Prove A contains all natural numbers ≥ n_0

[r0bHadz]

Homework Statement

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

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 [itex]n_0[/itex] is in the set A, then [itex]n_0[/itex] can = k, and from there we see that the set contains all natural numbers larger than [itex]n_0[/itex] including [itex]n_0[/itex]

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

 

[SammyS]

Homework Statement

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

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 [itex]n_0[/itex] is in the set A, then [itex]n_0[/itex] can = k, and from there we see that the set contains all natural numbers larger than [itex]n_0[/itex] including [itex]n_0[/itex]

My question is, the way this question is worded, "then A contains all natural numbers ≥ [itex]n_0[/itex]," this is not saying that the set A can't have natural numbers less than [itex]n_0[/itex] 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 [itex]n_0[/itex] and the well ordering principle

 

[r0bHadz]

Is this proof sufficient:

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

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

It becomes oblivious that the set contains all numbers including and greater than [itex] n_0[/itex]

 

[SammyS]

Is this proof sufficient:

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

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

It becomes oblivious that the set contains all numbers including and greater than [itex] n_0[/itex]

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 [itex]n_0[/itex] 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 [itex] n_0[/itex] I don't see how numbers smaller than [itex]n_0[/itex] 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