Question

Can anyone help me out with mathematical induction?

Answer 1

ask your question ..

Answer 2

i just started it and i don't get it

Answer 3

well you try to prove general rule based on private case

for example :

0 + 1 + 2 + 3 + ... + n = n(n+1)/2

now you check for n = 0 :
is 0 = 0(0+1)/2
yes

you can check also for another number : for example 2

0 + 1 + 2 = 2(2+1)/2
3=3
yes

now you assume that for every k this is hold (this is the assumption or Hypothesis)
so you say : 0 + 1 + 2+ ... + k = k(k+1)/2

now you want to verify that it holds also for k+1
so
0 + 1 + 2 + 3 + ... + k + k+1
now look it is the same as ::
(0 + 1 + 2 +3 ... +k) + (k+1)

what about the sum ? :
plug k+1 :
(k+1)(k+2)/2

so :
(0 + 1 + 2 +3 +... +k) + (k+1) = (k+1)(k+2)/2
but based on the hypothesis (0 + 1 + 2 +3 +... +k) = k(k+1)/2
so
k(k+1)/2 + (k+1) = (k+1)(k+2)/2
(k(k+1) + 2k + 2)/2 = (k+1)(k+2)/2
(k(k+1)+2(k+1))/2 = (k+1)(k+2)/2
(k+2)(k+1)/2 = (k+1)(k+2)/2

Answer 4

the philosophy of mathematical induction is just the same .. if it happens for one particular arbitrary number 'n' and if the rule can be induced to 'n+1' then it will happen for any integer.

Answer 5

1. Show it's true for a base case, like n=1.
2. Show that if it's true for n=k, then it's true for n=k+1.
Using those two steps, you've shown that it's true for n=1 (by step 1), and since it's true for n=1 it's true for n=2 (by step 2), and since it's true for n=2 it's true for n=3 (by step 2 again), etc.

However, notice that this has only proven it's true for the positive integers. If you want to prove it's true for some other set of numbers you have to tweak the steps appropriately.