Question

What is wrong with this strong mathematical induction proof?

Answer 1

whats wrong is that

Answer 2

I am a little lost with this second principle of induction :s

Answer 3

so you want help with this

Answer 4

yes please :3

Answer 5

I am thinking it has something to do with K = 0 being the first nonnegative interget

Answer 6

this is a good place to start, with the definition of strong induction
http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html

Answer 7

For strong induction, you need to know that all the rungs below the rung you are on are solid in order to step up

Answer 8

So it fails at 0!!! right?

Answer 9

well , not exactly

Answer 10

step 2 : Let k be any integer greater than or equal to zero. then for 0 <= j <= k
assume that P(k) is true

Answer 11

step 2 : Let k be any integer greater than or equal to zero.
then for all integers j such that 0 <= j <= k assume that P(k) is true

Answer 12

Fails step 2

Answer 13

i dont see how it fails step 2

Answer 14

j = -1 when k = 0

Answer 15

so step 3d) when it says a^(k-1) , when k = 0 we have a problem

Answer 16

yes. I am not too sure why tho

Answer 17

I think it is because it can't just be part of it

Answer 18

basically in this strong induction we have
suppose k > = 0 and that P(0), P(1),... P(k) is true
show that P(k+1) is true

Answer 19

when k = 0, thats not a problem since we already checked that a^0 = 1 , so we actually have

Answer 20

hmm so what is the problem then?

Answer 21

maybe you are right, that inductive step seems odd if we plug in k = 0