- #1
Seydlitz
- 263
- 4
With my current understanding, the ordinary principle of Mathematical Induction is a method to prove, whether some statement ##P(x)## is true for all ##x## in ##\mathbb{Z}##, considering the truth of ##P(1)## and then the truth of ##P(k+1)## assuming ##P(k)## is true.
For Complete Induction, Spivak mentioned that, "It sometimes happen than in order to prove ##P(k+1)## we must assume not only ##P(k)## but also ##P(l)## for all natural numbers ##l \leq k##." So that is then the only difference? The difference in assumption?
I cannot see yet the significance of it. When do we use ordinary principle and this stronger version? It's a bit more confusing when he says that we could prove the stronger version with the ordinary version. Can you guys give me perhaps simpler example or formulation?
Does it work like this? I want to know if ##A## contains ##\mathbb{Z}##. I assume this is so. To prove that, then ##A## must contain ##1##, which is true, and hence ##k## and ##k+1## is also in ##A##, if ##A## contains ##\mathbb{Z}##. But I've just assumed ##A## contains ##\mathbb{Z}##, so it is true then that ##A## contains ##\mathbb{Z}##. If we want to prove this using ordinary induction we can't, because we only assume ##k## is in ##A##, maybe there are other numbers that is in ##A## but not in ##\mathbb{Z}##.
(My last educational background is only that of high-school computational-styled Mathematics)
For Complete Induction, Spivak mentioned that, "It sometimes happen than in order to prove ##P(k+1)## we must assume not only ##P(k)## but also ##P(l)## for all natural numbers ##l \leq k##." So that is then the only difference? The difference in assumption?
I cannot see yet the significance of it. When do we use ordinary principle and this stronger version? It's a bit more confusing when he says that we could prove the stronger version with the ordinary version. Can you guys give me perhaps simpler example or formulation?
Does it work like this? I want to know if ##A## contains ##\mathbb{Z}##. I assume this is so. To prove that, then ##A## must contain ##1##, which is true, and hence ##k## and ##k+1## is also in ##A##, if ##A## contains ##\mathbb{Z}##. But I've just assumed ##A## contains ##\mathbb{Z}##, so it is true then that ##A## contains ##\mathbb{Z}##. If we want to prove this using ordinary induction we can't, because we only assume ##k## is in ##A##, maybe there are other numbers that is in ##A## but not in ##\mathbb{Z}##.
(My last educational background is only that of high-school computational-styled Mathematics)
Last edited: