the first n positive integers is n(n+1)/2, or that a statement preceded by n pairs of negative signs has the same truth value as the original statement.

Mathematical induction proceeds as follows: to prove that the statement P(n) is true, for n = 1, 2, ...:
1) Prove P(1) is true
2) Prove that for any natural number n, P(n) implies P(n+1).
Then, P(n) is true for every natural number n.

Think of it like an arrangement of dominos. For each n, P(n) is a domino. You want to prove all of them will fall. In part 1, you prove the first one will fall. In part 2, you prove that if any domino falls, it will knock down the next one. Put the two together, and you've shown that every domino will fall.

Example: show that 1+2+...+n = n(n+1)/2 for any natural number n.

1) 1=1(1+1)/2=1
2) assume 1+2+...+n = n(n+1)/2 for some integer n>=1. Then

1+2+...+n+(n+1)=
n(n+1)/2 + (n+1) =
(n+1)(n/2+1) =
(n+1)(n+2)/2 =
(n+1)((n+1)+1)/2

so it's true for all natural numbers.

Answer 2

induction is the concept of proof by repeated results. how would you prove the sun will come up tomorrow? By showing how often it came up before and with no other conflicting results. so you show two negatives as a positive, then four then eight. and so on. Induction is not regarded as strong logical proof..deduction is considered much stronger

Answer 3

what do you need to do with the even number of negative signs? add them?
multiply them?
what is true about that?

what i understand is the following

a_n= multiplication i=1 to 2n of (-1)
=(-1)(-1)...(-1), 2n-times
you want to prove that a_n=1 for every n.

start with n=1:
a_1 = multiplicacion i=1 to 2(1) of (-1)
=(-1)(-1) which is equal to 1.

now, assume that a_n=1,
prove that a_{n+1}=1:

a_{n+1}=(-1)...(-1) , 2(n+1)-times
we can group this multiplication in the following way: multiply first the 2n first -1's, and then multiply the last two -1's
a_{n+1}=a_n(-1)(-1)
= 1 (1), by the assumption a_n=1.
=1
QED

Answer 4

I'm not sure what you're asking for... something like how come the negative of the negative of a number equals that same number?

-(-a) + (-a) = 0

a + (-a) = 0

-(-a) + (-a) = a + (-a)
-(-a) = a

Answer 5

proofs suck