Prove A contains all natural numbers ≥ n_0

[r0bHadz]

Homework Statement

Prove that if a set A of natural numbers contains $n_0$ and contains k+1 whenever it contains k, then A contains all natural numbers ≥ $n_0$

Homework Equations

The Attempt at a Solution

I'm just confused by the question, please don't answer it.

Logically it makes sense that if $n_0$ is in the set A, then $n_0$ can = k, and from there we see that the set contains all natural numbers larger than $n_0$ including $n_0$

My question is, the way this question is worded, "then A contains all natural numbers ≥ $n_0,$" this is not saying that the set A can't have natural numbers less than $n_0$ though, correct?

 

[SammyS]

Homework Statement

Prove that if a set A of natural numbers contains $n_0$ and contains k+1 whenever it contains k, then A contains all natural numbers ≥ $n_0$

Homework Equations

The Attempt at a Solution

I'm just confused by the question, please don't answer it.

Logically it makes sense that if $n_0$ is in the set A, then $n_0$ can = k, and from there we see that the set contains all natural numbers larger than $n_0$ including $n_0$

My question is, the way this question is worded, "then A contains all natural numbers ≥ $n_0$," this is not saying that the set A can't have natural numbers less than $n_0$ though, correct?

Correct.

The set may contain other natural numbers.

( You missed a [/itex] code after the $n_0$.)

 

[r0bHadz]

Correct.

The set may contain other natural numbers.

( You missed a [/itex] code after the $n_0$.)

Ah, gotcha. The way the question was worded just kinda threw me off. I guess the proof just involves a mention of $n_0$ and the well ordering principle

 

[r0bHadz]

Is this proof sufficient:

Since $n_0$ is in the set, let's assume, by the well ordering principle that $n_0$ is the smallest natural number in the set.

Setting k=$n_0$ we now have $n_0$ and $n_0 + 1$ in the set. If we let k = $n_0 + 1$ then $n_0 +2$ is also in the set.

It becomes oblivious that the set contains all numbers including and greater than $n_0$

 

[SammyS]

Is this proof sufficient:

Since $n_0$ is in the set, let's assume, by the well ordering principle that $n_0$ is the smallest natural number in the set.

Setting k=$n_0$ we now have $n_0$ and $n_0 + 1$ in the set. If we let k = $n_0 + 1$ then $n_0 +2$ is also in the set.

It becomes oblivious that the set contains all numbers including and greater than $n_0$

As you concluded previously, there may be numbers in set A which are less than $n_0$ .

So, it's not correct to assume that $n_0$ is the smallest natural number in set A.

 

[FactChecker]

The form of the proof should be: Suppose natural number $n_1 \geq n_0$. Then prove $n_1 \in A$. It is not enough to just say that it is obvious. That may be true, but you should learn how to do a formal proof because there can be ugly surprises in things that look obvious.

 

[r0bHadz]

As you concluded previously, there may be numbers in set A which are less than $n_0$ .

So, it's not correct to assume that $n_0$ is the smallest natural number in set A.

Right but I thought in the context of this problem those numbers don't matter? The reason why I wrote to assume $n_0$ is the smallest is just to make the problem simpler. It's only asking me to prove that A contains all numbers equal to or greater than $n_0$ I don't see how numbers smaller than $n_0$ are relevant

The form of the proof should be: Suppose natural number $n_1 \geq n_0$. Then prove $n_1 \in A$. It is not enough to just say that it is obvious. That may be true, but you should learn how to do a formal proof because there can be ugly surprises in things that look obvious.

Gotcha I guess I'll have to work on it and report back