Vollständige Induktion 2?
Vollständige Induktion,
Hallo,
Ich habe mich wieder mit dem Prinzip der vollständigen Induktion befasst und habe folgenden Denkfehler: Am Anfang gibt es ja eine Behauptung, die man durch das Einsetzen einer Zahl beweist. Das bedeutet ja, dass man ab diesem Moment sagt, dass die Gleichung XY für eine bestimmte Variable (natürliche Zahl) gilt. Sagen wir Mal, die Voraussetzung ist, dass es beispielsweise für n 1 gilt. Dann setzt man für n, (n + 1) ein und beweist hiermit unter der Voraussetzung, dass die Gleichung für n=1 erfüllbar war, dass die Gleichung auch für den Nachfolger, also zwei erfüllbar ist.
Ab hier habe ich eine Frage: Üblicherweise hört man ja hier auf (die Annahme, dass es für alle natürlichen Zahlen gilt, ist bewiesen). Liegt es hierbei daran, dass wenn man zeigt (durch Umformungen etc), dass man n+1 auf der „anderen“ Seite rekonstruieren kann, dass auch n+2, also auch n+3 n+4 ..-…. rekonstruierbar ist? Oder wie versteht man das? Würde das also auch bedeuten, dass wenn ich (n-1) beweise, dass auch (n-2..) gilt?
Zudem: Verstehe ich das richtig, dass die vollständige Induktion also einer Art Beweissatz ist, der die „Gültigkeit“ einer Lösungsmenge darlegt?
Gibt es eine Möglichkeit nur mit der vollständigen Induktion zu beweisen, für welche Zahlenmengen eine Gleichung gilt (also ohne davor eine Voraussetzung zu haben, dass beispielsweise Gleichung XY für alle natürlichen Zahlen gilt)?
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You prove that if there is a number n, it also applies to a number n+1. With this, it is of the first number for which you have shown it concretely and proved it for all following.
If you can’t prove the beginning, then it’s potentially wrong for all n.
The opposite applies if you can prove that if you have proven N, it also applies to n-1. That’s the same in green.
You can also notice that it only applies to n+2, but also for each. And you already have a different amount for which this is valid.
The logical principle is the “domino effect”: You prove in the end in the induction step: If any The next stone will fall. In the beginning of induction, you will show that there is a first stone to fall around and thus it is proven that all fall. Because the next of the first stone is the second one that falls around, because the induction step had assumed any stone as falling. Thus, the second stone becomes the assumed falling stone and therefore the third (again because of the induction step) also falls. And that goes on indefinitely.
Hey,
Thank you for your answer. Exactly, I know that effect. I have just realized that it can also be that, for example, (n+2) can be filled, but (n+1) is not, how is it to be understood? Does that have anything to do with the quantity? For example, if you have an equation that only goes for odd numbers (n), then it is logical that only (n+2/4….) applies.
Back to the n+1 and the “infinity evidence” (no idea how to express it differently now, so that the equation is met for all natural numbers n): You put +1 and, by forming the termum, you get the expression you want. If it is assumed to be simpler to say that if I have got the same expression with n+1, then it must also apply to n+2 n+3 … (in the case of the forming, you will see the same steps as this) or how do you understand the mathematical background?
I absolutely don’t understand what you’re thinking now. My answer just says in the last sentence “And that goes infinity on. You can only add to the illustration: … because nε N
… I don’t understand how this should have a meaning.
No, one proves that the claim applies to the respective successor if it applies to n (or if it applies as in the induction assumption).
You therefore indicate any natural number that the statement applies to the successor if it applies to the number itself.
And then, on the basis of a single natural number, it is sufficient to show that the message is valid to prove that it also applies to all subsequent natural numbers (for n+1, for the successor of n+1, etc.).
Yes, that’s exactly what I meant: I show that for example n=1 applies.
If I now use n+1, I also show that it applies to the successor of n, i.e. 2 (if n is defined as 1 (in the beginning one meets a prerequisite that the statement applies to a specific value n) . However, if I can show by termumforming that it applies to n+1, then it must also apply to n+2 (n+3…).
That’s what you understood.
It becomes interesting if you have, for example, a case where you can not show n+1 but only n+2. Then it takes two induction beginnings.
With alternating consequences, for example.
I can’t give a concrete example, because I’m bad about examples and that’s something about me that I had to deal with dmait.
But maybe there’s something here:
https://www.emath.de/Referate/induction tasks-loesungen.pdf
Interesting topic here! And how can the case described by you occur? With the approach n+1 I try to show that no matter which n+ (number) is also found on the other “page”, it is possible to reconstruct. How can it be that n+2 does not apply n+1? Because perhaps every second number results in an event (for example an odd number)?